NCERT Solutions Class 7 Mathematics

Class 7 holds an important place in the academic journey of the student’s life. It lays the foundation of higher classes i.e. Class 8, 9 and 10. Since the competition is tough, students should try to understand all the basic concepts and should be able to apply the theories in real-life circumstances. Here, simply attending class won't be sufficient to  keep up with the competition.

 

NCERT Solutions For Class 7 Mathematics

Mathematics is one of the most challenging and scoring subjects. NCERT Solutions for Class 7 Mathematics from Extramarks help students in achieving their objectives and in improving their learning. These solutions by Extramarks are prepared only for students to grasp and understand both simple and difficult concepts. The language is easy to understand so that students can understand it better and get the maximum benefit.

Students must have the conceptual clarity to perform well on their exams. As a result, Extramarks offers NCERT Mathematics Class 7 Solutions to students. They can learn quickly and have a thorough understanding of the study material. The most important point in learning Mathematics is studying and learning concepts, as well as practicing questions based on them. If they don't have access to NCERT Solutions, they may find it difficult to get the right answer which can be time consuming and stressful.

 

NCERT Solutions For Class 7 Mathematics CBSE 2022-23 Edition

Students should practise and solve Class 7 Mathematics question papers regularly to perform better in exams. Later, they can access NCERT Solutions for Class 7 that are available on Extramarks website for free. The NCERT 7 Mathematics Solution by Extramarks will help students in getting clarity across different concepts. The solutions have been drafted in an organized and systematic manner. Students can understand concepts, difficult theorems, etc. The subject matter experts at Extramarks develop solutions as per the updated curriculum. Students can click on the link below and access NCERT Solutions for Class 7 Mathematics.

 

NCERT Solutions Class 7 Mathematics

Students of Class 7 can access chapter-wise NCERT Solutions Class 7 Mathematics.

NCERT Chapter-wise Exercise- Solutions for Class 7 Mathematics

After solving the NCERT Class 7th Mathematics exercises, students should go through the NCERT Solutions for Class 7 Mathematics. Going through the solutions might help students to understand and solve the exercises with better clarity and confidence.

Chapter 1: Integers (include a table)

Chapter 2: Fractions and Decimals 

Chapter 3: Data Handling 

Chapter 4: Simple Equations 

Chapter 5: Lines and Angles 

Chapter 6: The Triangle and its Properties 

Chapter 7: Congruence of Triangles 

Chapter 8:Comparing Quantities 

Chapter 9: Rational Numbers 

Chapter 10: Practical Geometry 

Chapter 11: Perimeter and Area 

Chapter 12: Algebraic Expressions 

Chapter 13: Exponents and Powers 

Chapter 14: Symmetry 

Chapter 15: Visualizing Solid Shapes 

CBSE Class 7 Mathematics Unit-wise Marks Weightage

Sr No Topic/Unit Marks %
1 Integer 15
2 Fractions and Decimals 17.5
3 Data Handling 17.5
4 Simple Equations 17.5
5 Lines and Angles 17.5
6 Triangle and its Properties 12
7 Congruence of a Triangle 12
8 Comparing Quantities 14
9 Rational Numbers 7
10 Practical Geometry 12
11 Perimeter and Area 14
12 Algebraic Expressions 13
13 Exponents and Powers 14
14 Symmetry 10
15 Visualising Solid Shapes 5

Note: As the weightage is provided unit-wise, the content cannot be shuffled.

 

Key Features of Extramarks NCERT Solutions for Class 7 Mathematics

  • The NCERT Solutions For Class 7th Mathematics can be easily downloaded and used offline or online.
  • The solutions developed by subject matter experts are easy to understand and given in a step-by-step format with detailed explanation.
  • The solutions are made available to the students chapter-wise  along with the exercises to save students’ time and give more time for revision..
  • These solutions are created by our highly qualified subject matter experts to provide students with the right study material.

 

Benefits of Class 7 Mathematics NCERT Solutions

  • Students will get an overall view of the whole concepts mentioned in the NCERT Class 7th Mathematics textbook.
  • The solutions have been prepared carefully for students who are good at grasping concepts as well as those who find it difficult to understand the basic concepts quickly. Extramarks NCERT Solutions for Class 7 Mathematics are easy to understand for the slow as well as quick learners.
  • The NCERT Solutions For Class 7th Mathematics also help you to understand the recent question paper pattern and give you an idea about how to analyse and solve the questions.
  • The NCERT Mathematics Class 7 solutions are written under the guidelines of the latest CBSE rules and regulations.
  • The NCERT Solutions For Class 7th Mathematics covers the whole syllabus unit-wise and exercise-wise.
  • Extramarks NCERT Solutions For Class 7th Mathematics are authentic,  and highly accurate. 

If students want to score good marks without referring to other study material, Extramarks provides students with NCERT Solutions For Class 7th Mathematics which covers the whole syllabus and can be easily accessed at their official website.

Q.1 ABCD is a quadrilateral. Is AB + BC + CD + DA>AC + BD?

Ans

Since, in a triangle, the sum of either two sides is always greater than the third side. So, in ΔABC AB+BC>CA ( i ) Similary in ΔBCD, we get BC+CD>DB ( ii ) In ΔCDA, we get CD+DA>AC ( iii ) In ΔDAB, we get DA+AB>DB ( iv ) Adding (i), (ii), (iii) and (iv) to get AB+BC+BC+CD+CD+DA+DA+AB>CA+DB+AC+DB 2( AB+BC+CD+DA )+2(AC+DB) AB+BC+CD+DA>AC+BD Therefore, the given expression is true.

Q.2

ABCD is a quadrilateral. IsAB+BC+CD+DA<2AC+BD?

Ans

Consider a quadrilateral ABCD.

Since, in a triangle, the sum of either two sides is always greater than the third side. So, in ΔOAB OA+OB>AB ( i )
Similary in
ΔOBC, we get
OC+OB>BC ( ii )
In ΔOCD, we get OD+OC>CD ( iii )
In
ΔODA, we get
OA+OD>DA ( iv )
Adding (i), (ii), (iii) and (iv) to get
OA+OB+OC+OB+OD+OC+OA+OD>AB+BC+CD+DA
2
( OA+OB+OC+OD )>2( AC+BD )
( OA+OB+OC+OD )>( AC+BD )
Therefore, the given expression is true
.

Q.3

The lengths of two sides of a triangle are 12cm and 15cm.Between what two measures should the length of the thirdside fall?

Ans

In a triangle, the sum of either two sides is always greater than the third side and also, the difference of the lengths of either two sides is always lesser than the third side. Here, the third side will be lesser than the sum of these two (i.e., 12+15=27) and also, it will be greater than the difference of these two (i.e., 1512=3). Therefore, those two measures are 27 cm and 3 cm.

Q.4

PQR is a triangle right angled at P. If PQ =10 cm and PR=24cm,findQR.

Ans

By applying Pythagoras theorem in ΔPQR to get

(PQ)2 + (PR)2 = (QR)2
102 + 242 = QR2
100
+ 576 = QR2

676
= QR2

QR
=

676

= 26cm

Q.5

ABC is a triangle right angled at C. If AB = 25 cm andAC = 7 cm, find BC.

Ans

By Pythagoras theorem in ΔABC, we get ( AC ) 2 + ( BC ) 2 = ( AB ) 2 ( BC ) 2 = ( AB ) 2 ( AC ) 2 = 24 2 7 2 =62549=576 BC=24cm

Q.6

A15m long ladder reached a window 12m high from theground on placing it against a wall at a distancea‘. Find thedistance of the foot of the ladder from the wall.

Ans

By Pythagoras theorem ( 15 ) 2 = ( 12 ) 2 + a 2 225144= a 2 81= a 2 a=9cm Therefore, the distance of the foot of the ladder from the wall is 9 cm.

Q.7

Which of the following can be the sides of a right triangle?i2.5cm,6.5cm,6cm.ii2cm,2cm,5cm.iii1.5cm,2cm,2.5cm

Ans

( i ) 2.5 cm, 6.5 cm, 6 cm. 2.5 2 =6.25, 6.5 2 =42.25 and 6 2 =36 Here, 2 .5 2 + 6 2 = 6.5 2 So, the square of the length of one side is the sum of the squares of the lengths of remaining two sides. Hence, these are the sides of a rightangled triangle. ( ii ) 2 cm, 2 cm, 5 cm. 2 2 =4, 2 2 =4 and 5 2 =25 Here, 2 2 + 2 2 5 2 So, the square of the length of one side is not the sum of the squares of the lengths of remaining two sides. Hence, these are not the sides of a rightangled triangle. ( iii )1.5 cm, 2cm, 2.5 cm 1.5 2 =2.25, 2 2 =4 and 2 .5 2 =6.25 Here, 1 .5 2 + 2 2 = 2.5 2 So, the square of the length of one side is the sum of the squares of the lengths of remaining two sides. Hence, these are the sides of a rightangled triangle. 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Q.8

A tree is broken at a height of 5 m from the ground and itstop touches the ground at a distance of 12 m from theBase of the tree.Find the original height of the tree.

Ans

In above figure, BC represents the unbroken part of the tree. Point C represents the point where the tree broke and CA represents the broken part of the tree. Triangle ABC thus formed is a right-angled triangle. So, applying Pythagoras theorewm to get AC 2 = AB 2 + BC 2 = 12 2 + 5 2 =144+25 =169=13cm Thus,the original height of the tree is=AC+CB =13 cm+5 cm = 18 cm MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGjbGaaeOBaiaabccacaqGHbGaaeOy aiaab+gacaqG2bGaaeyzaiaabccacaqGMbGaaeyAaiaabEgacaqG1b GaaeOCaiaabwgacaqGSaGaaeiiaiaabkeacaqGdbGaaeiiaiaabkha caqGLbGaaeiCaiaabkhacaqGLbGaae4CaiaabwgacaqGUbGaaeiDai aabohacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabwhacaqGUbGa aeOyaiaabkhacaqGVbGaae4AaiaabwgacaqGUbGaaeiiaiaabchaca qGHbGaaeOCaiaabshacaqGGaGaae4BaiaabAgacaqGGaGaaeiDaiaa bIgacaqGLbGaaeiiaiaabshacaqGYbGaaeyzaiaabwgacaqGUaaaba Gaaeiuaiaab+gacaqGPbGaaeOBaiaabshacaqGGaGaae4qaiaabcca caqGYbGaaeyzaiaabchacaqGYbGaaeyzaiaabohacaqGLbGaaeOBai aabshacaqGZbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGWbGa ae4BaiaabMgacaqGUbGaaeiDaiaabccacaqG3bGaaeiAaiaabwgaca qGYbGaaeyzaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeiDaiaa bkhacaqGLbGaaeyzaiaabccacaqGIbGaaeOCaiaab+gacaqGRbGaae yzaiaabccacaqGHbGaaeOBaiaabsgacaqGGaGaae4qaiaabgeaaeaa caqGYbGaaeyzaiaabchacaqGYbGaaeyzaiaabohacaqGLbGaaeOBai aabshacaqGZbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGIbGa aeOCaiaab+gacaqGRbGaaeyzaiaab6gacaqGGaGaaeiCaiaabggaca qGYbGaaeiDaiaabccacaqGVbGaaeOzaiaabccacaqG0bGaaeiAaiaa bwgacaqGGaGaaeiDaiaabkhacaqGLbGaaeyzaiaab6caaeaacaqGub GaaeOCaiaabMgacaqGHbGaaeOBaiaabEgacaqGSbGaaeyzaiaabcca caqGbbGaaeOqaiaaboeacaqGGaGaaeiDaiaabIgacaqG1bGaae4Cai aabccacaqGMbGaae4BaiaabkhacaqGTbGaaeyzaiaabsgacaqGGaGa aeyAaiaabohacaqGGaGaaeyyaiaabccacaqGYbGaaeyAaiaabEgaca qGObGaaeiDaiaab2cacaqGHbGaaeOBaiaabEgacaqGSbGaaeyzaiaa bsgacaqGGaGaaeiDaiaabkhacaqGPbGaaeyyaiaab6gacaqGNbGaae iBaiaabwgacaqGUaaabaGaae4uaiaab+gacaqGSaGaaeiiaiaabgga caqGWbGaaeiCaiaabYgacaqG5bGaaeyAaiaab6gacaqGNbGaaeiiai aabcfacaqG5bGaaeiDaiaabIgacaqGHbGaae4zaiaab+gacaqGYbGa aeyyaiaabohacaqGGaGaaeiDaiaabIgacaqGLbGaae4Baiaabkhaca qGLbGaae4Daiaab2gacaqGGaGaaeiDaiaab+gacaqGGaGaae4zaiaa bwgacaqG0baabaGaaeyqaiaaboeadaahaaWcbeqaaiaaikdaaaGccq GH9aqpcaqGbbGaaeOqamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaa bkeacaqGdbWaaWbaaSqabeaacaaIYaaaaaGcbaGaaeiiaiaabccaca qGGaGaaeiiaiaabccacqGH9aqpcaaIXaGaaGOmamaaCaaaleqabaGa aGOmaaaakiabgUcaRiaaiwdadaahaaWcbeqaaiaaikdaaaaakeaaca qGGaGaaeiiaiaabccacaqGGaGaaeiiaiabg2da9iaaigdacaaI0aGa aGinaiabgUcaRiaaikdacaaI1aaabaGaaeiiaiaabccacaqGGaGaae iiaiaabccacqGH9aqpcaaIXaGaaGOnaiaaiMdacqGH9aqpcaaIXaGa aG4maiaaysW7caqGJbGaaeyBaaqaaiaabsfacaqGObGaaeyDaiaabo hacaqGSaGaaeiDaiaabIgacaqGLbGaaeiiaiaab+gacaqGYbGaaeyA aiaabEgacaqGPbGaaeOBaiaabggacaqGSbGaaeiiaiaabIgacaqGLb GaaeyAaiaabEgacaqGObGaaeiDaiaabccacaqGVbGaaeOzaiaabcca caqG0bGaaeiAaiaabwgacaqGGaGaaeiDaiaabkhacaqGLbGaaeyzai aabccacaqGPbGaae4Caiabg2da9iaabgeacaqGdbGaey4kaSIaae4q aiaabkeaaeaacaWLjaGaaCzcaiaaxMaacaWLjaGaaCzcaiaaxMaaca WLjaGaaCzcaiaaysW7caaMe8Uaeyypa0JaaeymaiaabodacaqGGaGa ae4yaiaab2gacqGHRaWkcaqG1aGaaeiiaiaabogacaqGTbaabaGaaC zcaiaaxMaacaWLjaGaaCzcaiaaxMaacaWLjaGaaCzcaiaaxMaacaaM e8UaaGjbVlabg2da9maaL4babaGaaeymaiaabIdacaqGGaGaae4yai aab2gaaaaaaaa@82B5@

Q.9

Angles Q and R of a ΔPQR are 25º and 65º.Write which of the following is true:iPQ2+QR2=RP2iiPQ2+RP2=QR2iiiRP2+QR2=PQ2

Ans

Since, the sum of all interior angle is 180°. So, 25° + 65°+QPR=180° QPR=180°90° =90° Therefore, ΔPQR is a rigfht angled triangle at P. Thus, PQ 2 + RP 2 = QR 2 So, (ii) is true. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGtbGaaeyAaiaab6gacaqGJbGaaeyz aiaabYcacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabohacaqG1b GaaeyBaiaabccacaqGVbGaaeOzaiaabccacaqGHbGaaeiBaiaabYga caqGGaGaaeyAaiaab6gacaqG0bGaaeyzaiaabkhacaqGPbGaae4Bai aabkhacaqGGaGaaeyyaiaab6gacaqGNbGaaeiBaiaabwgacaqGGaGa aeyAaiaabohacaqGGaGaaeymaiaabIdacaqGWaGaeyiSaaRaaeOlaa qaaiaabofacaqGVbGaaeilaiaabccacaqGYaGaaeynaiabgclaWkaa bccacaqGRaGaaeiiaiaabAdacaqG1aGaeyiSaaRaey4kaSIaeyiiIa TaaeyuaiaabcfacaqGsbGaeyypa0JaaGymaiaaiIdacaaIWaGaeyiS aalabaGaeyiiIaTaaeyuaiaabcfacaqGsbGaeyypa0JaaGymaiaaiI dacaaIWaGaeyiSaaRaeyOeI0IaaGyoaiaaicdacqGHWcaSaeaacaWL jaGaaGjbVlaaysW7cqGH9aqpcaaI5aGaaGimaiabgclaWcqaaiaabs facaqGObGaaeyzaiaabkhacaqGLbGaaeOzaiaab+gacaqGYbGaaeyz aiaabYcacaqGGaGaeyiLdqKaaeiuaiaabgfacaqGsbGaaeiiaiaabM gacaqGZbGaaeiiaiaabggacaqGGaGaaeOCaiaabMgacaqGNbGaaeOz aiaabIgacaqG0bGaaeiiaiaabggacaqGUbGaae4zaiaabYgacaqGLb GaaeizaiaabccacaqG0bGaaeOCaiaabMgacaqGHbGaaeOBaiaabEga caqGSbGaaeyzaiaabccacaqGHbGaaeiDaiaabccacaqGqbGaaeOlaa qaaiaabsfacaqGObGaaeyDaiaabohacaqGSaGaaeiiamaaL4babaGa aeiuaiaabgfadaahaaWcbeqaaiaabkdaaaGccqGHRaWkcaqGsbGaae iuamaaCaaaleqabaGaaeOmaaaakiabg2da9iaabgfacaqGsbWaaWba aSqabeaacaqGYaaaaaaaaOqaaiaabofacaqGVbGaaeilaiaabccaca qGOaGaaeyAaiaabMgacaqGPaGaaeiiaiaabMgacaqGZbGaaeiiaiaa bshacaqGYbGaaeyDaiaabwgacaqGUaaaaaa@D7AA@

Q.10

Find the perimeter of the rectangle whose length is 40 cmand a diagonal is 41cm.

Ans

In a rectangle, all interior angles are of 90º measure. Therefore,Pythagoras theorem can be applied here. ( 41 ) 2 = ( 40 ) 2 + x 2 1681 =1600 + x 2 x 2 = 1681 1600 = 81 x= 9 cm So, Perimeter =2( Length + Breadth ) = 2( x+ 40 ) = 2 ( 9 + 40 ) = 98 cm MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGjbGaaeOBaiaabccacaqGHbGaaeii aiaabkhacaqGLbGaae4yaiaabshacaqGHbGaaeOBaiaabEgacaqGSb GaaeyzaiaacYcacaqGGaGaaeyyaiaabYgacaqGSbGaaeiiaiaabMga caqGUbGaaeiDaiaabwgacaqGYbGaaeyAaiaab+gacaqGYbGaaeiiai aabggacaqGUbGaae4zaiaabYgacaqGLbGaae4CaiaabccacaqGHbGa aeOCaiaabwgacaqGGaGaae4BaiaabAgacaqGGaGaaeyoaiaaicdaca GG6cGaaeiiaiaab2gacaqGLbGaaeyyaiaabohacaqG1bGaaeOCaiaa bwgacaGGUaaabaGaaeivaiaabIgacaqGLbGaaeOCaiaabwgacaqGMb Gaae4BaiaabkhacaqGLbGaaiilaiaabcfacaqG5bGaaeiDaiaabIga caqGHbGaae4zaiaab+gacaqGYbGaaeyyaiaabohacaqGGaGaaeiDai aabIgacaqGLbGaae4BaiaabkhacaqGLbGaaeyBaiaabccacaqGJbGa aeyyaiaab6gacaqGGaGaaeOyaiaabwgacaqGGaGaaeyyaiaabchaca qGWbGaaeiBaiaabMgacaqGLbGaaeizaiaabccacaqGObGaaeyzaiaa bkhacaqGLbGaaiOlaaqaaiaaxMaacaWLjaWaaeWaaeaacaqG0aGaae ymaaGaayjkaiaawMcaamaaCaaaleqabaGaaeOmaaaakiaaysW7caaM e8Uaeyypa0ZaaeWaaeaacaqG0aGaaGimaaGaayjkaiaawMcaamaaCa aaleqabaGaaeOmaaaakiabgUcaRiaadIhadaahaaWcbeqaaiaabkda aaaakeaacaWLjaGaaCzcaiaabgdacaqG2aGaaeioaiaabgdacaqGGa Gaeyypa0JaaeymaiaabAdacaaIWaGaaGimaiaabccacqGHRaWkcaWG 4bWaaWbaaSqabeaacaqGYaaaaaGcbaGaaCzcaiaaxMaacaaMe8UaaG jbVlaaysW7caaMe8UaamiEamaaCaaaleqabaGaaeOmaaaakiabg2da 9iaabccacaqGXaGaaeOnaiaabIdacaqGXaGaaeiiaiabgkHiTiaabc cacaqGXaGaaeOnaiaaicdacaaIWaaabaGaaCzcaiaaxMaacaaMe8Ua aGjbVlaabccacaaMe8UaaCzcaiaaysW7caaMe8Uaeyypa0Jaaeiiai aabIdacaqGXaaabaGaaCzcaiaaxMaacaaMe8UaaGjbVlaaysW7caaM e8UaaGjbVlaadIhacqGH9aqpcaqGGaGaaeyoaiaabccacaqGJbGaae yBaaqaaiaabofacaqGVbGaaeilaiaabccacaqGqbGaaeyzaiaabkha caqGPbGaaeyBaiaabwgacaqG0bGaaeyzaiaabkhacaqGGaGaeyypa0 JaaeOmamaabmaabaGaaeitaiaabwgacaqGUbGaae4zaiaabshacaqG ObGaaeiiaiabgUcaRiaabccacaqGcbGaaeOCaiaabwgacaqGHbGaae izaiaabshacaqGObaacaGLOaGaayzkaaaabaGaaCzcaiaaxMaacaWL jaGaaGjbVlaaysW7cqGH9aqpcaqGGaGaaeOmamaabmaabaGaamiEai abgUcaRiaabccacaqG0aGaaGimaaGaayjkaiaawMcaaaqaaiaaxMaa caWLjaGaaCzcaiaaysW7caaMe8Uaeyypa0JaaeiiaiaabkdacaqGGa WaaeWaaeaacaqG5aGaaeiiaiabgUcaRiaabccacaqG0aGaaGimaaGa ayjkaiaawMcaaaqaaiaaxMaacaWLjaGaaCzcaiaaysW7caaMe8Uaey ypa0JaaeiiamaaL4babaGaaeyoaiaabIdacaqGGaGaae4yaiaab2ga aaaaaaa@2661@

Q.11

The diagonals of a rhombus measure 16 cm and 30 cm.Find its perimeter.

Ans

Let ABCD be a rhombus( all sides are of equal length )and its diagonals, AC and BD, are intersecting each other at point O. Since, diagonals of a rhombus bisect each other at 90°. So, By applying Pythagoras theorem in ΔAOB, OA 2 + OB 2 = AB 2 8 2 + 15 2 = AB 2 64+225 = AB 2 289 = AB 2 AB = 17 cm Therefore,the length of the side of rhombus is 17 cm. Perimeter of rhombus =4×Side of the rhombus = 4×17 = 68 cm MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGmbGaaeyzaiaabshacaqGGaGaaeyq aiaabkeacaqGdbGaaeiraiaabccacaqGIbGaaeyzaiaabccacaqGHb GaaeiiaiaabkhacaqGObGaae4Baiaab2gacaqGIbGaaeyDaiaaboha daqadaqaaiaabggacaqGSbGaaeiBaiaabccacaqGZbGaaeyAaiaabs gacaqGLbGaae4CaiaabccacaqGHbGaaeOCaiaabwgacaqGGaGaae4B aiaabAgacaqGGaGaaeyzaiaabghacaqG1bGaaeyyaiaabYgacaqGGa GaaeiBaiaabwgacaqGUbGaae4zaiaabshacaqGObaacaGLOaGaayzk aaGaaeyyaiaab6gacaqGKbGaaeiiaiaabMgacaqG0bGaae4Caaqaai aabsgacaqGPbGaaeyyaiaabEgacaqGVbGaaeOBaiaabggacaqGSbGa ae4CaiaacYcacaqGGaGaaeyqaiaaboeacaqGGaGaaeyyaiaab6gaca qGKbGaaeiiaiaabkeacaqGebGaaiilaiaabccacaqGHbGaaeOCaiaa bwgacaqGGaGaaeyAaiaab6gacaqG0bGaaeyzaiaabkhacaqGZbGaae yzaiaabogacaqG0bGaaeyAaiaab6gacaqGNbGaaeiiaiaabwgacaqG HbGaae4yaiaabIgacaqGGaGaae4BaiaabshacaqGObGaaeyzaiaabk hacaqGGaGaaeyyaiaabshacaqGGaGaaeiCaiaab+gacaqGPbGaaeOB aiaabshacaqGGaGaae4taiaac6caaeaacaqGtbGaaeyAaiaab6gaca qGJbGaaeyzaiaabYcacaqGGaGaaeizaiaabMgacaqGHbGaae4zaiaa b+gacaqGUbGaaeyyaiaabYgacaqGZbGaaeiiaiaab+gacaqGMbGaae iiaiaabggacaqGGaGaaeOCaiaabIgacaqGVbGaaeyBaiaabkgacaqG 1bGaae4CaiaabccacaqGIbGaaeyAaiaabohacaqGLbGaae4yaiaabs hacaqGGaGaaeyzaiaabggacaqGJbGaaeiAaiaabccacaqGVbGaaeiD aiaabIgacaqGLbGaaeOCaiaabccacaqGHbGaaeiDaiaabccacaqG5a GaaGimaiabgclaWkaac6caaeaacaqGtbGaae4BaiaabYcaaeaacaqG cbGaaeyEaiaabccacaqGHbGaaeiCaiaabchacaqGSbGaaeyEaiaabM gacaqGUbGaae4zaiaabccacaqGqbGaaeyEaiaabshacaqGObGaaeyy aiaabEgacaqGVbGaaeOCaiaabggacaqGZbGaaeiiaiaabshacaqGOb Gaaeyzaiaab+gacaqGYbGaaeyzaiaab2gacaqGGaGaaeyAaiaab6ga caqGGaGaeuiLdqKaaeyqaiaab+eacaqGcbGaaiilaaqaaiaab+eaca qGbbWaaWbaaSqabeaacaqGYaaaaOGaey4kaSIaae4taiaabkeadaah aaWcbeqaaiaabkdaaaGccqGH9aqpcaqGGaGaaeyqaiaabkeadaahaa WcbeqaaiaabkdaaaaakeaacaaMe8UaaGjbVlaaysW7caqG4aWaaWba aSqabeaacaqGYaaaaOGaey4kaSIaaeymaiaabwdadaahaaWcbeqaai aabkdaaaGccqGH9aqpcaqGGaGaaeyqaiaabkeadaahaaWcbeqaaiaa bkdaaaaakeaacaaMe8UaaeOnaiaabsdacqGHRaWkcaqGYaGaaeOmai aabwdacaqGGaGaeyypa0JaaeiiaiaabgeacaqGcbWaaWbaaSqabeaa caqGYaaaaaGcbaGaaCzcaiaaysW7caaMe8UaaeOmaiaabIdacaqG5a Gaaeiiaiabg2da9iaabccacaqGbbGaaeOqamaaCaaaleqabaGaaeOm aaaaaOqaaiaaxMaacaaMe8UaaGjbVlaaysW7caqGbbGaaeOqaiaabc cacqGH9aqpcaqGGaGaaeymaiaabEdacaqGGaGaae4yaiaab2gaaeaa caqGubGaaeiAaiaabwgacaqGYbGaaeyzaiaabAgacaqGVbGaaeOCai aabwgacaGGSaGaaeiDaiaabIgacaqGLbGaaeiiaiaabYgacaqGLbGa aeOBaiaabEgacaqG0bGaaeiAaiaabccacaqGVbGaaeOzaiaabccaca qG0bGaaeiAaiaabwgacaqGGaGaae4CaiaabMgacaqGKbGaaeyzaiaa bccacaqGVbGaaeOzaiaabccacaqGYbGaaeiAaiaab+gacaqGTbGaae OyaiaabwhacaqGZbGaaeiiaiaabMgacaqGZbGaaeiiaiaabgdacaqG 3aGaaeiiaiaabogacaqGTbGaaiOlaaqaaiaabcfacaqGLbGaaeOCai aabMgacaqGTbGaaeyzaiaabshacaqGLbGaaeOCaiaabccacaqGVbGa aeOzaiaabccacaqGYbGaaeiAaiaab+gacaqGTbGaaeOyaiaabwhaca qGZbGaaeiiaiabg2da9iaabsdacqGHxdaTcaqGtbGaaeyAaiaabsga caqGLbGaaeiiaiaab+gacaqGMbGaaeiiaiaabshacaqGObGaaeyzai aabccacaqGYbGaaeiAaiaab+gacaqGTbGaaeOyaiaabwhacaqGZbaa baGaaeiiaiaaxMaacaWLjaGaaCzcaiaaxMaacaWLjaGaeyypa0Jaae iiaiaabsdacqGHxdaTcaqGXaGaae4naaqaaiaabccacaWLjaGaaCzc aiaaxMaacaWLjaGaaCzcaiabg2da9iaabccadaqjEaqaaiaabAdaca qG4aGaaeiiaiaabogacaqGTbaaaaaaaa@9BE4@

Q.12

Complete the following statements:a) Two line segments are congruent if ________.b) Among two congruent angles, one has a measure of 70°; the measure of the other angle is _________.c) When we writeA=B, we actually mean _________.

Ans

Complete the following statements:(a) Two line segments are congruent if they have the same length_.(b) Among two congruent angles, one has a measure of 70°;the measure of the other angle is 70°_.(c) When we writeA =B, we actually mean m A = m B_.

Q.13

Give any two reallife examples for congruent shapes.

Ans

(i) Sheets of same letter pad(ii) Biscuits in the same packet

Q.14

If ΔABC ΔFED under the correspondence ABC FED,write all the corresponding congruent parts of the triangles.

Ans

If these triangles are congruent,then the corresponding anglesand sides will be equal to each other.So,A FB EC DAB¯FE¯BC¯ED¯CA¯DF¯

Q.15

If ΔDEF ΔBCA, write the part(s) of ΔBCA thatcorrespond to(i)E (ii)EF ¯(iii)F (iv) DF¯

Ans

(i) C(ii) CA¯(iii) A(iv) BA¯

Q.16

Which congruence criterion do you use in the following?(a)Given:AC=DFAB=DEBC=EF


S
o,ΔABCΔDEF

(b)Given:ZX=RP  RQ=ZY    PRQ=XZYSo,ΔPQRΔXYZ

(c)Given:MLN=FGH  NML=GFHML=FG   So,ΔLMNΔGFH

(d)Given:EB=DB  AE=BC  A=C=90°So,ΔABEΔCDB

Ans

(a) SSS, as the sides of ΔABC are equal to the sides of ΔDEF.(b) SAS, as two sides and the angle included between these sides of ΔPQR are equal to two sides and the angle included between these sides of ΔXYZ.(c) ASA, as two angles and the side included between these angles of ΔLMN are equal to two angles and the side included between these angles of ΔGFH.(d) RHS, as in the given two right-angled triangles,one side and the hypotenuse are respectively equal.

Q.17

You want to show that ΔART@ΔPEN,a If you have to use SSS criterion, then you need to showi AR = ii RT = iii AT =b If it is given that ∠T =N and you are to use SAS criterion, you need to havei RT = andii PN =c If it is given that AT=PN and you are to use ASA criterion, you need to havei ? ii?

Ans

(a)(i) AR=PE (ii) RT=EN (iii) AT = PN(b)(i) RT=EN (ii) PN=AT(c)(i) ATR=PNE (ii) RAT=EPN

Q.18

You have to show that AMPAMQ.In the following


p
roof,supply the missing reasons.

Steps Reasons
(i) PM = QM (i)…
( ii )PMA =QMA MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGcbaWaaeWaaeaaieqacaWFPbGaa8xAaaGaayjkaiaa wMcaaiabgcIiqlaa=bfacaWFnbGaa8xqaiaa=bcacqGH9aqpcqGHGi c0caWFrbGaa8xtaiaa=feaaaa@4A4C@ (ii)…
(iii) AM = AM (iii)…
( iv )ΔAMPΔAMQ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGcbaWaaeWaaeaaieqacaWFPbGaa8NDaaGaayjkaiaa wMcaaiaa=r5acaWFbbGaa8xtaiaa=bfacqGHfjcqcaWFuoGaa8xqai aa=1eacaWFrbaaaa@48D9@ (iv)…

Ans

Steps Reasons
(i) PM = QM (i)…Given
(ii)PMA=QMA (ii)…Given
(iii) AM = AM (iii)…Common
(iv)ΔAMPΔAMQ (iv) SAS,as the two sides and the angle included between these sides of ΔAMP are equal to two sides and the angleincluded between these sides of ΔAMQ

Q.19

In ΔABC,A = 30°,B = 40° andC = 110°InΔPQR,P = 30° ,Q = 40° andR = 110°A student says that ΔABCΔPQR by AAA congruencecriterion. Is he justified? Why or why not?

Ans

No. This property represents that these triangles have theirrespective angles of equal measure. However, this gives noinformation about their sides. The sides of these triangles havea ratio somewhat different than 1:1.Therefore, AAA property does not prove the twotriangles are congruent.

Q.20

In the figure, the two triangles are congruent.The corresponding parts are marked.


W
e can write ΔRAT?

Ans

From the figure, we observe that:RAT = WONART = OWN   AR = OWTherefore, ΔRATΔWON, by ASA criterion.

Q.21

Complete the congruence statement:

Ans

Given that,BC = BTTA = CABA is common.Therefore, ΔBCAΔBTASimilarly,PQ = RSTQ = QSPT = RQTherefore, ΔQRSΔTPQ.

Q.22

In a squared sheet,draw two triangles of equalareas such that(i) the triangles are congruent.(ii) the triangles are not congruent.(iii) What can you say about the irperimeters?

Ans

(i)

Here
,ΔABC and ΔPQR have the same area and are congruent to each other also. Also, the perimeter of both the triangles
will be the same
.
(ii)

Here,the two triangles have the same height and base.Thus, their areas are equal.However, these triangles are not congruent to each other.Also, the perimeter of both the triangles will not be the same.

Q.23

Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent.

Ans

A pair of triangles with threee equal angles, and two equalsides are non congruent.


Below is the example:

Q.24

If ΔABC and ΔPQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

Ans

Here,BC = QRSo, ΔABCΔPQR (ASA criterion)

Q.25

Explain, whyΔABC ΔFED

Ans

Given that, ABC = FED ( 1 ) BAC = EFD (2) The two angles of ΔABC are equal to the two respective angles of ΔFED. Also, the sum of all interior angles of a triangle is 180º. Therefore, third angle of both triangles will also be equal in measure. BCA = EDF( 3 ) Also, given that, BC = ED( 4 ) By using equation ( 1 ), ( 3 ), and ( 4 ),we obtain ΔABC ΔFED (ASA criterion) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGhbGaaeyAaiaabAhacaqGLbGaaeOB aiaabccacaqG0bGaaeiAaiaabggacaqG0bGaaiilaaqaaiabgcIiql aabgeacaqGcbGaae4qaiaabccacqGH9aqpcaqGGaGaeyiiIaTaaeOr aiaabweacaqGebGaaeiiaiablAciljablAciljablAciljablAciln aabmaabaGaaeymaaGaayjkaiaawMcaaaqaaiabgcIiqlaabkeacaqG bbGaae4qaiaabccacqGH9aqpcaqGGaGaeyiiIaTaaeyraiaabAeaca qGebGaeSOjGSKaeSOjGSKaeSOjGSKaeSOjGSKaaeiiaiaacIcacaqG YaGaaiykaaqaaiaabsfacaqGObGaaeyzaiaabccacaqG0bGaae4Dai aab+gacaqGGaGaaeyyaiaab6gacaqGNbGaaeiBaiaabwgacaqGZbGa aeiiaiaab+gacaqGMbGaaeiiaiabfs5aejaabgeacaqGcbGaae4qai aabccacaqGHbGaaeOCaiaabwgacaqGGaGaaeyzaiaabghacaqG1bGa aeyyaiaabYgacaqGGaGaaeiDaiaab+gacaqGGaGaaeiDaiaabIgaca qGLbGaaeiiaiaabshacaqG3bGaae4BaiaabccacaqGYbGaaeyzaiaa bohacaqGWbGaaeyzaiaabogacaqG0bGaaeyAaiaabAhacaqGLbaaba Gaaeyyaiaab6gacaqGNbGaaeiBaiaabwgacaqGZbGaaeiiaiaab+ga caqGMbGaaeiiaiabfs5aejaabAeacaqGfbGaaeiraiaac6cacaqGGa GaaeyqaiaabYgacaqGZbGaae4BaiaacYcacaqGGaGaaeiDaiaabIga caqGLbGaaeiiaiaabohacaqG1bGaaeyBaiaabccacaqGVbGaaeOzai aabccacaqGHbGaaeiBaiaabYgacaqGGaGaaeyAaiaab6gacaqG0bGa aeyzaiaabkhacaqGPbGaae4BaiaabkhacaqGGaGaaeyyaiaab6gaca qGNbGaaeiBaiaabwgacaqGZbGaaeiiaiaab+gacaqGMbGaaeiiaiaa bggaaeaacaqG0bGaaeOCaiaabMgacaqGHbGaaeOBaiaabEgacaqGSb GaaeyzaiaabccacaqGPbGaae4CaiaabccacaqGXaGaaeioaiaaicda caGG6cGaaiOlaiaabccacaqGubGaaeiAaiaabwgacaqGYbGaaeyzai aabAgacaqGVbGaaeOCaiaabwgacaGGSaGaaeiiaiaabshacaqGObGa aeyAaiaabkhacaqGKbGaaeiiaiaabggacaqGUbGaae4zaiaabYgaca qGLbGaaeiiaiaab+gacaqGMbGaaeiiaiaabkgacaqGVbGaaeiDaiaa bIgacaqGGaGaaeiDaiaabkhacaqGPbGaaeyyaiaab6gacaqGNbGaae iBaiaabwgacaqGZbGaaeiiaiaabEhacaqGPbGaaeiBaiaabYgaaeaa caqGHbGaaeiBaiaabohacaqGVbGaaeiiaiaabkgacaqGLbGaaeiiai aabwgacaqGXbGaaeyDaiaabggacaqGSbGaaeiiaiaabMgacaqGUbGa aeiiaiaab2gacaqGLbGaaeyyaiaabohacaqG1bGaaeOCaiaabwgaca GGUaaabaGaeyiiIaTaaeOqaiaaboeacaqGbbGaaeiiaiabg2da9iaa bccacqGHGic0caqGfbGaaeiraiaabAeacqWIMaYscqWIMaYscqWIMa YscqWIMaYsdaqadaqaaiaabodaaiaawIcacaGLPaaaaeaacaqGbbGa aeiBaiaabohacaqGVbGaaiilaiaabccacaqGNbGaaeyAaiaabAhaca qGLbGaaeOBaiaabccacaqG0bGaaeiAaiaabggacaqG0bGaaiilaaqa aiaabccacaqGcbGaae4qaiaabccacqGH9aqpcaqGGaGaaeyraiaabs eacqWIMaYscqWIMaYscqWIMaYscqWIMaYscqWIMaYscqWIMaYsdaqa daqaaiaabsdaaiaawIcacaGLPaaaaeaacaqGcbGaaeyEaiaabccaca qG1bGaae4CaiaabMgacaqGUbGaae4zaiaabccacaqGLbGaaeyCaiaa bwhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBaiaabccadaqadaqaai aabgdaaiaawIcacaGLPaaacaGGSaGaaeiiamaabmaabaGaae4maaGa ayjkaiaawMcaaiaacYcacaqGGaGaaeyyaiaab6gacaqGKbGaaeiiam aabmaabaGaaeinaaGaayjkaiaawMcaaiaacYcacaqG3bGaaeyzaiaa bccacaqGVbGaaeOyaiaabshacaqGHbGaaeyAaiaab6gaaeaadaqjEa qaaiabfs5aejaabgeacaqGcbGaae4qaiaabccacqGHfjcqcaqGGaGa euiLdqKaaeOraiaabweacaqGebaaaiaabccacaGGOaGaaeyqaiaabo facaqGbbGaaeiiaiaabogacaqGYbGaaeyAaiaabshacaqGLbGaaeOC aiaabMgacaqGVbGaaeOBaiaacMcaaaaa@8308@

Q.26

Find the ratio of:a Rs 5 to 50 paise b 15 kg to 210 gc 9 m to 27 cm d 30 days to 36 hour

Ans

( a ) Rs 5 to 50 paise 1 ruppe=100 paise So, 5 rupee=500 paise So, Rs 5 50 paise = 500 50 = 10 1 Thus, the required ratio is 10:1 . ( b ) 15 kg to 210 g 1 kg=1000 g So, 15 kg=15000 g So, 15 kg 210 g = 15000 210 = 500 7 Thus, the required ratio is 500:7 . ( c ) 9 m to 27 cm 1 m=100 cm So, 9 m=900 cm So, 9 m 27 cm = 900 27 = 100 3 Thus, the required ratio is 100:3 . ( d ) 30 days to 36 hour 1 day=24 hours So, 30 days=720 hours So, 30 days 36 hour = 720 36 = 20 1 Thus, the required ratio is 20:1 . MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabggaaiaawIcacaGLPaaa caqGGaGaaeOuaiaabohacaqGGaGaaeynaiaabccacaqG0bGaae4Bai aabccacaqG1aGaaGimaiaabccacaqGWbGaaeyyaiaabMgacaqGZbGa aeyzaaqaaiaaxMaacaaIXaGaaeiiaiaabkhacaqG1bGaaeiCaiaabc hacaqGLbGaeyypa0JaaeymaiaabcdacaqGWaGaaeiiaiaabchacaqG HbGaaeyAaiaabohacaqGLbaabaGaae4uaiaab+gacaqGSaGaaeiiai aaysW7caqG1aGaaeiiaiaabkhacaqG1bGaaeiCaiaabwgacaqGLbGa eyypa0JaaeynaiaabcdacaqGWaGaaeiiaiaabchacaqGHbGaaeyAai aabohacaqGLbaabaGaae4uaiaab+gacaqGSaGaaeiiamaalaaabaGa aeOuaiaabohacaqGGaGaaGynaaqaaiaaiwdacaaIWaGaaeiiaiaabc hacaqGHbGaaeyAaiaabohacaqGLbaaaiabg2da9maalaaabaGaaGyn aiaaicdacaaIWaaabaGaaGynaiaaicdaaaGaeyypa0ZaaSaaaeaaca aIXaGaaGimaaqaaiaaigdaaaaabaGaaeivaiaabIgacaqG1bGaae4C aiaabYcacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabkhacaqGLb GaaeyCaiaabwhacaqGPbGaaeOCaiaabwgacaqGKbGaaeiiaiaabkha caqGHbGaaeiDaiaabMgacaqGVbGaaeiiaiaabMgacaqGZbGaaeiiam aaL4babaGaaeymaiaabcdacaqG6aGaaeymaaaacaGGUaaabaWaaeWa aeaacaqGIbaacaGLOaGaayzkaaGaaeiiaiaabgdacaqG1aGaaeiiai aabUgacaqGNbGaaeiiaiaabshacaqGVbGaaeiiaiaabkdacaqGXaGa aGimaiaabccacaqGNbaabaGaaeymaiaabccacaqGRbGaae4zaiabg2 da9iaabgdacaqGWaGaaeimaiaabcdacaqGGaGaae4zaaqaaiaabofa caqGVbGaaeilaiaabccacaqGXaGaaeynaiaabccacaqGRbGaae4zai abg2da9iaabgdacaqG1aGaaeimaiaabcdacaqGWaGaaeiiaiaabEga aeaacaqGtbGaae4BaiaabYcacaqGGaWaaSaaaeaacaaIXaGaaGynai aabccacaqGRbGaae4zaaqaaiaaikdacaaIXaGaaGimaiaabccacaqG Nbaaaiabg2da9maalaaabaGaaGymaiaaiwdacaaIWaGaaGimaiaaic daaeaacaaIYaGaaGymaiaaicdaaaGaeyypa0ZaaSaaaeaacaaI1aGa aGimaiaaicdaaeaacaaI3aaaaaqaaiaabsfacaqGObGaaeyDaiaabo hacaqGSaGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGYbGaaeyz aiaabghacaqG1bGaaeyAaiaabkhacaqGLbGaaeizaiaabccacaqGYb GaaeyyaiaabshacaqGPbGaae4BaiaabccacaqGPbGaae4Caiaabcca daqjEaqaaiaabwdacaqGWaGaaeimaiaabQdacaqG3aaaaiaac6caae aadaqadaqaaiaabogaaiaawIcacaGLPaaacaqGGaGaaeyoaiaabcca caqGTbGaaeiiaiaabshacaqGVbGaaeiiaiaabkdacaqG3aGaaeiiai aabogacaqGTbaabaGaaeymaiaabccacaqGTbGaeyypa0Jaaeymaiaa bcdacaqGWaGaaeiiaiaabogacaqGTbaabaGaae4uaiaab+gacaqGSa GaaeiiaiaabMdacaqGGaGaaeyBaiabg2da9iaabMdacaqGWaGaaeim aiaabccacaqGJbGaaeyBaaqaaiaabofacaqGVbGaaeilaiaabccada WcaaqaaiaaiMdacaqGGaGaaeyBaaqaaiaaikdacaaI3aGaaeiiaiaa bogacaqGTbaaaiabg2da9maalaaabaGaaGyoaiaaicdacaaIWaaaba GaaGOmaiaaiEdaaaGaeyypa0ZaaSaaaeaacaaIXaGaaGimaiaaicda aeaacaaIZaaaaaqaaiaabsfacaqGObGaaeyDaiaabohacaqGSaGaae iiaiaabshacaqGObGaaeyzaiaabccacaqGYbGaaeyzaiaabghacaqG 1bGaaeyAaiaabkhacaqGLbGaaeizaiaabccacaqGYbGaaeyyaiaabs hacaqGPbGaae4BaiaabccacaqGPbGaae4CaiaabccadaqjEaqaaiaa bgdacaqGWaGaaeimaiaabQdacaqGZaaaaiaac6caaeaadaqadaqaai aabsgaaiaawIcacaGLPaaacaqGGaGaae4maiaaicdacaqGGaGaaeiz aiaabggacaqG5bGaae4CaiaabccacaqG0bGaae4BaiaabccacaqGZa GaaeOnaiaabccacaqGObGaae4BaiaabwhacaqGYbaabaGaaeymaiaa bccacaqGKbGaaeyyaiaabMhacqGH9aqpcaqGYaGaaeinaiaabccaca qGObGaae4BaiaabwhacaqGYbGaae4CaaqaaiaabofacaqGVbGaaeil aiaabccacaqGZaGaaeimaiaabccacaqGKbGaaeyyaiaabMhacaqGZb Gaeyypa0Jaae4naiaabkdacaqGWaGaaeiiaiaabIgacaqGVbGaaeyD aiaabkhacaqGZbaabaGaae4uaiaab+gacaqGSaGaaeiiamaalaaaba GaaG4maiaaicdacaqGGaGaaeizaiaabggacaqG5bGaae4Caaqaaiaa iodacaaI2aGaaeiiaiaabIgacaqGVbGaaeyDaiaabkhaaaGaeyypa0 ZaaSaaaeaacaaI3aGaaGOmaiaaicdaaeaacaaIZaGaaGOnaaaacqGH 9aqpdaWcaaqaaiaaikdacaaIWaaabaGaaGymaaaaaeaacaqGubGaae iAaiaabwhacaqGZbGaaeilaiaabccacaqG0bGaaeiAaiaabwgacaqG GaGaaeOCaiaabwgacaqGXbGaaeyDaiaabMgacaqGYbGaaeyzaiaabs gacaqGGaGaaeOCaiaabggacaqG0bGaaeyAaiaab+gacaqGGaGaaeyA aiaabohacaqGGaWaauIhaeaacaqGYaGaaeimaiaabQdacaqGXaaaai aac6caaaaa@A847@

Q.27

In a computer lab, there are 3 computers for every 6students.How many computers will be needed for 24students?

Ans

For 6 students, number of computers required=3 So, for 1 student, number of computers required= 3 6 = 1 2 Thus, for 24 students, number of computers required= 1 2 ×24 = 12 Therefore, 12 computers are needed for 24 students.

Q.28

Population of Rajasthan=570 lakhs andpopulation of UP=1660 lakhArea of Rajasthan= 3 lakh km2 andarea of UP = 2 lakh km2.i How many people are there per km2 in both these States?iiWhich State is less populated?

Ans

(i) Population of Rajasthan in 3 lakh km 2 =570 lakhs Population of Rajasthan in 1 lakh km 2 = 570 3 = 190 Population of UP in 2 lakh km 2 =1660 lakhs Population of UP in 1 lakh km 2 = 1660 2 = 830 (ii) From above data, clearly Rajasthan is less populated.

Q.29

Convert the given fractional numbers to per cents.(a)19 b54 c340 d27

Ans

(a) 1 8 = 1 8 × 100 100 = 1 8 ×100%=12.5% ( b ) 5 4 = 5 4 × 100 100 = 5 4 ×100%=125% ( c ) 3 40 = 3 40 × 100 100 = 3 40 ×100%=7.5% ( d ) 2 7 = 2 7 × 100 100 = 2 7 ×100%=28 4 7 %

Q.30 Convert the given decimal fractions to percents.
a. 0.65 b. 2.1 c. 0.02 d. 12.35

Ans

( a ) 0.65 =0.65×100%= 65×100 100 %=65% ( b ) 2.1 =2.1×100%= 21×100 10 %=210% ( c ) 0.02 =0.02×100%= 2×100 100 %=2% ( d ) 12.35 =12.35×100%= 1235×100 100 %=1235% MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabggaaiaawIcacaGLPaaa caqGGaGaaGimaiaac6cacaqG2aGaaeynaaqaaiabg2da9iaaicdaca GGUaGaaGOnaiaaiwdacqGHxdaTcaaIXaGaaGimaiaaicdacaGGLaGa eyypa0ZaaSaaaeaacaaI2aGaaGynaiabgEna0kaaigdacaaIWaGaaG imaaqaaiaaigdacaaIWaGaaGimaaaacaGGLaGaeyypa0JaaGOnaiaa iwdacaGGLaaabaWaaeWaaeaacaqGIbaacaGLOaGaayzkaaGaaeiiai aabkdacaGGUaGaaeymaaqaaiabg2da9iaaikdacaGGUaGaaGymaiab gEna0kaaigdacaaIWaGaaGimaiaacwcacqGH9aqpdaWcaaqaaiaaik dacaaIXaGaey41aqRaaGymaiaaicdacaaIWaaabaGaaGymaiaaicda aaGaaiyjaiabg2da9iaaikdacaaIXaGaaGimaiaacwcaaeaadaqada qaaiaabogaaiaawIcacaGLPaaacaqGGaGaaGimaiaac6cacaaIWaGa aeOmaaqaaiabg2da9iaaicdacaGGUaGaaGimaiaaikdacqGHxdaTca aIXaGaaGimaiaaicdacaGGLaGaeyypa0ZaaSaaaeaacaaIYaGaey41 aqRaaGymaiaaicdacaaIWaaabaGaaGymaiaaicdacaaIWaaaaiaacw cacqGH9aqpcaaIYaGaaiyjaaqaamaabmaabaGaaeizaaGaayjkaiaa wMcaaiaabccacaqGXaGaaeOmaiaac6cacaqGZaGaaeynaaqaaiabg2 da9iaaigdacaaIYaGaaiOlaiaaiodacaaI1aGaey41aqRaaGymaiaa icdacaaIWaGaaiyjaiabg2da9maalaaabaGaaGymaiaaikdacaaIZa GaaGynaiabgEna0kaaigdacaaIWaGaaGimaaqaaiaaigdacaaIWaGa aGimaaaacaGGLaGaeyypa0JaaGymaiaaikdacaaIZaGaaGynaiaacw caaaaa@AD33@

Q.31 Estimate what part of the figures is coloured and hence find the percent which is coloured.

Ans

(i) From the figure, we observe that 1 part out of 4 is shaded, so its 1 4 . 1 4 × 100 100 = 1 4 ×100%=25% (ii) From the figure, we observe that 3 part out of 5 is shaded, so its 3 5 . 3 5 × 100 100 = 3 5 ×100%=60% (iii) From the figure, we observe that 3 part out of 8 is shaded, so its 3 8 . 3 8 × 100 100 = 3 8 ×100%=37.5% MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGOaGaaeyAaiaabMcacaqGGaaabaGa aeOraiaabkhacaqGVbGaaeyBaiaabccacaqG0bGaaeiAaiaabwgaca qGGaGaaeOzaiaabMgacaqGNbGaaeyDaiaabkhacaqGLbGaaeilaiaa bccacaqG3bGaaeyzaiaabccacaqGVbGaaeOyaiaabohacaqGLbGaae OCaiaabAhacaqGLbGaaeiiaiaabshacaqGObGaaeyyaiaabshacaqG GaGaaeymaiaabccacaqGWbGaaeyyaiaabkhacaqG0bGaaeiiaiaab+ gacaqG1bGaaeiDaiaabccacaqGVbGaaeOzaiaabccacaqG0aGaaeii aiaabMgacaqGZbGaaeiiaiaabohacaqGObGaaeyyaiaabsgacaqGLb GaaeizaiaabYcaaeaacaqGZbGaae4BaiaabccacaqGPbGaaeiDaiaa bohacaqGGaWaaSaaaeaacaaIXaaabaGaaGinaaaacaGGUaaabaWaaS aaaeaacaaIXaaabaGaaGinaaaacqGHxdaTdaWcaaqaaiaaigdacaaI WaGaaGimaaqaaiaaigdacaaIWaGaaGimaaaacqGH9aqpdaWcaaqaai aaigdaaeaacaaI0aaaaiabgEna0kaaigdacaaIWaGaaGimaiaacwca cqGH9aqpcaaIYaGaaGynaiaacwcaaeaacaqGOaGaaeyAaiaabMgaca qGPaGaaeiiaaqaaiaabAeacaqGYbGaae4Baiaab2gacaqGGaGaaeiD aiaabIgacaqGLbGaaeiiaiaabAgacaqGPbGaae4zaiaabwhacaqGYb GaaeyzaiaabYcacaqGGaGaae4DaiaabwgacaqGGaGaae4Baiaabkga caqGZbGaaeyzaiaabkhacaqG2bGaaeyzaiaabccacaqG0bGaaeiAai aabggacaqG0bGaaeiiaiaabodacaqGGaGaaeiCaiaabggacaqGYbGa aeiDaiaabccacaqGVbGaaeyDaiaabshacaqGGaGaae4BaiaabAgaca qGGaGaaeynaiaabccacaqGPbGaae4CaiaabccacaqGZbGaaeiAaiaa bggacaqGKbGaaeyzaiaabsgacaqGSaaabaGaae4Caiaab+gacaqGGa GaaeyAaiaabshacaqGZbGaaeiiamaalaaabaGaaG4maaqaaiaaiwda aaGaaiOlaaqaamaalaaabaGaaG4maaqaaiaaiwdaaaGaey41aq7aaS aaaeaacaaIXaGaaGimaiaaicdaaeaacaaIXaGaaGimaiaaicdaaaGa eyypa0ZaaSaaaeaacaaIZaaabaGaaGynaaaacqGHxdaTcaaIXaGaaG imaiaaicdacaGGLaGaeyypa0JaaGOnaiaaicdacaGGLaaabaGaaeik aiaabMgacaqGPbGaaeyAaiaabMcacaaMe8oabaGaaeOraiaabkhaca qGVbGaaeyBaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeOzaiaa bMgacaqGNbGaaeyDaiaabkhacaqGLbGaaeilaiaabccacaqG3bGaae yzaiaabccacaqGVbGaaeOyaiaabohacaqGLbGaaeOCaiaabAhacaqG LbGaaeiiaiaabshacaqGObGaaeyyaiaabshacaqGGaGaae4maiaabc cacaqGWbGaaeyyaiaabkhacaqG0bGaaeiiaiaab+gacaqG1bGaaeiD aiaabccacaqGVbGaaeOzaiaabccacaqG4aGaaeiiaiaabMgacaqGZb GaaeiiaiaabohacaqGObGaaeyyaiaabsgacaqGLbGaaeizaiaabYca aeaacaqGZbGaae4BaiaabccacaqGPbGaaeiDaiaabohacaqGGaWaaS aaaeaacaaIZaaabaGaaGioaaaacaGGUaaabaWaaSaaaeaacaaIZaaa baGaaGioaaaacqGHxdaTdaWcaaqaaiaaigdacaaIWaGaaGimaaqaai aaigdacaaIWaGaaGimaaaacqGH9aqpdaWcaaqaaiaaiodaaeaacaaI 4aaaaiabgEna0kaaigdacaaIWaGaaGimaiaacwcacqGH9aqpcaaIZa GaaG4naiaac6cacaaI1aGaaiyjaaaaaa@3382@

Q.32

Find:a 15% of 250 b 1% of 1 hourc 20% of Rs 2500 d 75% of 1 k

Ans

( a ) 15% of 250 = 15 100 ×250= 75 2 =37.5 ( b ) 1% of 1 hour 1 hour=60 minutes. So, 1% of 1 hour = 1 100 ×60= 6 10 = 3 5 ( c ) 20% of Rs 2500 = 20 100 ×2500=20×25=500 ( d ) 75% of 1 kg 1 kg=1000 g. So, 75% of 1 kg = 75 100 ×1000=75×10=750

Q.33

Find the whole quantity ifa 5% of it is 600. b 12% of it is 1080.c 40% of it is 500 km. d 70% of it is 14 minutese 8% of it is 40 litre

Ans

( a ) 5% of it is 600 is same as 5% of x is 600. 5 100 ×x=600 x= 600×100 5 =12000 ( b ) 12% of it is 1080 is same as 12% of x is 1080. 12 100 ×x=1080 x= 1080×100 12 =9000 ( c ) 40% of it is 500 km is same as 40% of x is 500km. 40 100 ×x=500 x= 500×100 40 =1250 ( d ) 70% of it is 14 minutes is same as 70% of x is 14minutes 70 100 ×x=14 x= 14×100 70 =20 ( e ) 8% of it is 40 litres is same as 8% of x is 40litres 8 100 ×x=40 x= 40×100 8 =500

Q.34

Convert given per cents to decimal fractions and also tofractions in simplest forms:a 25% b 150% c 20% d 5%

Ans

( a ) 25% = 25 100 = 1 4 =0.25 ( b ) 150% = 150 100 = 15 10 = 3 2 =1.5 ( c ) 20% = 20 100 = 2 10 = 1 5 =0.2 ( d ) 5% = 5 100 = 1 20 =0.05

Q.35

In a city, 30% are females, 40% are males and remainingare children. What percent are children?

Ans

Children=( 100( 30+40 ) )% =( 10070 )% =30%

Q.36

Out of 15,000 voters in a constituency, 60% voted.Find the percentage of voters who did not vote. Can younow find how many actually did not vote?

Ans

Percentage of voters who voted =60% Percentage of voters who did not vote=100%-60% =40% Number of voters who did not vote=40% of 15,000 = 40 100 ×15000 =6000 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGqbGaaeyzaiaabkhacaqGJbGaaeyz aiaab6gacaqG0bGaaeyyaiaabEgacaqGLbGaaeiiaiaab+gacaqGMb GaaeiiaiaabAhacaqGVbGaaeiDaiaabwgacaqGYbGaae4Caiaabcca caqG3bGaaeiAaiaab+gacaqGGaGaaeODaiaab+gacaqG0bGaaeyzai aabsgacaWLjaGaaGjbVlaaysW7caaMe8UaaGjbVlabg2da9iaaiAda caaIWaGaaiyjaiaaxMaaaeaacaqGqbGaaeyzaiaabkhacaqGJbGaae yzaiaab6gacaqG0bGaaeyyaiaabEgacaqGLbGaaeiiaiaab+gacaqG MbGaaeiiaiaabAhacaqGVbGaaeiDaiaabwgacaqGYbGaae4Caiaabc cacaqG3bGaaeiAaiaab+gacaqGGaGaaeizaiaabMgacaqGKbGaaeii aiaab6gacaqGVbGaaeiDaiaabccacaqG2bGaae4BaiaabshacaqGLb GaaGjbVlabg2da9iaabgdacaqGWaGaaeimaiaabwcacaqGTaGaaeOn aiaabcdacaqGLaaabaGaaCzcaiaaxMaacaWLjaGaaCzcaiaaxMaaca WLjaGaaCzcaiaaxMaacaaMe8UaaGjbVlaaysW7cqGH9aqpcaqG0aGa aeimaiaabwcaaeaacaqGobGaaeyDaiaab2gacaqGIbGaaeyzaiaabk hacaqGGaGaae4BaiaabAgacaqGGaGaaeODaiaab+gacaqG0bGaaeyz aiaabkhacaqGZbGaaeiiaiaabEhacaqGObGaae4BaiaabccacaqGKb GaaeyAaiaabsgacaqGGaGaaeOBaiaab+gacaqG0bGaaeiiaiaabAha caqGVbGaaeiDaiaabwgacqGH9aqpcaqG0aGaaeimaiaabwcacaqGGa Gaae4BaiaabAgacaqGGaGaaeymaiaabwdacaqGSaGaaeimaiaabcda caqGWaaabaGaaCzcaiaaxMaacaWLjaGaaCzcaiaaxMaacaWLjaGaaC zcaiaaysW7caaMe8UaaGjbVlaaysW7cqGH9aqpdaWcaaqaaiaaisda caaIWaaabaGaaGymaiaaicdacaaIWaaaaiabgEna0kaaigdacaaI1a GaaGimaiaaicdacaaIWaaabaGaaCzcaiaaxMaacaWLjaGaaCzcaiaa xMaacaWLjaGaaCzcaiaaysW7caaMe8UaaGjbVlaaysW7cqGH9aqpca aI2aGaaGimaiaaicdacaaIWaaaaaa@E379@

Q.37

Meeta saves 400 from her salary. If this is 10% of hersalary. What is her salay?

Ans

Let her salary be x. Then according to the question, we have 10% of x= 400 10 100 ×x=400 x= 400×100 10 =4000 Therefore Meeta’s salary is 4000

Q.38 Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using ruler and compasses only.

Ans

The steps of construction are as follows: (i) Draw a line AB. Take a point P on it. Take a point C outsidethis line. Join C to P. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@DC25@

(ii) Taking P as centre and with a convenient radius, draw arc intersecting line AB at point D and PC at point E.

(iii) Taking C as centre and with the same radius as before, draw an arc FG intersecting PC at H. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@B714@

(iv) Adjust the compasses up to the length of DE. Without changing the opening of the compasses and taking H as the cenre, draw an arc to intersect the previously draw arc FG at point I. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@26CC@

(v) Join the points C and I to draw a line l. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqG2bGDcqqGPaqkcqqGGaaicqqGkbGscqqGVbWBcqqGPbqAcqqGUbGBcqqGGaaicqqG0baDcqqGObaAcqqGLbqzcqqGGaaicqqGWbaCcqqGVbWBcqqGPbqAcqqGUbGBcqqG0baDcqqGZbWCcqqGGaaicqqGdbWqcqqGGaaicqqGHbqycqqGUbGBcqqGKbazcqqGGaaicqqGjbqscqqGGaaicqqG0baDcqqGVbWBcqqGGaaicqqGKbazcqqGYbGCcqqGHbqycqqG3bWDcqqGGaaicqqGHbqycqqGGaaicqqGSbaBcqqGPbqAcqqGUbGBcqqGLbqzcqqGGaaicqWGSbaBcqGGUaGlaaa@76FB@

This is the required line which is parallel to AB. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGubavcqqGObaAcqqGPbqAcqqGZbWCcqqGGaaicqqGPbqAcqqGZbWCcqqGGaaicqqG0baDcqqGObaAcqqGLbqzcqqGGaaicqqGYbGCcqqGLbqzcqqGXbqCcqqG1bqDcqqGPbqAcqqGYbGCcqqGLbqzcqqGKbazcqqGGaaicqqGSbaBcqqGPbqAcqqGUbGBcqqGLbqzcqqGGaaicqqG3bWDcqqGObaAcqqGPbqAcqqGJbWycqqGObaAcqqGGaaicqqGPbqAcqqGZbWCcqqGGaaicqqGWbaCcqqGHbqycqqGYbGCcqqGHbqycqqGSbaBcqqGSbaBcqqGLbqzcqqGSbaBcqqGGaaicqqG0baDcqqGVbWBcqqGGaaicqqGbbqqcqqGcbGqcqqGUaGlaaa@7FE7@

Q.39

A local cricket team played 20 matches in one season.It won 25% of them. How many matches did they win?

Ans

Since team played 20 matches, So, number of matches won=25% of 20 = 25 100 ×20 = 500 100 =5 So, number they won 5 matchces

Q.40

Tell what is the profit or loss in the following transactions.Also find profit per cent or loss per cent in each case.a Gardening shears bought for 250 and sold for 325.b A refrigerater bought for 12,000 and sold at 13,500.c A cupboard bought for 2,500 and sold at 3,000.d A skirt bought for 250 and sold at 150.

Ans

(a) Cost price= 250 Selling price= 325 Profit=325250 =75 Profit% = Profit Cost Price ×100 = 75 250 ×100 =30% (b) Cost price= 12000 Selling price= 13,500 Profit=135001200 =1500 Profit% = Profit Cost Price ×100 = 1500 12000 ×100 =12.5% (c) Cost price= 2500 Selling price= 3,000 Profit=30002500 =500 Profit% = Profit Cost Price ×100 = 500 2500 ×100 =20% (d) Cost price= 250 Selling price= 150 Loss=250150 =1000 Loss% = Loss Cost Price ×100 = 100 250 ×100 =40% MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqedu uDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wD YLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYf gasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9 q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff 0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbIcaOiab bggaHjabbMcaPiabbccaGiabboeadjabb+gaVjabbohaZjabbsha0j abbccaGiabbchaWjabbkhaYjabbMgaPjabbogaJjabbwgaLjabg2da 9iabbcgaGjabbccaGiabbkdaYiabbwda1iabbcdaWaqaaiabbccaGi 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Q.41

Convert each part of the ratio to percentage:a 3:1 b2:3:5 c 1:4 d1: 2:5.

Ans

( a ) 3:1 Here total parts =3+1=4 1 st part= 3 4 = 3 4 ×100%=75% 2 nd part= 1 4 = 1 4 ×100%=25% ( b )2:3:5 Here total parts =2+3+5=10 1 st part= 2 10 = 2 10 ×100%=20% 2 nd part= 3 10 = 3 10 ×100%=30% 3 rd part= 5 10 = 5 10 ×100%=50% ( c ) 1:4 Here total parts =1+4=5 1 st part= 1 5 = 1 5 ×100%=20% 2 nd part= 4 5 = 4 5 ×100%=80% ( d )1: 2:5 Here total parts =1+2+5=8 1 st part= 1 8 = 1 8 ×100%=12.5% 2 nd part= 2 8 = 2 8 ×100%=25% 3 rd part= 5 8 = 5 8 ×100%=62.5% MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqedu uDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wD YLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYf gasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9 q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff 0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaamaabmaabaGa eeyyaegacaGLOaGaayzkaaGaeeiiaaIaee4mamJaeiOoaOJaeeymae dabaGaeeisaGKaeeyzauMaeeOCaiNaeeyzauMaeeiiaaIaeeiDaqNa ee4Ba8MaeeiDaqNaeeyyaeMaeeiBaWMaeeiiaaIaeeiCaaNaeeyyae MaeeOCaiNaeeiDaqNaee4CamNaeeiiaaIaeyypa0JaeG4mamJaey4k aSIaeGymaeJaeyypa0JaeGinaqdabaGaeGymaeZaaWbaaSqabeaacq qGZbWCcqqG0baDaaGccqqGGaaicqqGWbaCcqqGHbqycqqGYbGCcqqG 0baDcqGH9aqpdaWcaaqaaiabiodaZaqaaiabisda0aaacqGH9aqpda WcaaqaaiabiodaZaqaaiabisda0aaacqGHxdaTcqaIXaqmcqaIWaam cqaIWaamcqGGLaqjcqGH9aqpcqaI3aWncqaI1aqncqGGLaqjaeaacq aIYaGmdaahaaWcbeqaaiabb6gaUjabbsgaKbaakiabbccaGiabbcha WjabbggaHjabbkhaYjabbsha0jabg2da9maalaaabaGaeGymaedaba GaeGinaqdaaiabg2da9maalaaabaGaeGymaedabaGaeGinaqdaaiab gEna0kabigdaXiabicdaWiabicdaWiabcwcaLiabg2da9iabikdaYi abiwda1iabcwcaLaqaamaabmaabaGaeeOyaigacaGLOaGaayzkaaGa eeOmaiJaeiOoaOJaee4mamJaeiOoaOJaeeynaudabaGaeeisaGKaee yzauMaeeOCaiNaeeyzauMaeeiiaaIaeeiDaqNaee4Ba8MaeeiDaqNa eeyyaeMaeeiBaWMaeeiiaaIaeeiCaaNaeeyyaeMaeeOCaiNaeeiDaq Naee4CamNaeeiiaaIaeyypa0JaeGOmaiJaey4kaSIaeG4mamJaey4k aSIaeGynauJaeyypa0JaeGymaeJaeGimaadabaGaeGymaeZaaWbaaS qabeaacqqGZbWCcqqG0baDaaGccqqGGaaicqqGWbaCcqqGHbqycqqG YbGCcqqG0baDcqGH9aqpdaWcaaqaaiabikdaYaqaaiabigdaXiabic daWaaacqGH9aqpdaWcaaqaaiabikdaYaqaaiabigdaXiabicdaWaaa cqGHxdaTcqaIXaqmcqaIWaamcqaIWaamcqGGLaqjcqGH9aqpcqaIYa GmcqaIWaamcqGGLaqjaeaacqaIYaGmdaahaaWcbeqaaiabb6gaUjab bsgaKbaakiabbccaGiabbchaWjabbggaHjabbkhaYjabbsha0jabg2 da9maalaaabaGaeG4mamdabaGaeGymaeJaeGimaadaaiabg2da9maa laaabaGaeG4mamdabaGaeGymaeJaeGimaadaaiabgEna0kabigdaXi abicdaWiabicdaWiabcwcaLiabg2da9iabiodaZiabicdaWiabcwca LaqaaiabiodaZmaaCaaaleqabaGaeeOCaiNaeeizaqgaaOGaeeiiaa IaeeiCaaNaeeyyaeMaeeOCaiNaeeiDaqNaeyypa0ZaaSaaaeaacqaI 1aqnaeaacqaIXaqmcqaIWaamaaGaeyypa0ZaaSaaaeaacqaI1aqnae aacqaIXaqmcqaIWaamaaGaey41aqRaeGymaeJaeGimaaJaeGimaaJa eiyjauIaeyypa0JaeGynauJaeGimaaJaeiyjaucabaWaaeWaaeaacq qGJbWyaiaawIcacaGLPaaacqqGGaaicqqGXaqmcqGG6aGocqqG0aan cqqGGaaiaeaacqqGibascqqGLbqzcqqGYbGCcqqGLbqzcqqGGaaicq qG0baDcqqGVbWBcqqG0baDcqqGHbqycqqGSbaBcqqGGaaicqqGWbaC cqqGHbqycqqGYbGCcqqG0baDcqqGZbWCcqqGGaaicqGH9aqpcqaIXa qmcqGHRaWkcqaI0aancqGH9aqpcqaI1aqnaeaacqaIXaqmdaahaaWc beqaaiabbohaZjabbsha0baakiabbccaGiabbchaWjabbggaHjabbk haYjabbsha0jabg2da9maalaaabaGaeGymaedabaGaeGynaudaaiab g2da9maalaaabaGaeGymaedabaGaeGynaudaaiabgEna0kabigdaXi abicdaWiabicdaWiabcwcaLiabg2da9iabikdaYiabicdaWiabcwca LaqaaiabikdaYmaaCaaaleqabaGaeeOBa4MaeeizaqgaaOGaeeiiaa IaeeiCaaNaeeyyaeMaeeOCaiNaeeiDaqNaeyypa0ZaaSaaaeaacqaI 0aanaeaacqaI1aqnaaGaeyypa0ZaaSaaaeaacqaI0aanaeaacqaI1a qnaaGaey41aqRaeGymaeJaeGimaaJaeGimaaJaeiyjauIaeyypa0Ja eGioaGJaeGimaaJaeiyjaucabaWaaeWaaeaacqqGKbazaiaawIcaca GLPaaacqqGXaqmcqGG6aGocqqGGaaicqqGYaGmcqGG6aGocqqG1aqn aeaacqqGibascqqGLbqzcqqGYbGCcqqGLbqzcqqGGaaicqqG0baDcq qGVbWBcqqG0baDcqqGHbqycqqGSbaBcqqGGaaicqqGWbaCcqqGHbqy cqqGYbGCcqqG0baDcqqGZbWCcqqGGaaicqGH9aqpcqaIXaqmcqGHRa WkcqaIYaGmcqGHRaWkcqaI1aqncqGH9aqpcqaI4aaoaeaacqaIXaqm daahaaWcbeqaaiabbohaZjabbsha0baakiabbccaGiabbchaWjabbg gaHjabbkhaYjabbsha0jabg2da9maalaaabaGaeGymaedabaGaeGio aGdaaiabg2da9maalaaabaGaeGymaedabaGaeGioaGdaaiabgEna0k abigdaXiabicdaWiabicdaWiabcwcaLiabg2da9iabigdaXiabikda Yiabc6caUiabiwda1iabcwcaLaqaaiabikdaYmaaCaaaleqabaGaee OBa4MaeeizaqgaaOGaeeiiaaIaeeiCaaNaeeyyaeMaeeOCaiNaeeiD aqNaeyypa0ZaaSaaaeaacqaIYaGmaeaacqaI4aaoaaGaeyypa0ZaaS aaaeaacqaIYaGmaeaacqaI4aaoaaGaey41aqRaeGymaeJaeGimaaJa eGimaaJaeiyjauIaeyypa0JaeGOmaiJaeGynauJaeiyjaucabaGaeG 4mamZaaWbaaSqabeaacqqGYbGCcqqGKbazaaGccqqGGaaicqqGWbaC cqqGHbqycqqGYbGCcqqG0baDcqGH9aqpdaWcaaqaaiabiwda1aqaai abiIda4aaacqGH9aqpdaWcaaqaaiabiwda1aqaaiabiIda4aaacqGH xdaTcqaIXaqmcqaIWaamcqaIWaamcqGGLaqjcqGH9aqpcqaI2aGncq aIYaGmcqGGUaGlcqaI1aqncqGGLaqjaaaa@DB5C@

Q.42

The population of a city decreased from 25,000 to 24,500.Find the percentage decrease.

Ans

Initial population=25000 Final populaton=24500 Decrease=2500024500 =500 Decrease%= 500 25000 ×100 =2%

Q.43

Arun bought a car for 3,50,000. The next year, the pricewent up to 3,70,000. What was the Percentage of priceincrease?

Ans

Initial Price= 3,50,000 Final Price = 3,70,000 Increase =3,70,0003,50,000 = 20,000 Increase%= 20,000 3,50,000 ×100 =5 5 7 % MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqedu uDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wD YLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYf gasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9 q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff 0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbMeajjab b6gaUjabbMgaPjabbsha0jabbMgaPjabbggaHjabbYgaSjabbccaGi abbcfaqjabbkhaYjabbMgaPjabbogaJjabbwgaLjabg2da9iabbcga GjabbccaGiabbodaZiabbYcaSiabbwda1iabbcdaWiabbYcaSiabbc daWiabbcdaWiabbcdaWaqaaiabbAeagjabbMgaPjabb6gaUjabbgga HjabbYgaSjabbccaGiabbcfaqjabbkhaYjabbMgaPjabbogaJjabbw gaLjabbccaGiabg2da9iabbcgaGjabbccaGiabbodaZiabbYcaSiab bEda3iabbcdaWiabbYcaSiabbcdaWiabbcdaWiabbcdaWaqaaiabbM eajjabb6gaUjabbogaJjabbkhaYjabbwgaLjabbggaHjabbohaZjab bwgaLjabbccaGiabbccaGiabbccaGiabbccaGiabg2da9iabbcgaGj aaysW7cqqGZaWmcqqGSaalcqqG3aWncqqGWaamcqqGSaalcqqGWaam cqqGWaamcqqGWaamcqGHsislcqqGGbaycaaMe8Uaee4mamJaeeilaW IaeeynauJaeeimaaJaeeilaWIaeeimaaJaeeimaaJaeeimaadabaGa aCzcaiaaxMaacqqGGaaicqqGGaaicqqGGaaicqGH9aqpcqqGGbayca aMe8UaeeOmaiJaeeimaaJaeeilaWIaeeimaaJaeeimaaJaeeimaada baGaeeiiaaIaeeysaKKaeeOBa4Maee4yamMaeeOCaiNaeeyzauMaee yyaeMaee4CamNaeeyzauMaeeyjauIaeyypa0ZaaSaaaeaacqaIYaGm cqaIWaamcqGGSaalcqaIWaamcqaIWaamcqaIWaamaeaacqaIZaWmcq GGSaalcqaI1aqncqaIWaamcqGGSaalcqaIWaamcqaIWaamcqaIWaam aaGaey41aqRaeGymaeJaeGimaaJaeGimaadabaGaeeiiaaIaeeiiaa IaeeiiaaIaeeiiaaIaeeiiaaIaeeiiaaIaeeiiaaIaeeiiaaIaeeii aaIaeeiiaaIaeeiiaaIaeeiiaaIaeeiiaaIaeeiiaaIaeeiiaaIaee iiaaIaeeiiaaIaeyypa0JaeGynauZaaSaaaeaacqaI1aqnaeaacqaI 3aWnaaGaeiyjaucaaaa@D7A1@

Q.44

Draw a line l. Draw a perpendicular to l at any point on l.On this perpendicular choose a point X, 4 cm away from l.Through X, draw a line m parallel to l.

Ans

The steps of construction are as follows: (i) Draw a line l and take a point P on the line l. Then draw a perpendicular at point P. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@E131@

(ii) Adjusting the compasses up to the length of 4 cm, draw an arc to intersect this perpendicular at point X. Choose any point Y on the line l. Join X to Y. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@002A@

(iii) Taking Y as centre and with a convenient radius, draw an arc intersecting l at A and XY at B. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@B928@

(iv) Taking X as centre and with same radius as before, draw an arc CD cutting XY at E. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@AB54@

(v) Adjust the compasses upto the length of AB. Wothout changing the opening of the compasses and taking E as the centre, draw an arc intersecting the previous arc CD t F. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@1455@

(vi) Join the point X and F to draw a line m. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqG2bGDcqqGPbqAcqqGPaqkcqqGGaaicqqGkbGscqqGVbWBcqqGPbqAcqqGUbGBcqqGGaaicqqG0baDcqqGObaAcqqGLbqzcqqGGaaicqqGWbaCcqqGVbWBcqqGPbqAcqqGUbGBcqqG0baDcqqGGaaicqqGybawcqqGGaaicqqGHbqycqqGUbGBcqqGKbazcqqGGaaicqqGgbGrcqqGGaaicqqG0baDcqqGVbWBcqqGGaaicqqGKbazcqqGYbGCcqqGHbqycqqG3bWDcqqGGaaicqqGHbqycqqGGaaicqqGSbaBcqqGPbqAcqqGUbGBcqqGLbqzcqqGGaaicqWGTbqBcqqGUaGlaaa@770C@ Line m is the required line parallel to l. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGmbatcqqGPbqAcqqGUbGBcqqGLbqzcqqGGaaicqWGTbqBcqqGGaaicqqGPbqAcqqGZbWCcqqGGaaicqqG0baDcqqGObaAcqqGLbqzcqqGGaaicqqGYbGCcqqGLbqzcqqGXbqCcqqG1bqDcqqGPbqAcqqGYbGCcqqGLbqzcqqGKbazcqqGGaaicqqGSbaBcqqGPbqAcqqGUbGBcqqGLbqzcqqGGaaicqqGWbaCcqqGHbqycqqGYbGCcqqGHbqycqqGSbaBcqqGSbaBcqqGLbqzcqqGSbaBcqqGGaaicqqG0baDcqqGVbWBcqqGGaaicqWGSbaBcqqGUaGlaaa@7621@

Q.45

I buy a T.V. for 10,000 and sell it at a profit of 20%. Howmuch money do I get for it?

Ans

We know that Profit%= Profit Cost price ×100 So, 20= Profit 10,000 ×100 = Profit 100× 100 × 100 Profit=20×100 =2000 Now, Profit=Selling priceCost price 2000=Selling price10000 Selling price=10000+2000 =12000 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqedu uDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wD YLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYf gasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9 q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff 0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbEfaxjab bwgaLjabbccaGiabbUgaRjabb6gaUjabb+gaVjabbEha3jabbccaGi abbsha0jabbIgaOjabbggaHjabbsha0bqaaiabbcfaqjabbkhaYjab b+gaVjabbAgaMjabbMgaPjabbsha0jabbwcaLiabg2da9maalaaaba GaeeiuaaLaeeOCaiNaee4Ba8MaeeOzayMaeeyAaKMaeeiDaqhabaGa ee4qamKaee4Ba8Maee4CamNaeeiDaqNaeeiiaaIaeeiCaaNaeeOCai NaeeyAaKMaee4yamMaeeyzaugaaiabgEna0kabigdaXiabicdaWiab icdaWaqaaiabbofatjabb+gaVjabbYcaSaqaaiaaxMaacqaIYaGmcq aIWaamcqGH9aqpdaWcaaqaaiabbcfaqjabbkhaYjabb+gaVjabbAga MjabbMgaPjabbsha0bqaaiabigdaXiabicdaWiabcYcaSiabicdaWi abicdaWiabicdaWaaacqGHxdaTcqaIXaqmcqaIWaamcqaIWaamaeaa caWLjaGaaGjbVlaaysW7caaMe8UaaGjbVlabg2da9maalaaabaGaee iuaaLaeeOCaiNaee4Ba8MaeeOzayMaeeyAaKMaeeiDaqhabaGaeGym aeJaeGimaaJaeGimaaJaey41aq7aaqIaaeaacqaIXaqmcqaIWaamcq aIWaamaaaaaiabgEna0oaaKiaabaGaeGymaeJaeGimaaJaeGimaada aaqaaiaaysW7caaMe8UaeeiuaaLaeeOCaiNaee4Ba8MaeeOzayMaee yAaKMaeeiDaqNaeyypa0JaeeOmaiJaeeimaaJaey41aqRaeeymaeJa eeimaaJaeeimaadabaGaaCzcaiaaysW7caaMe8UaaGjbVlaaysW7cq GH9aqpcqqGGbaycaaMe8UaeeOmaiJaeeimaaJaeeimaaJaeeimaada baGaeeOta4Kaee4Ba8Maee4DaCNaeeilaWcabaGaaCzcaiaaysW7ca aMe8UaaGjbVlaaysW7caaMe8UaaGjbVlabbcfaqjabbkhaYjabb+ga VjabbAgaMjabbMgaPjabbsha0jabg2da9iabbofatjabbwgaLjabbY gaSjabbYgaSjabbMgaPjabb6gaUjabbEgaNjabbccaGiabbchaWjab bkhaYjabbMgaPjabbogaJjabbwgaLjabgkHiTiabboeadjabb+gaVj abbohaZjabbsha0jabbccaGiabbchaWjabbkhaYjabbMgaPjabboga JjabbwgaLbqaaiaaxMaacaaMe8UaaGjbVlaaysW7caaMe8Uaeeiyaa MaaGjbVlabbkdaYiabbcdaWiabbcdaWiabbcdaWiabg2da9iabbofa tjabbwgaLjabbYgaSjabbYgaSjabbMgaPjabb6gaUjabbEgaNjabbc caGiabbchaWjabbkhaYjabbMgaPjabbogaJjabbwgaLjabgkHiTiab bcgaGjaaysW7cqqGXaqmcqqGWaamcqqGWaamcqqGWaamcqqGWaamae aacaaMe8UaaGjbVlabbofatjabbwgaLjabbYgaSjabbYgaSjabbMga Pjabb6gaUjabbEgaNjabbccaGiabbchaWjabbkhaYjabbMgaPjabbo gaJjabbwgaLjabg2da9iabbcgaGjaaysW7cqqGXaqmcqqGWaamcqqG WaamcqqGWaamcqqGWaamcqGHRaWkcqqGGbaycaaMe8UaeeOmaiJaee imaaJaeeimaaJaeeimaadabaGaaCzcaiaaxMaacaaMe8UaaGjbVlaa ysW7caaMe8UaaGjbVlaaykW7caaMe8UaaGjbVlabg2da9iabbcgaGj aaysW7cqqGXaqmcqqGYaGmcqqGWaamcqqGWaamcqqGWaamaaaa@647E@

Q.46

Let l be a line and P be a point not on l. Through P, drawa line m parallel to l. Now join P to any point Q on l. Chooseany other point R on m. Through R, draw a line parallel to PQ.Let this meet l at S. What shape do the two sets of parallellines enclose?

Ans

The steps of construction are as follows: (i) Draw a line l and take a point A ​on it. Take a point P not on l and join A to P. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@D8ED@

(ii) Taking A as centre and with a convenient radius, draw an arc cutting l at B and AP at C. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@B09E@

(iii) Taking P as centre and with the same radius as before, draw an arc DE to intersect AP at F. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@B5E8@

(iv) Adust the compasses up to the length of BC. Without changing the opening of compasses and taking F as the centre, draw an arc to intersect the previously draw arc DE at point G. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@21E5@

(v) Join P to G to draw a line m. Line m will be parallel to line l. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@9225@

(vi) Join P to any point Q on line . Choose another point R on line m. Similarly, a line can be drawn through point R and parallel to PQ. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@E6FB@

Let it meet line l at point S. In quadrilateral PQRS, opposite lines are parallel to each other. PQ∥RS and PR∥QS. Thus PQRS is a parallelogram.MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@EE36@

Q.47

Juhi sells a washing machine for 13,500. She loses 20% inthe bargain. What was the price at which she bought it?

Ans

Selling price = 13500 Loss % = 20% Let the cost price be x. Loss = 20% of x Cost price Loss = Selling price x20% of x=13500 x 20 100 ×x=13500 100x20x 100 =13500 80x 100 =13500 x= 13500×100 80 =16875 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqedu uDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wD YLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYf gasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9 q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff 0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbofatjab bwgaLjabbYgaSjabbYgaSjabbMgaPjabb6gaUjabbEgaNjabbccaGi abbchaWjabbkhaYjabbMgaPjabbogaJjabbwgaLjabbccaGiabg2da 9iabbccaGiabbcgaGjabbccaGiabbgdaXiabbodaZiabbwda1iabic daWiabicdaWaqaaiabbYeamjabb+gaVjabbohaZjabbohaZjabbcca GiabcwcaLiabbccaGiabg2da9iabbccaGiabbkdaYiabicdaWiabcw caLaqaaiabbYeamjabbwgaLjabbsha0jabbccaGiabbsha0jabbIga OjabbwgaLjabbccaGiabbogaJjabb+gaVjabbohaZjabbsha0jabbc caGiabbchaWjabbkhaYjabbMgaPjabbogaJjabbwgaLjabbccaGiab bkgaIjabbwgaLjabbccaGiabdIha4jabc6caUaqaaiabgsJiCjabbc caGiabbYeamjabb+gaVjabbohaZjabbohaZjabbccaGiabg2da9iab bccaGiabbkdaYiabicdaWiabcwcaLiabbccaGiabb+gaVjabbAgaMj abbccaGiabdIha4bqaaiabboeadjabb+gaVjabbohaZjabbsha0jab bccaGiabbchaWjabbkhaYjabbMgaPjabbogaJjabbwgaLjabbccaGi abgkHiTiabbccaGiabbYeamjabb+gaVjabbohaZjabbohaZjabbcca Giabg2da9iabbccaGiabbofatjabbwgaLjabbYgaSjabbYgaSjabbM gaPjabb6gaUjabbEgaNjabbccaGiabbchaWjabbkhaYjabbMgaPjab bogaJjabbwgaLbqaaiaaxMaacaaMe8UaaGjbVlaaysW7caaMe8Uaee iEaGNaeyOeI0IaeGOmaiJaeGimaaJaeiyjauIaeeiiaaIaee4Ba8Ma eeOzayMaeeiiaaIaeeiEaGNaeyypa0JaeeiyaaMaaGjbVlabbgdaXi abbodaZiabbwda1iabbcdaWiabbcdaWaqaaiaaxMaacaWLjaGaeeiE aGNaeyOeI0YaaSaaaeaacqaIYaGmcqaIWaamaeaacqaIXaqmcqaIWa amcqaIWaamaaGaey41aqRaemiEaGNaeyypa0JaeGymaeJaeG4mamJa eGynauJaeGimaaJaeGimaadabaGaaCzcaiaaxMaadaWcaaqaaiabig daXiabicdaWiabicdaWiabdIha4jabgkHiTiabikdaYiabicdaWiab dIha4bqaaiabigdaXiabicdaWiabicdaWaaacqGH9aqpcqaIXaqmcq aIZaWmcqaI1aqncqaIWaamcqaIWaamaeaacaWLjaGaaCzcaiaaxMaa caaMe8UaaGjbVlaaysW7caaMe8+aaSaaaeaacqaI4aaocqaIWaamcq WG4baEaeaacqaIXaqmcqaIWaamcqaIWaamaaGaeyypa0JaeGymaeJa eG4mamJaeGynauJaeGimaaJaeGimaadabaGaaCzcaiaaxMaacaWLja GaaCzcaiaaysW7cqWG4baEcqGH9aqpdaWcaaqaaiabigdaXiabioda Ziabiwda1iabicdaWiabicdaWiabgEna0kabigdaXiabicdaWiabic daWaqaaiabiIda4iabicdaWaaaaeaacaWLjaGaaCzcaiaaxMaacaWL jaGaaGjbVlaaysW7caaMe8UaaGjbVlabg2da9iabbcgaGjaaysW7cq aIXaqmcqaI2aGncqaI4aaocqaI3aWncqaI1aqnaaaa@3634@

Q.48

i Chalk contains calcium, carbon and oxygen in the ratio 10:3:12. Find the percentage of carbon in chalk.ii If in a stick of chalk, carbon is 3g, what is the weight of the chalk stick.

Ans

(i) Given ratio=10:3:12 Total=10+3+12 =25 Percentage of Carbon= 3 25 ×100 =12% (ii) Let the weight of the stick be x g So, 12% of x=3 12 100 ×x=3 x= 300 12 =25g MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqedu uDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wD YLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYf gasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9 q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff 0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbIcaOiab bMgaPjabbMcaPiabbccaGiabbEeahjabbMgaPjabbAha2jabbwgaLj abb6gaUjabbccaGiabbkhaYjabbggaHjabbsha0jabbMgaPjabb+ga Vjabg2da9iabbgdaXiabbcdaWiabbQda6iabbodaZiabbQda6iabbg daXiabbkdaYaqaaiaaxMaacaaMe8UaaGjbVlaaysW7caaMe8UaaGjb VlaaysW7cqqGubavcqqGVbWBcqqG0baDcqqGHbqycqqGSbaBcqGH9a qpcqqGXaqmcqqGWaamcqGHRaWkcqqGZaWmcqGHRaWkcqqGXaqmcqqG YaGmaeaacaWLjaGaaCzcaiaaxMaacqGH9aqpcqqGYaGmcqqG1aqnae aacqqGqbaucqqGLbqzcqqGYbGCcqqGJbWycqqGLbqzcqqGUbGBcqqG 0baDcqqGHbqycqqGNbWzcqqGLbqzcqqGGaaicqqGVbWBcqqGMbGzcq qGGaaicqqGdbWqcqqGHbqycqqGYbGCcqqGIbGycqqGVbWBcqqGUbGB cqGH9aqpdaWcaaqaaiabiodaZaqaaiabikdaYiabiwda1aaacqGHxd aTcqaIXaqmcqaIWaamcqaIWaamaeaacaWLjaGaaCzcaiaaxMaacaWL jaGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8Uaeyypa0JaeG ymaeJaeGOmaiJaeiyjaucabaGaeeikaGIaeeyAaKMaeeyAaKMaeeyk aKIaeeiiaaIaeeitaWKaeeyzauMaeeiDaqNaeeiiaaIaeeiDaqNaee iAaGMaeeyzauMaeeiiaaIaee4DaCNaeeyzauMaeeyAaKMaee4zaCMa eeiAaGMaeeiDaqNaeeiiaaIaee4Ba8MaeeOzayMaeeiiaaIaeeiDaq NaeeiAaGMaeeyzauMaeeiiaaIaee4CamNaeeiDaqNaeeyAaKMaee4y amMaee4AaSMaeeiiaaIaeeOyaiMaeeyzauMaeeiiaaIaeeiEaGNaee iiaaIaee4zaCgabaGaee4uamLaee4Ba8MaeeilaWIaeeiiaaIaeeym aeJaeeOmaiJaeeyjauIaeeiiaaIaee4Ba8MaeeOzayMaeeiiaaIaee iEaGNaeyypa0Jaee4mamdabaGaaCzcamaalaaabaGaeGymaeJaeGOm aidabaGaeGymaeJaeGimaaJaeGimaadaaiabgEna0kabdIha4jabg2 da9iabiodaZaqaaiaaxMaacaWLjaGaemiEaGNaeyypa0ZaaSaaaeaa cqaIZaWmcqaIWaamcqaIWaamaeaacqaIXaqmcqaIYaGmaaaabaGaaC zcaiaaxMaacaaMe8UaaGjbVlabg2da9iabikdaYiabiwda1iaaysW7 cqqGNbWzaaaa@052E@

Q.49 Amina buys a book for 275 and sells it at a loss of 15%. How much does she sell it for?

Ans

Cost price of book= 275 Loss%=15 We have, Loss%= Loss Cost price ×100 15= Loss 275 ×100 Loss= 15×275 100 =41.25 Selling price=Cost priceLoss =27541.25 = 233.75 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqedu uDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wD YLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYf gasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9 q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff 0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabboeadjab b+gaVjabbohaZjabbsha0jabbccaGiabbchaWjabbkhaYjabbMgaPj abbogaJjabbwgaLjabbccaGiabb+gaVjabbAgaMjabbccaGiabbkga Ijabb+gaVjabb+gaVjabbUgaRjabg2da9iabbcgaGjabbccaGiabbk daYiabbEda3iabbwda1aqaaiabbYeamjabb+gaVjabbohaZjabboha ZjabbwcaLiabg2da9iabbgdaXiabbwda1aqaaiabbEfaxjabbwgaLj abbccaGiabbIgaOjabbggaHjabbAha2jabbwgaLjabbYcaSaqaaiab bYeamjabb+gaVjabbohaZjabbohaZjabbwcaLiabg2da9maalaaaba GaeeitaWKaee4Ba8Maee4CamNaee4CamhabaGaee4qamKaee4Ba8Ma ee4CamNaeeiDaqNaeeiiaaIaeeiCaaNaeeOCaiNaeeyAaKMaee4yam MaeeyzaugaaiabgEna0kabigdaXiabicdaWiabicdaWaqaaiaaysW7 caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlabigdaXiabiwda1iabg2 da9maalaaabaGaeeitaWKaee4Ba8Maee4CamNaee4CamhabaGaeGOm aiJaeG4naCJaeGynaudaaiabgEna0kabigdaXiabicdaWiabicdaWa qaaiaaxMaacaaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7cqqG mbatcqqGVbWBcqqGZbWCcqqGZbWCcqGH9aqpdaWcaaqaaiabigdaXi abiwda1iabgEna0kabikdaYiabiEda3iabiwda1aqaaiabigdaXiab icdaWiabicdaWaaaaeaacaWLjaGaaGjbVlaaysW7caaMe8UaaCzcai aaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlabg2da9iabisda 0iabigdaXiabc6caUiabikdaYiabiwda1aqaaiabbofatjabbwgaLj abbYgaSjabbYgaSjabbMgaPjabb6gaUjabbEgaNjabbccaGiabbcha WjabbkhaYjabbMgaPjabbogaJjabbwgaLjabg2da9iabboeadjabb+ gaVjabbohaZjabbsha0jabbccaGiabbchaWjabbkhaYjabbMgaPjab bogaJjabbwgaLjabgkHiTiabbYeamjabb+gaVjabbohaZjabbohaZb qaaiaaxMaacaWLjaGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaM c8Uaeyypa0JaeeiyaaMaaGjbVlabbkdaYiabbEda3iabbwda1iabgk HiTiabbcgaGjaaysW7cqqG0aancqqGXaqmcqqGUaGlcqqGYaGmcqqG 1aqnaeaacaWLjaGaaCzcaiaaysW7caaMe8UaaGjbVlaaysW7caaMe8 UaaGPaVlabg2da9iabbcgaGjabbccaGiabbkdaYiabbodaZiabboda Ziabb6caUiabbEda3iabbwda1aaaaa@2A0A@

Q.50 Find the amount to be paid at the end of 3 years in each case:
(a) Principal = 1,200 at 12% p.a.
(b) Principal = 7,500 at 5% p.a.

Ans

(a) Principal (P)= 1200 Rate (R)=12% Time (T)=3 years S.I.= P×R×T 100 = 1200×12×3 100 = 432 Amount =P+S.I. =1200+432= 1632 (b) Principal (P) = 7500 Rate (R)=5% Time (T)=3 years S.I.= P×R×T 100 = 7500×5×3 100 = 1125 Amount =P+S.I. =7500+1125 = 8625 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqedu uDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wD YLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYf gasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9 q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff 0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbIcaOiab bggaHjabbMcaPaqaaiaaysW7cqqGqbaucqqGYbGCcqqGPbqAcqqGUb GBcqqGJbWycqqGPbqAcqqGWbaCcqqGHbqycqqGSbaBcqqGGaaicqqG OaakcqqGqbaucqqGPaqkcqGH9aqpcqqGGbaycqqGGaaicqqGXaqmcq qGYaGmcqqGWaamcqqGWaamaeaacaWLjaGaeeOuaiLaeeyyaeMaeeiD aqNaeeyzauMaeeiiaaIaeeikaGIaeeOuaiLaeeykaKIaeyypa0Jaee ymaeJaeeOmaiJaeeyjaucabaGaaCzcaiabbsfaujabbMgaPjabb2ga TjabbwgaLjabbccaGiabbIcaOiabbsfaujabbMcaPiabg2da9iabio daZiabbccaGiabbMha5jabbwgaLjabbggaHjabbkhaYjabbohaZbqa aiaaxMaacaWLjaGaaGjbVlaaykW7cqqGtbWucqqGUaGlcqqGjbqscq qGUaGlcqGH9aqpdaWcaaqaaiabbcfaqjabgEna0kabbkfasjabgEna 0kabbsfaubqaaiabigdaXiabicdaWiabicdaWaaaaeaacaWLjaGaaG jbVlaaykW7caWLjaGaaCzcaiabg2da9maalaaabaGaeGymaeJaeGOm aiJaeGimaaJaeGimaaJaey41aqRaeGymaeJaeGOmaiJaey41aqRaeG 4mamdabaGaeGymaeJaeGimaaJaeGimaadaaaqaaiaaxMaacaaMe8Ua aGPaVlaaxMaacaWLjaGaeyypa0JaeeiyaaMaeeiiaaIaeGinaqJaeG 4mamJaeGOmaidabaGaaCzcaiabbgeabjabb2gaTjabb+gaVjabbwha 1jabb6gaUjabbsha0jabbccaGiabg2da9iabbcfaqjabgUcaRiabbo fatjabb6caUiabbMeajjabb6caUaqaaiaaxMaacaWLjaGaaGjbVlaa ykW7caWLjaGaeyypa0JaeeiyaaMaaGjbVlabbgdaXiabbkdaYiabbc daWiabbcdaWiabgUcaRiabbcgaGjaaysW7cqqG0aancqqGZaWmcqqG YaGmcqGH9aqpcqqGGbaycqqGGaaicqqGXaqmcqqG2aGncqqGZaWmcq qGYaGmaeaacqqGOaakcqqGIbGycqqGPaqkaeaacqqGqbaucqqGYbGC cqqGPbqAcqqGUbGBcqqGJbWycqqGPbqAcqqGWbaCcqqGHbqycqqGSb aBcqqGGaaicqqGOaakcqqGqbaucqqGPaqkcqqGGaaicqGH9aqpcqqG GbaycqqGGaaicqqG3aWncqqG1aqncqqGWaamcqqGWaamaeaacaWLja GaeeOuaiLaeeyyaeMaeeiDaqNaeeyzauMaeeiiaaIaeeikaGIaeeOu aiLaeeykaKIaeyypa0JaeeynauJaeeyjaucabaGaaCzcaiabbsfauj abbMgaPjabb2gaTjabbwgaLjabbccaGiabbIcaOiabbsfaujabbMca Piabg2da9iabbodaZiabbccaGiabbMha5jabbwgaLjabbggaHjabbk haYjabbohaZbqaaiaaxMaacaWLjaGaaGjbVlaaysW7cqqGtbWucqqG UaGlcqqGjbqscqqGUaGlcqGH9aqpdaWcaaqaaiabbcfaqjabgEna0k abbkfasjabgEna0kabbsfaubqaaiabigdaXiabicdaWiabicdaWaaa aeaacaWLjaGaaCzcaiaaxMaacqGH9aqpdaWcaaqaaiabiEda3iabiw da1iabicdaWiabicdaWiabgEna0kabiwda1iabgEna0kabiodaZaqa aiabigdaXiabicdaWiabicdaWaaaaeaacaWLjaGaaCzcaiaaxMaacq GH9aqpcqqGGbaycqqGGaaicqaIXaqmcqaIXaqmcqaIYaGmcqaI1aqn aeaacaWLjaGaeeyqaeKaeeyBa0Maee4Ba8MaeeyDauNaeeOBa4Maee iDaqNaeeiiaaIaeyypa0JaeeiuaaLaey4kaSIaee4uamLaeeOla4Ia eeysaKKaeeOla4cabaGaaCzcaiaaxMaacaWLjaGaeyypa0Jaeeiyaa MaaGjbVlabbEda3iabbwda1iabbcdaWiabbcdaWiabgUcaRiabbcga GjaaysW7cqqGXaqmcqqGXaqmcqqGYaGmcqqG1aqnaeaacaWLjaGaaC zcaiaaxMaacqGH9aqpcqqGGbaycqqGGaaicqqG4aaocqqG2aGncqqG YaGmcqqG1aqnaaaa@6A86@

Q.51 What rate gives 280 as interest on a sum of 56000 in 2 years?

Ans

We know thatS.I=P×R×T100280=56000×R×2100R=280×10056000×2=0.25Therefore, the rate is 0.25%.

Q.52

If Meena gives an interest of 45 for one year at 9%rate p.a. What is the sum she has borrowed?

Ans

We know that S.I= P×R×T 100 So, we get 45= P×9×1 100 P= 45×100 9 =500 Therefore, she borrowed 500. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqedu uDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wD YLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYf gasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9 q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff 0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbEfaxjab bwgaLjabbccaGiabbUgaRjabb6gaUjabb+gaVjabbEha3jabbccaGi abbsha0jabbIgaOjabbggaHjabbsha0bqaaiabbofatjabb6caUiab bMeajjabg2da9maalaaabaGaeeiuaaLaey41aqRaeeOuaiLaey41aq RaeeivaqfabaGaeGymaeJaeGimaaJaeGimaadaaaqaaiabbofatjab b+gaVjabbYcaSiabbccaGiabbEha3jabbwgaLjabbccaGiabbEgaNj abbwgaLjabbsha0bqaaiabbsda0iabbwda1iabb2da9maalaaabaGa eeiuaaLaee41aCTaeeyoaKJaee41aCTaeeymaedabaGaeeymaeJaee imaaJaeeimaadaaaqaaiabbcfaqjabb2da9maalaaabaGaeeinaqJa eeynauJaee41aCTaeeymaeJaeeimaaJaeeimaadabaGaeeyoaKdaaa qaaiabb2da9iabbcgaGjaaysW7cqqG1aqncqqGWaamcqqGWaamaeaa cqqGubavcqqGObaAcqqGLbqzcqqGYbGCcqqGLbqzcqqGMbGzcqqGVb WBcqqGYbGCcqqGLbqzcqqGSaalcqqGGaaicqqGZbWCcqqGObaAcqqG LbqzcqqGGaaicqqGIbGycqqGVbWBcqqGYbGCcqqGYbGCcqqGVbWBcq qG3bWDcqqGLbqzcqqGKbazcqqGGaaicqqGGbaycqqGGaaicqqG1aqn cqqGWaamcqqGWaamcqqGUaGlaaaa@ADA9@

Q.53

Construct ΔXYZ in which XY=4.5 cm, YZ=5 cm and ZX=6 cm.

Ans

The steps of construction are as follows: (i) Draw a line segment YZ of length 5 cm. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbsfaujabbIgaOjabbwgaLjabbccaGiabbohaZjabbsha0jabbwgaLjabbchaWjabbohaZjabbccaGiabb+gaVjabbAgaMjabbccaGiabbogaJjabb+gaVjabb6gaUjabbohaZjabbsha0jabbkhaYjabbwha1jabbogaJjabbsha0jabbMgaPjabb+gaVjabb6gaUjabbccaGiabbggaHjabbkhaYjabbwgaLjabbccaGiabbggaHjabbohaZjabbccaGiabbAgaMjabb+gaVjabbYgaSjabbYgaSjabb+gaVjabbEha3jabbohaZjabbQda6aqaaiabbIcaOiabbMgaPjabbMcaPiabbccaGiabbseaejabbkhaYjabbggaHjabbEha3jabbccaGiabbggaHjabbccaGiabbYgaSjabbMgaPjabb6gaUjabbwgaLjabbccaGiabbohaZjabbwgaLjabbEgaNjabb2gaTjabbwgaLjabb6gaUjabbsha0jabbccaGiabbMfazjabbQfaAjabbccaGiabb+gaVjabbAgaMjabbccaGiabbYgaSjabbwgaLjabb6gaUjabbEgaNjabbsha0jabbIgaOjabbccaGiabbwda1iabbccaGiabbogaJjabb2gaTjabb6caUaaaaa@A7F3@

(ii) Point X is at a distance of 4.5 cm from point Y. Therefore taking Y as centre, draw an arc of 4.5 cm radius. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@C74C@

(iii) Point X is at a distance of 6 cm from point Z.Therefore taking Z as centre, draw an arc of 6 cm radius. Mark the point of intersection of the arcs as X. Join XY and XZ. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@12A8@

Therefore, XYZ is the required triangle.MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGubavcqqGObaAcqqGLbqzcqqGYbGCcqqGLbqzcqqGMbGzcqqGVbWBcqqGYbGCcqqGLbqzcqqGSaalcqqGGaaicqqGybawcqqGzbqwcqqGAbGwcqqGGaaicqqGPbqAcqqGZbWCcqqGGaaicqqG0baDcqqGObaAcqqGLbqzcqqGGaaicqqGYbGCcqqGLbqzcqqGXbqCcqqG1bqDcqqGPbqAcqqGYbGCcqqGLbqzcqqGKbazcqqGGaaicqqG0baDcqqGYbGCcqqGPbqAcqqGHbqycqqGUbGBcqqGNbWzcqqGSbaBcqqGLbqzcqqGUaGlaaa@743F@

Q.54

Construct an equilateral triangle of side 5.5 cm.

Ans

Since, we need to construct an equilateral triangle so all sides should be equal. So, AB=BC=CA=5.5 cm The steps of construction are as follows: (i) Draw a line segment BC of length 5.5 cm. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@2301@

(ii) Taking B as centre, draw an arc of 5.5 cm radius. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqGPbqAcqqGPaqkcqqGGaaicqqGubavcqqGHbqycqqGRbWAcqqGPbqAcqqGUbGBcqqGNbWzcqqGGaaicqqGcbGqcqqGGaaicqqGHbqycqqGZbWCcqqGGaaicqqGJbWycqqGLbqzcqqGUbGBcqqG0baDcqqGYbGCcqqGLbqzcqqGSaalcqqGGaaicqqGKbazcqqGYbGCcqqGHbqycqqG3bWDcqqGGaaicqqGHbqycqqGUbGBcqqGGaaicqqGHbqycqqGYbGCcqqGJbWycqqGGaaicqqGVbWBcqqGMbGzcqqGGaaicqqG1aqncqqGUaGlcqqG1aqncqqGGaaicqqGJbWycqqGTbqBcqqGGaaicqqGYbGCcqqGHbqycqqGKbazcqqGPbqAcqqG1bqDcqqGZbWCcqqGUaGlaaa@813B@

(iii) Taking C as centre, draw an arc of 5.5 cm radius to meet previous arc at point A. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbIcaOiabbMgaPjabbMgaPjabbMgaPjabbMcaPiabbccaGiabbsfaujabbggaHjabbUgaRjabbMgaPjabb6gaUjabbEgaNjabbccaGiabboeadjabbccaGiabbggaHjabbohaZjabbccaGiabbogaJjabbwgaLjabb6gaUjabbsha0jabbkhaYjabbwgaLjabbYcaSiabbccaGiabbsgaKjabbkhaYjabbggaHjabbEha3jabbccaGiabbggaHjabb6gaUjabbccaGiabbggaHjabbkhaYjabbogaJjabbccaGiabb+gaVjabbAgaMjabbccaGiabbwda1iabb6caUiabbwda1iabbccaGiabbogaJjabb2gaTjabbccaGiabbkhaYjabbggaHjabbsgaKjabbMgaPjabbwha1jabbohaZjabbccaGiabbsha0jabb+gaVbqaaiabb2gaTjabbwgaLjabbwgaLjabbsha0jabbccaGiabbchaWjabbkhaYjabbwgaLjabbAha2jabbMgaPjabb+gaVjabbwha1jabbohaZjabbccaGiabbggaHjabbkhaYjabbogaJjabbccaGiabbggaHjabbsha0jabbccaGiabbchaWjabb+gaVjabbMgaPjabb6gaUjabbsha0jabbccaGiabbgeabjabb6caUaaaaa@A978@

(iv) Join A​ to B and C.

Therefore, ABC is the required triangle.MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGubavcqqGObaAcqqGLbqzcqqGYbGCcqqGLbqzcqqGMbGzcqqGVbWBcqqGYbGCcqqGLbqzcqqGSaalcqqGGaaicqqGbbqqcqqGcbGqcqqGdbWqcqqGGaaicqqGPbqAcqqGZbWCcqqGGaaicqqG0baDcqqGObaAcqqGLbqzcqqGGaaicqqGYbGCcqqGLbqzcqqGXbqCcqqG1bqDcqqGPbqAcqqGYbGCcqqGLbqzcqqGKbazcqqGGaaicqqG0baDcqqGYbGCcqqGPbqAcqqGHbqycqqGUbGBcqqGNbWzcqqGSbaBcqqGLbqzcqqGUaGlaaa@73B5@

Q.55

Draw ΔPQR with PQ=4 cm, QR=3.5 cm and PR=4cm.What type of triangle is this?

Ans

The steps of construction are as follows: (i) Draw a line segment QR of length 3.5 cm. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A9A3@

(ii) Taking Q as centre, draw an arc of 4 cm radius. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqGPbqAcqqGPaqkcqqGGaaicqqGubavcqqGHbqycqqGRbWAcqqGPbqAcqqGUbGBcqqGNbWzcqqGGaaicqqGrbqucqqGGaaicqqGHbqycqqGZbWCcqqGGaaicqqGJbWycqqGLbqzcqqGUbGBcqqG0baDcqqGYbGCcqqGLbqzcqqGSaalcqqGGaaicqqGKbazcqqGYbGCcqqGHbqycqqG3bWDcqqGGaaicqqGHbqycqqGUbGBcqqGGaaicqqGHbqycqqGYbGCcqqGJbWycqqGGaaicqqGVbWBcqqGMbGzcqqGGaaicqqG0aancqqGGaaicqqGJbWycqqGTbqBcqqGGaaicqqGYbGCcqqGHbqycqqGKbazcqqGPbqAcqqG1bqDcqqGZbWCcqqGUaGlaaa@7F83@

(iii) Taking R as centre, draw an arc of 4 cm radius to meet previous arc at point P. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A7DE@

(iv) Join P to Q and R. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqG2bGDcqqGPaqkcqqGGaaicqqGkbGscqqGVbWBcqqGPbqAcqqGUbGBcqqGGaaicqqGqbaucqqGGaaicqqG0baDcqqGVbWBcqqGGaaicqqGrbqucqqGGaaicqqGHbqycqqGUbGBcqqGKbazcqqGGaaicqqGsbGucqqGUaGlaaa@5BBC@

Therefore, PQR is the required triangle. Since two sides of required triangle are equal, so it is an isosceles triangle. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@D681@

Q.56

Construct ΔABC such that AB=2.5cm, BC=6 cm andAC=6.5 cm. Measure B.

Ans

The steps of construction are as follows: (i) Draw a line segment BC of length 6 cm. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A799@

(ii) Taking C as centre, draw an arc of 6.5 cm radius. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqGPbqAcqqGPaqkcqqGGaaicqqGubavcqqGHbqycqqGRbWAcqqGPbqAcqqGUbGBcqqGNbWzcqqGGaaicqqGdbWqcqqGGaaicqqGHbqycqqGZbWCcqqGGaaicqqGJbWycqqGLbqzcqqGUbGBcqqG0baDcqqGYbGCcqqGLbqzcqqGSaalcqqGGaaicqqGKbazcqqGYbGCcqqGHbqycqqG3bWDcqqGGaaicqqGHbqycqqGUbGBcqqGGaaicqqGHbqycqqGYbGCcqqGJbWycqqGGaaicqqGVbWBcqqGMbGzcqqGGaaicqqG2aGncqqGUaGlcqqG1aqncqqGGaaicqqGJbWycqqGTbqBcqqGGaaicqqGYbGCcqqGHbqycqqGKbazcqqGPbqAcqqG1bqDcqqGZbWCcqqGUaGlaaa@813F@

(iii) Taking B as centre, draw an arc of 2.5 cm radius to meet previous arc at point A. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A970@

(iv) Join A​ to B and C. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqG2bGDcqqGPaqkcqqGGaaicqqGkbGscqqGVbWBcqqGPbqAcqqGUbGBcqqGGaaicqqGbbqqcaaMb8UaeeiiaaIaeeiDaqNaee4Ba8MaeeiiaaIaeeOqaiKaeeiiaaIaeeyyaeMaeeOBa4MaeeizaqMaeeiiaaIaee4qamKaeeOla4caaa@5CEC@

Therefore, ABC is the required triangle. B can be measured with the help of a protractor and it comes out to be 90°. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@CF15@

Q.57

Construct ΔDEF such that DE=5 cm, DF=3 cm and m EDF=90°.

Ans

The steps of construction are as follows: (i) Draw a line segment DE of length 5 cm. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A79F@

(ii) At point D, draw a ray DX making an angle of 90° with DE. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@8A92@

(iii) Taking D as centre, draw an arc of 3 cm radius. It will intersect DX at point F MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A6EE@

(iv) Join F to E MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqG2bGDcqqGPaqkcqqGGaaicqqGkbGscqqGVbWBcqqGPbqAcqqGUbGBcqqGGaaicqqGgbGrcqqGGaaicqqG0baDcqqGVbWBcqqGGaaicqqGfbqraaa@53F9@

Thus, DEF is the required triangle. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGubavcqqGObaAcqqG1bqDcqqGZbWCcqqGSaalcqqGGaaicqqGebarcqqGfbqrcqqGgbGrcqqGGaaicqqGPbqAcqqGZbWCcqqGGaaicqqG0baDcqqGObaAcqqGLbqzcqqGGaaicqqGYbGCcqqGLbqzcqqGXbqCcqqG1bqDcqqGPbqAcqqGYbGCcqqGLbqzcqqGKbazcqqGGaaicqqG0baDcqqGYbGCcqqGPbqAcqqGHbqycqqGUbGBcqqGNbWzcqqGSbaBcqqGLbqzcqqGUaGlaaa@6D24@

Q.58

Construct an isosceles triangle in which the lengths ofeach of its equal sides is 6.5 cm and the angle betweenthem is 110°.

Ans

An isosceles triangle PQR has to be constructed with PQ=QR=6.5 cm. The steps of construction are as follows: (i) Draw a line segment QR of length 6.5 cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbgeabjabb6gaUjabbccaGiabbMgaPjabbohaZjabb+gaVjabbohaZjabbogaJjabbwgaLjabbYgaSjabbwgaLjabbohaZjabbccaGiabbsha0jabbkhaYjabbMgaPjabbggaHjabb6gaUjabbEgaNjabbYgaSjabbwgaLjabbccaGiabbcfaqjabbgfarjabbkfasjabbccaGiabbIgaOjabbggaHjabbohaZjabbccaGiabbsha0jabb+gaVjabbccaGiabbkgaIjabbwgaLjabbccaGiabbogaJjabb+gaVjabb6gaUjabbohaZjabbsha0jabbkhaYjabbwha1jabbogaJjabbsha0jabbwgaLjabbsgaKjabbccaGiabbEha3jabbMgaPjabbsha0jabbIgaObqaaiabbcfaqjabbgfarjabg2da9iabbgfarjabbkfasjabg2da9iabbAda2iabb6caUiabbwda1iabbccaGiabbogaJjabb2gaTjabb6caUaqaaiabbsfaujabbIgaOjabbwgaLjabbccaGiabbohaZjabbsha0jabbwgaLjabbchaWjabbohaZjabbccaGiabb+gaVjabbAgaMjabbccaGiabbogaJjabb+gaVjabb6gaUjabbohaZjabbsha0jabbkhaYjabbwha1jabbogaJjabbsha0jabbMgaPjabb+gaVjabb6gaUjabbccaGiabbggaHjabbkhaYjabbwgaLjabbccaGiabbggaHjabbohaZjabbccaGiabbAgaMjabb+gaVjabbYgaSjabbYgaSjabb+gaVjabbEha3jabbohaZjabbQda6aqaaiabbIcaOiabbMgaPjabbMcaPiabbccaGiabbseaejabbkhaYjabbggaHjabbEha3jabbccaGiabbggaHjabbccaGiabbYgaSjabbMgaPjabb6gaUjabbwgaLjabbccaGiabbohaZjabbwgaLjabbEgaNjabb2gaTjabbwgaLjabb6gaUjabbsha0jabbccaGiabbgfarjabbkfasjabbccaGiabb+gaVjabbAgaMjabbccaGiabbYgaSjabbwgaLjabb6gaUjabbEgaNjabbsha0jabbIgaOjabbccaGiabbAda2iabb6caUiabbwda1iabbccaGiabbogaJjabb2gaTbaaaa@F827@

(ii) At point Q, draw a ray QX making an angle 110°with QR. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@878D@

(iii) Taking Q as centre, draw an arc of 6.5 cm radius. It intersects QX at point P. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A5D4@

(iv) Join P to R to obtain the required triangle PQR.

Q.59

Construct ΔABC with BC=7.5 cm, AC=5 cm and mC=60°.

Ans

The steps of constructions are as follows: (i) Draw a line segment BC of length 7.5 cm. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@AADC@

(ii) At point C, draw a ray CX making 60 with BC. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqGPbqAcqqGPaqkcqqGGaaicqqGbbqqcqqG0baDcqqGGaaicqqGWbaCcqqGVbWBcqqGPbqAcqqGUbGBcqqG0baDcqqGGaaicqqGdbWqcqqGSaalcqqGGaaicqqGKbazcqqGYbGCcqqGHbqycqqG3bWDcqqGGaaicqqGHbqycqqGGaaicqqGYbGCcqqGHbqycqqG5bqEcqqGGaaicqqGdbWqcqqGybawcqqGGaaicqqGTbqBcqqGHbqycqqGRbWAcqqGPbqAcqqGUbGBcqqGNbWzcqqGGaaicqqG2aGncqqGWaamcqqGGaaicqqG3bWDcqqGPbqAcqqG0baDcqqGObaAcqqGGaaicqqGcbGqcqqGdbWqcqqGUaGlaaa@7A2A@

(iii) Taking C as centre, draw an arc of 5 cm radius. It intersect CX at point A. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A23B@

(iv) Join A to B to obtain the required triangle ABC. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqG2bGDcqqGPaqkcqqGGaaicqqGkbGscqqGVbWBcqqGPbqAcqqGUbGBcqqGGaaicqqGbbqqcqqGGaaicqqG0baDcqqGVbWBcqqGGaaicqqGcbGqcqqGGaaicqqG0baDcqqGVbWBcqqGGaaicqqGVbWBcqqGIbGycqqG0baDcqqGHbqycqqGPbqAcqqGUbGBcqqGGaaicqqG0baDcqqGObaAcqqGLbqzcqqGGaaicqqGYbGCcqqGLbqzcqqGXbqCcqqG1bqDcqqGPbqAcqqGYbGCcqqGLbqzcqqGKbazcqqGGaaicqqG0baDcqqGYbGCcqqGPbqAcqqGHbqycqqGUbGBcqqGNbWzcqqGSbaBcqqGLbqzcqqGGaaicqqGbbqqcqqGcbGqcqqGdbWqcqqGUaGlaaa@8184@

Q.60

Construct ΔABC, given m A=60°, m B=30° and AB=5.8 cm.

Ans

(i) Draw a line segment AB of length 5.8 cm. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqGPaqkcqqGGaaicqqGebarcqqGYbGCcqqGHbqycqqG3bWDcqqGGaaicqqGHbqycqqGGaaicqqGSbaBcqqGPbqAcqqGUbGBcqqGLbqzcqqGGaaicqqGZbWCcqqGLbqzcqqGNbWzcqqGTbqBcqqGLbqzcqqGUbGBcqqG0baDcqqGGaaicqqGbbqqcqqGcbGqcqqGGaaicqqGVbWBcqqGMbGzcqqGGaaicqqGSbaBcqqGLbqzcqqGUbGBcqqGNbWzcqqG0baDcqqGObaAcqqGGaaicqqG1aqncqqGUaGlcqqG4aaocqqGGaaicqqGJbWycqqGTbqBcqqGUaGlaaa@750F@

(ii) At point A, draw a ray AX making 60° angle with AB. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@8382@

(iii) At point B, draw a ray AX making 30° angle with AB. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@84D7@

(iv) Point C has to lie on both the rays, AX and BY. Therefore, C is the point of intersection of these two rays. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@CAC3@

Thus, ABC is the required triangle. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGubavcqqGObaAcqqG1bqDcqqGZbWCcqqGSaalcqqGGaaicqqGbbqqcqqGcbGqcqqGdbWqcqqGGaaicqqGPbqAcqqGZbWCcqqGGaaicqqG0baDcqqGObaAcqqGLbqzcqqGGaaicqqGYbGCcqqGLbqzcqqGXbqCcqqG1bqDcqqGPbqAcqqGYbGCcqqGLbqzcqqGKbazcqqGGaaicqqG0baDcqqGYbGCcqqGPbqAcqqGHbqycqqGUbGBcqqGNbWzcqqGSbaBcqqGLbqzcqqGUaGlaaa@6D12@

Q.61

Construct ΔPQR if PQ=5 cm, m PQR=105° and m QRP=40°.(Hint: Recallanglesum property of a triangle).

Ans

To construct triangle PQR, we need to find RPQ. Using angle sum property of triangles to get PQR+PRQ+RPQ=180° 105°+40°+RPQ=180° RPQ=180°145° =35° The steps of construction are as follows: (i) Draw a line segment PQ of length 5 cm. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@79C4@

(ii) At P, draw a ray PX making an angle of 35° with PQ. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@8332@

(iii) At Q, draw a ray QY making an angle of 105° with PQ. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@8574@

(iv) Point R has to lie on both the ray, PX and QY. Therefore, R is the point of intersectin of these two rays. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@C7A2@

Thus, PQR is the required triangle. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGubavcqqGObaAcqqG1bqDcqqGZbWCcqqGSaalcqqGGaaicqqGqbaucqqGrbqucqqGsbGucqqGGaaicqqGPbqAcqqGZbWCcqqGGaaicqqG0baDcqqGObaAcqqGLbqzcqqGGaaicqqGYbGCcqqGLbqzcqqGXbqCcqqG1bqDcqqGPbqAcqqGYbGCcqqGLbqzcqqGKbazcqqGGaaicqqG0baDcqqGYbGCcqqGPbqAcqqGHbqycqqGUbGBcqqGNbWzcqqGSbaBcqqGLbqzcqqGUaGlaaa@6D6C@

Q.62

Examine whether you can construct ΔDEF such thatEF=7.2 cm, m E=110° and m F=80°. Justify your answer.

Ans

We are given, E=110° and F=80° Using angle sum property to get D+E+F=180° D+110°+80°=180° D+190°=180° Since Dcan not be zero and it does not satisfy angle sum property. Thus, we can not construct ΔDEF. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbEfaxjabbwgaLjabbccaGiabbggaHjabbkhaYjabbwgaLjabbccaGiabbEgaNjabbMgaPjabbAha2jabbwgaLjabb6gaUjabbYcaSaqaaiabgcIiqlabbweafjabg2da9iabbgdaXiabbgdaXiabbcdaWiabgclaWkabbccaGiabbggaHjabb6gaUjabbsgaKjabbccaGiabgcIiqlabbAeagjabg2da9iabbIda4iabbcdaWiabgclaWcqaaiabbwfavjabbohaZjabbMgaPjabb6gaUjabbEgaNjabbccaGiabbggaHjabb6gaUjabbEgaNjabbYgaSjabbwgaLjabbccaGiabbohaZjabbwha1jabb2gaTjabbccaGiabbchaWjabbkhaYjabb+gaVjabbchaWjabbwgaLjabbkhaYjabbsha0jabbMha5jabbccaGiabbsha0jabb+gaVjabbccaGiabbEgaNjabbwgaLjabbsha0bqaaiabgcIiqlabbseaejabbUcaRiabgcIiqlabbweafjabbUcaRiabgcIiqlabbAeagjabb2da9iabbgdaXiabbIda4iabbcdaWiabgclaWcqaaiabgcIiqlabdseaejabgUcaRiabigdaXiabigdaXiabicdaWiabgclaWkabgUcaRiabiIda4iabicdaWiabgclaWkabg2da9iabigdaXiabiIda4iabicdaWiabgclaWcqaaiabgcIiqlabdseaejabgUcaRiabigdaXiabiMda5iabicdaWiabgclaWkabg2da9iabigdaXiabiIda4iabicdaWiabgclaWcqaaiabbofatjabbMgaPjabb6gaUjabbogaJjabbwgaLjabbccaGiabgcIiqlabbseaejaaysW7cqqGJbWycqqGHbqycqqGUbGBcqqGGaaicqqGUbGBcqqGVbWBcqqG0baDcqqGGaaicqqGIbGycqqGLbqzcqqGGaaicqqG6bGEcqqGLbqzcqqGYbGCcqqGVbWBcqqGGaaicqqGHbqycqqGUbGBcqqGKbazcqqGGaaicqqGPbqAcqqG0baDcqqGGaaicqqGKbazcqqGVbWBcqqGLbqzcqqGZbWCcqqGGaaicqqGUbGBcqqGVbWBcqqG0baDcqqGGaaicqqGZbWCcqqGHbqycqqG0baDcqqGPbqAcqqGZbWCcqqGMbGzcqqG5bqEcqqGGaaicqqGHbqycqqGUbGBcqqGNbWzcqqGSbaBcqqGLbqzcqqGGaaicqqGZbWCcqqG1bqDcqqGTbqBaeaacqqGWbaCcqqGYbGCcqqGVbWBcqqGWbaCcqqGLbqzcqqGYbGCcqqG0baDcqqG5bqEcqqGUaGlaeaacqqGubavcqqGObaAcqqG1bqDcqqGZbWCcqqGSaalcqqGGaaicqqG3bWDcqqGLbqzcqqGGaaicqqGJbWycqqGHbqycqqGUbGBcqqGGaaicqqGUbGBcqqGVbWBcqqG0baDcqqGGaaicqqGJbWycqqGVbWBcqqGUbGBcqqGZbWCcqqG0baDcqqGYbGCcqqG1bqDcqqGJbWycqqG0baDcqqGGaaicqGHuoarcqqGebarcqqGfbqrcqqGgbGrcqqGUaGlaaaa@3BA0@

Also, it can be observed that point D should lie on both rays, EX and FY, for the constructing the required triangle. However, both rays are not intersecting each other. Therefore, the required triangle can not be formed. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbgeabjabbYgaSjabbohaZjabb+gaVjabbYcaSiabbccaGiabbMgaPjabbsha0jabbccaGiabbogaJjabbggaHjabb6gaUjabbccaGiabbkgaIjabbwgaLjabbccaGiabb+gaVjabbkgaIjabbohaZjabbwgaLjabbkhaYjabbAha2jabbwgaLjabbsgaKjabbccaGiabbsha0jabbIgaOjabbggaHjabbsha0jabbccaGiabbchaWjabb+gaVjabbMgaPjabb6gaUjabbsha0jabbccaGiabbseaejabbccaGiabbohaZjabbIgaOjabb+gaVjabbwha1jabbYgaSjabbsgaKjabbccaGiabbYgaSjabbMgaPjabbwgaLjabbccaGiabb+gaVjabb6gaUjabbccaGiabbkgaIjabb+gaVjabbsha0jabbIgaOjabbccaGiabbkhaYjabbggaHjabbMha5jabbohaZjabbYcaSaqaaiabbweafjabbIfayjabbccaGiabbggaHjabb6gaUjabbsgaKjabbccaGiabbAeagjabbMfazjabbYcaSiabbccaGiabbAgaMjabb+gaVjabbkhaYjabbccaGiabbsha0jabbIgaOjabbwgaLjabbccaGiabbogaJjabb+gaVjabb6gaUjabbohaZjabbsha0jabbkhaYjabbwha1jabbogaJjabbsha0jabbMgaPjabb6gaUjabbEgaNjabbccaGiabbsha0jabbIgaOjabbwgaLjabbccaGiabbkhaYjabbwgaLjabbghaXjabbwha1jabbMgaPjabbkhaYjabbwgaLjabbsgaKjabbccaGiabbsha0jabbkhaYjabbMgaPjabbggaHjabb6gaUjabbEgaNjabbYgaSjabbwgaLjabb6caUaqaaiabbIeaijabb+gaVjabbEha3jabbwgaLjabbAha2jabbwgaLjabbkhaYjabbYcaSiabbccaGiabbkgaIjabb+gaVjabbsha0jabbIgaOjabbccaGiabbkhaYjabbggaHjabbMha5jabbohaZjabbccaGiabbggaHjabbkhaYjabbwgaLjabbccaGiabb6gaUjabb+gaVjabbsha0jabbccaGiabbMgaPjabb6gaUjabbsha0jabbwgaLjabbkhaYjabbohaZjabbwgaLjabbogaJjabbsha0jabbMgaPjabb6gaUjabbEgaNjabbccaGiabbwgaLjabbggaHjabbogaJjabbIgaOjabbccaGiabb+gaVjabbsha0jabbIgaOjabbwgaLjabbkhaYjabb6caUaqaaiabbsfaujabbIgaOjabbwgaLjabbkhaYjabbwgaLjabbAgaMjabb+gaVjabbkhaYjabbwgaLjabbYcaSiabbccaGiabbsha0jabbIgaOjabbwgaLjabbccaGiabbkhaYjabbwgaLjabbghaXjabbwha1jabbMgaPjabbkhaYjabbwgaLjabbsgaKjabbccaGiabbsha0jabbkhaYjabbMgaPjabbggaHjabb6gaUjabbEgaNjabbYgaSjabbwgaLjabbccaGiabbogaJjabbggaHjabb6gaUjabbccaGiabb6gaUjabb+gaVjabbsha0jabbccaGiabbkgaIjabbwgaLjabbccaGiabbAgaMjabb+gaVjabbkhaYjabb2gaTjabbwgaLjabbsgaKjabb6caUaaaaa@5270@

Q.63

Construct the right angled ΔPQR, where m Q=90°, QR=8 cm and PR=10 cm.

Ans

The steps of construction are as follows: (i) Draw a line segment QR of length 8 cm. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A7D9@

(ii) At point Q, draw a ray QX making 90° with QR. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqGPbqAcqqGPaqkcqqGGaaicqqGbbqqcqqG0baDcqqGGaaicqqGWbaCcqqGVbWBcqqGPbqAcqqGUbGBcqqG0baDcqqGGaaicqqGrbqucqqGSaalcqqGGaaicqqGKbazcqqGYbGCcqqGHbqycqqG3bWDcqqGGaaicqqGHbqycqqGGaaicqqGYbGCcqqGHbqycqqG5bqEcqqGGaaicqqGrbqucqqGybawcqqGGaaicqqGTbqBcqqGHbqycqqGRbWAcqqGPbqAcqqGUbGBcqqGNbWzcqqGGaaicqqG5aqocqqGWaamcqGHWcaScqqGGaaicqqG3bWDcqqGPbqAcqqG0baDcqqGObaAcqqGGaaicqqGrbqucqqGsbGucqqGUaGlaaa@7C90@

(iii) Taking R as centre, draw an arc of 10 cm radius to intersect ray QX at point P. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A7CF@

(iv) Join P to R. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqG2bGDcqqGPaqkcqqGGaaicqqGkbGscqqGVbWBcqqGPbqAcqqGUbGBcqqGGaaicqqGqbaucqqGGaaicqqG0baDcqqGVbWBcqqGGaaicqqGsbGucqqGUaGlaaa@550A@

Thus, ΔPQR is the required right-angled triangle.MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGubavcqqGObaAcqqG1bqDcqqGZbWCcqqGSaalcqqGGaaicqGHuoarcqqGqbaucqqGrbqucqqGsbGucqqGGaaicqqGPbqAcqqGZbWCcqqGGaaicqqG0baDcqqGObaAcqqGLbqzcqqGGaaicqqGYbGCcqqGLbqzcqqGXbqCcqqG1bqDcqqGPbqAcqqGYbGCcqqGLbqzcqqGKbazcqqGGaaicqqGYbGCcqqGPbqAcqqGNbWzcqqGObaAcqqG0baDcqqGTaqlcqqGHbqycqqGUbGBcqqGNbWzcqqGSbaBcqqGLbqzcqqGKbazcqqGGaaicqqG0baDcqqGYbGCcqqGPbqAcqqGHbqycqqGUbGBcqqGNbWzcqqGSbaBcqqGLbqzcqqGUaGlaaa@7F5A@

Q.64 Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.

Ans

A right-angled triangle ABC with hypotenuse 6 cm and one of the legs as 4 cm has to be constructed. The steps of construction are as follows: (i) Draw a line segment BC of length 4 cm. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@1FB7@

(ii) At point B, draw a ray BX making 90° with BC. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqGPbqAcqqGPaqkcqqGGaaicqqGbbqqcqqG0baDcqqGGaaicqqGWbaCcqqGVbWBcqqGPbqAcqqGUbGBcqqG0baDcqqGGaaicqqGcbGqcqqGSaalcqqGGaaicqqGKbazcqqGYbGCcqqGHbqycqqG3bWDcqqGGaaicqqGHbqycqqGGaaicqqGYbGCcqqGHbqycqqG5bqEcqqGGaaicqqGcbGqcqqGybawcqqGGaaicqqGTbqBcqqGHbqycqqGRbWAcqqGPbqAcqqGUbGBcqqGNbWzcqqGGaaicqqG5aqocqqGWaamcqGHWcaScqqGGaaicqqG3bWDcqqGPbqAcqqG0baDcqqGObaAcqqGGaaicqqGcbGqcqqGdbWqcqqGUaGlaaa@7C18@

(iii) Taking C as centre, draw an arc of 6 cm radius to intersect ray BX at point A. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbIcaOiabbMgaPjabbMgaPjabbMgaPjabbMcaPiabbccaGiabbsfaujabbggaHjabbUgaRjabbMgaPjabb6gaUjabbEgaNjabbccaGiabboeadjabbccaGiabbggaHjabbohaZjabbccaGiabbogaJjabbwgaLjabb6gaUjabbsha0jabbkhaYjabbwgaLjabbYcaSiabbccaGiabbsgaKjabbkhaYjabbggaHjabbEha3jabbccaGiabbggaHjabb6gaUjabbccaGiabbggaHjabbkhaYjabbogaJjabbccaGiabb+gaVjabbAgaMjabbccaGiabbAda2iabbccaGiabbogaJjabb2gaTjabbccaGiabbkhaYjabbggaHjabbsgaKjabbMgaPjabbwha1jabbohaZjabbccaGiabbsha0jabb+gaVjabbccaGaqaaiabbMgaPjabb6gaUjabbsha0jabbwgaLjabbkhaYjabbohaZjabbwgaLjabbogaJjabbsha0jabbccaGiabbkhaYjabbggaHjabbMha5jabbccaGiabbkeacjabbIfayjabbccaGiabbggaHjabbsha0jabbccaGiabbchaWjabb+gaVjabbMgaPjabb6gaUjabbsha0jabbccaGiabbgeabjabb6caUaaaaa@A698@

(iv) Join A to C.

athType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqG2bGDcqqGPaqkcqqGGaaicqqGkbGscqqGVbWBcqqGPbqAcqqGUbGBcqqGGaaicqqGbbqqcqqGGaaicqqG0baDcqqGVbWBcqqGGaaicqqGdbWqcqqGUaGlaaa@54CE@ Thus, ΔABC is the required right-angled triangle. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGubavcqqGObaAcqqG1bqDcqqGZbWCcqqGSaalcqqGGaaicqGHuoarcqqGbbqqcqqGcbGqcqqGdbWqcqqGGaaicqqGPbqAcqqGZbWCcqqGGaaicqqG0baDcqqGObaAcqqGLbqzcqqGGaaicqqGYbGCcqqGLbqzcqqGXbqCcqqG1bqDcqqGPbqAcqqGYbGCcqqGLbqzcqqGKbazcqqGGaaicqqGYbGCcqqGPbqAcqqGNbWzcqqGObaAcqqG0baDcqqGTaqlcqqGHbqycqqGUbGBcqqGNbWzcqqGSbaBcqqGLbqzcqqGKbazcqqGGaaicqqG0baDcqqGYbGCcqqGPbqAcqqGHbqycqqGUbGBcqqGNbWzcqqGSbaBcqqGLbqzcqqGUaGlaaa@7F00@

Q.65

Construct an isosceles rightangled triangle ABC,where m ACB=90° and AC=6cm.

Ans

Since, in a isosceles triangle, the length of any two side are equal. So, AB=BC=6cm The steps of construction are as follows: (i) Draw a line segment AC of length 6 cm. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@0B0B@

(ii) At point C, draw a ray CX making 90° with AC. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqGPbqAcqqGPaqkcqqGGaaicqqGbbqqcqqG0baDcqqGGaaicqqGWbaCcqqGVbWBcqqGPbqAcqqGUbGBcqqG0baDcqqGGaaicqqGdbWqcqqGSaalcqqGGaaicqqGKbazcqqGYbGCcqqGHbqycqqG3bWDcqqGGaaicqqGHbqycqqGGaaicqqGYbGCcqqGHbqycqqG5bqEcqqGGaaicqqGdbWqcqqGybawcqqGGaaicqqGTbqBcqqGHbqycqqGRbWAcqqGPbqAcqqGUbGBcqqGNbWzcqqGGaaicqqG5aqocqqGWaamcqGHWcaScqqGGaaicqqG3bWDcqqGPbqAcqqG0baDcqqGObaAcqqGGaaicqqGbbqqcqqGdbWqcqqGUaGlaaa@7C1A@

(iii) Taking C as centre, draw an arc of 6 cm radius to intersect ray CX at point B. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbIcaOiabbMgaPjabbMgaPjabbMgaPjabbMcaPiabbccaGiabbsfaujabbggaHjabbUgaRjabbMgaPjabb6gaUjabbEgaNjabbccaGiabboeadjabbccaGiabbggaHjabbohaZjabbccaGiabbogaJjabbwgaLjabb6gaUjabbsha0jabbkhaYjabbwgaLjabbYcaSiabbccaGiabbsgaKjabbkhaYjabbggaHjabbEha3jabbccaGiabbggaHjabb6gaUjabbccaGiabbggaHjabbkhaYjabbogaJjabbccaGiabb+gaVjabbAgaMjabbccaGiabbAda2iabbccaGiabbogaJjabb2gaTjabbccaGiabbkhaYjabbggaHjabbsgaKjabbMgaPjabbwha1jabbohaZjabbccaGiabbsha0jabb+gaVjabbccaGaqaaiabbMgaPjabb6gaUjabbsha0jabbwgaLjabbkhaYjabbohaZjabbwgaLjabbogaJjabbsha0jabbccaGiabbkhaYjabbggaHjabbMha5jabbccaGiabboeadjabbIfayjabbccaGiabbggaHjabbsha0jabbccaGiabbchaWjabb+gaVjabbMgaPjabb6gaUjabbsha0jabbccaGiabbkeacjabb6caUaaaaa@A69C@

(iv) Join A to B.

MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqG2bGDcqqGPaqkcqqGGaaicqqGkbGscqqGVbWBcqqGPbqAcqqGUbGBcqqGGaaicqqGbbqqcqqGGaaicqqG0baDcqqGVbWBcqqGGaaicqqGcbGqcqqGUaGlaaa@54CC@ Thus, ΔABC is the required triangle. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGubavcqqGObaAcqqG1bqDcqqGZbWCcqqGSaalcqqGGaaicqGHuoarcqqGbbqqcqqGcbGqcqqGdbWqcqqGGaaicqqGPbqAcqqGZbWCcqqGGaaicqqG0baDcqqGObaAcqqGLbqzcqqGGaaicqqGYbGCcqqGLbqzcqqGXbqCcqqG1bqDcqqGPbqAcqqGYbGCcqqGLbqzcqqGKbazcqqGGaaicqqG0baDcqqGYbGCcqqGPbqAcqqGHbqycqqGUbGBcqqGNbWzcqqGSbaBcqqGLbqzcqqGUaGlaaa@6E79@

Q.66

Identify the nets which can be used to make cubes(cut out copies of the nets and try it):

Ans

(i)The given net can be folded as:

When the faces that are in sky blue colur and in pink colour are folded to make cube, they will be overlaping each other. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@D794@ (ii) The given net can be folded as: MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqGPbqAcqqGPaqkcqqGGaaicqqGubavcqqGObaAcqqGLbqzcqqGGaaicqqGNbWzcqqGPbqAcqqG2bGDcqqGLbqzcqqGUbGBcqqGGaaicqqGUbGBcqqGLbqzcqqG0baDcqqGGaaicqqGJbWycqqGHbqycqqGUbGBcqqGGaaicqqGIbGycqqGLbqzcqqGGaaicqqGMbGzcqqGVbWBcqqGSbaBcqqGKbazcqqGLbqzcqqGKbazcqqGGaaicqqGHbqycqqGZbWCcqqG6aGoaaa@6D1B@

Thus, a cube can thus be formed in above way. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGubavcqqGObaAcqqG1bqDcqqGZbWCcqqGSaalcqqGGaaicqqGHbqycqqGGaaicqqGJbWycqqG1bqDcqqGIbGycqqGLbqzcqqGGaaicqqGJbWycqqGHbqycqqGUbGBcqqGGaaicqqG0baDcqqGObaAcqqG1bqDcqqGZbWCcqqGGaaicqqGIbGycqqGLbqzcqqGGaaicqqGMbGzcqqGVbWBcqqGYbGCcqqGTbqBcqqGLbqzcqqGKbazcqqGGaaicqqGPbqAcqqGUbGBcqqGGaaicqqGHbqycqqGIbGycqqGVbWBcqqG2bGDcqqGLbqzcqqGGaaicqqG3bWDcqqGHbqycqqG5bqEcqqGUaGlaaa@7902@ (iii) The given net can be folded as: MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqG2bGDcqqGPaqkcqqGGaaicqqGubavcqqGObaAcqqGLbqzcqqGGaaicqqGNbWzcqqGPbqAcqqG2bGDcqqGLbqzcqqGUbGBcqqGGaaicqqGUbGBcqqGLbqzcqqG0baDcqqGGaaicqqGJbWycqqGHbqycqqGUbGBcqqGGaaicqqGIbGycqqGLbqzcqqGGaaicqqGMbGzcqqGVbWBcqqGSbaBcqqGKbazcqqGLbqzcqqGKbazcqqGGaaicqqGHbqycqqGZbWCcqqG6aGoaaa@6D35@

Thus, a cube can thus be formed in above way. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGubavcqqGObaAcqqG1bqDcqqGZbWCcqqGSaalcqqGGaaicqqGHbqycqqGGaaicqqGJbWycqqG1bqDcqqGIbGycqqGLbqzcqqGGaaicqqGJbWycqqGHbqycqqGUbGBcqqGGaaicqqG0baDcqqGObaAcqqG1bqDcqqGZbWCcqqGGaaicqqGIbGycqqGLbqzcqqGGaaicqqGMbGzcqqGVbWBcqqGYbGCcqqGTbqBcqqGLbqzcqqGKbazcqqGGaaicqqGPbqAcqqGUbGBcqqGGaaicqqGHbqycqqGIbGycqqGVbWBcqqG2bGDcqqGLbqzcqqGGaaicqqG3bWDcqqGHbqycqqG5bqEcqqGUaGlaaa@7902@ (iv) The given net can be folded as: MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqG2bGDcqqGPaqkcqqGGaaicqqGubavcqqGObaAcqqGLbqzcqqGGaaicqqGNbWzcqqGPbqAcqqG2bGDcqqGLbqzcqqGUbGBcqqGGaaicqqGUbGBcqqGLbqzcqqG0baDcqqGGaaicqqGJbWycqqGHbqycqqGUbGBcqqGGaaicqqGIbGycqqGLbqzcqqGGaaicqqGMbGzcqqGVbWBcqqGSbaBcqqGKbazcqqGLbqzcqqGKbazcqqGGaaicqqGHbqycqqGZbWCcqqG6aGoaaa@6D35@

Thus, a cube can thus be formed in above way. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGubavcqqGObaAcqqG1bqDcqqGZbWCcqqGSaalcqqGGaaicqqGHbqycqqGGaaicqqGJbWycqqG1bqDcqqGIbGycqqGLbqzcqqGGaaicqqGJbWycqqGHbqycqqGUbGBcqqGGaaicqqG0baDcqqGObaAcqqG1bqDcqqGZbWCcqqGGaaicqqGIbGycqqGLbqzcqqGGaaicqqGMbGzcqqGVbWBcqqGYbGCcqqGTbqBcqqGLbqzcqqGKbazcqqGGaaicqqGPbqAcqqGUbGBcqqGGaaicqqGHbqycqqGIbGycqqGVbWBcqqG2bGDcqqGLbqzcqqGGaaicqqG3bWDcqqGHbqycqqG5bqEcqqGUaGlaaa@7902@ (v) The given net can be folded as: MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqG2bGDcqqGPaqkcqqGGaaicqqGubavcqqGObaAcqqGLbqzcqqGGaaicqqGNbWzcqqGPbqAcqqG2bGDcqqGLbqzcqqGUbGBcqqGGaaicqqGUbGBcqqGLbqzcqqG0baDcqqGGaaicqqGJbWycqqGHbqycqqGUbGBcqqGGaaicqqGIbGycqqGLbqzcqqGGaaicqqGMbGzcqqGVbWBcqqGSbaBcqqGKbazcqqGLbqzcqqGKbazcqqGGaaicqqGHbqycqqGZbWCcqqG6aGoaaa@6BDC@

When the faces are in blue colour and in red colour are folded to make a cube, they will be overlapping each other. (vi) The given net can be folded as: MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@FAB5@

Thus, a cube can thus be formed in above way.MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGubavcqqGObaAcqqG1bqDcqqGZbWCcqqGSaalcqqGGaaicqqGHbqycqqGGaaicqqGJbWycqqG1bqDcqqGIbGycqqGLbqzcqqGGaaicqqGJbWycqqGHbqycqqGUbGBcqqGGaaicqqG0baDcqqGObaAcqqG1bqDcqqGZbWCcqqGGaaicqqGIbGycqqGLbqzcqqGGaaicqqGMbGzcqqGVbWBcqqGYbGCcqqGTbqBcqqGLbqzcqqGKbazcqqGGaaicqqGPbqAcqqGUbGBcqqGGaaicqqGHbqycqqGIbGycqqGVbWBcqqG2bGDcqqGLbqzcqqGGaaicqqG3bWDcqqGHbqycqqG5bqEcqqGUaGlaaa@7902@

Q.67

Dice are cubes with dots on each face. Opposite faces ofa die always have a total of seven dots on them.

Here are two nets to make dice cubes; the numbersinserted in each square indicate the number of dots inthat box.

Insert suitable numbers in the blanks, remembering thatthe number on the opposite faces should total to 7.

Ans

(i) The numbers can be inserted as follows so as to make the given net into a net of a dice. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@B167@

It can be seen that the sum of opposite faces is 7. (ii) The numbers can be inserted as follows so as to make the given net into a net of a dice. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@F0CD@

It can be seen that the sum of opposite faces is 7. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGjbqscqqG0baDcqqGGaaicqqGJbWycqqGHbqycqqGUbGBcqqGGaaicqqGIbGycqqGLbqzcqqGGaaicqqGZbWCcqqGLbqzcqqGLbqzcqqGUbGBcqqGGaaicqqG0baDcqqGObaAcqqGHbqycqqG0baDcqqGGaaicqqG0baDcqqGObaAcqqGLbqzcqqGGaaicqqGZbWCcqqG1bqDcqqGTbqBcqqGGaaicqqGVbWBcqqGMbGzcqqGGaaicqqGVbWBcqqGWbaCcqqGWbaCcqqGVbWBcqqGZbWCcqqGPbqAcqqG0baDcqqGLbqzcqqGGaaicqqGMbGzcqqGHbqycqqGJbWycqqGLbqzcqqGZbWCcqqGGaaicqqGPbqAcqqGZbWCcqqGGaaicqqG3aWncqqGUaGlaaa@8038@

Q.68

Can this be a net for a die? Explain your answer.

Ans

The given net can be folded as: MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGubavcqqGObaAcqqGLbqzcqqGGaaicqqGNbWzcqqGPbqAcqqG2bGDcqqGLbqzcqqGUbGBcqqGGaaicqqGUbGBcqqGLbqzcqqG0baDcqqGGaaicqqGJbWycqqGHbqycqqGUbGBcqqGGaaicqqGIbGycqqGLbqzcqqGGaaicqqGMbGzcqqGVbWBcqqGSbaBcqqGKbazcqqGLbqzcqqGKbazcqqGGaaicqqGHbqycqqGZbWCcqqG6aGoaaa@67F2@

It can be observed that the opposite face of the dice so formed have 2 and 5, 1 and 4, 3 and 6 on them. The sum of numbers on the opposite faces comes to 7, 5 and 9 respectively. However, in case of a dice, the sum of numbers on the opposite faces should be7. Therefore, this net is not of a dice. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A39A@

Q.69

Here is an incomplete net for making a cube. Completeit in at least two different ways. Remember that a cubehas six faces. How many are there in the net here?(Give two separate diagrams. If you like, you may use a squared sheet for easy manipulation.)

Ans

There are 3 faces given in the net. The given net can be completed as: MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@9737@

Q.70

Match the nets with appropriate solids:

Ans

Consider net (i) It can be folded as:

Consider net (ii) It can be folded as: MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGdbWqcqqGVbWBcqqGUbGBcqqGZbWCcqqGPbqAcqqGKbazcqqGLbqzcqqGYbGCcqqGGaaicqqGUbGBcqqGLbqzcqqG0baDcqqGGaaicqqGOaakcqqGPbqAcqqGPbqAcqqGPaqkcqqGGaaicqqGjbqscqqG0baDcqqGGaaicqqGJbWycqqGHbqycqqGUbGBcqqGGaaicqqGIbGycqqGLbqzcqqGGaaicqqGMbGzcqqGVbWBcqqGSbaBcqqGKbazcqqGLbqzcqqGKbazcqqGGaaicqqGHbqycqqGZbWCcqqG6aGoaaa@6F9D@

It is a net of cube. Hence, (a) is the correct option. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGjbqscqqG0baDcqqGGaaicqqGPbqAcqqGZbWCcqqGGaaicqqGHbqycqqGGaaicqqGUbGBcqqGLbqzcqqG0baDcqqGGaaicqqGVbWBcqqGMbGzcqqGGaaicqqGJbWycqqG1bqDcqqGIbGycqqGLbqzcqqGUaGlcqqGGaaicqqGibascqqGLbqzcqqGUbGBcqqGJbWycqqGLbqzcqqGSaalcqqGGaaicqqGOaakcqqGHbqycqqGPaqkcqqGGaaicqqGPbqAcqqGZbWCcqqGGaaicqqG0baDcqqGObaAcqqGLbqzcqqGGaaicqqGJbWycqqGVbWBcqqGYbGCcqqGYbGCcqqGLbqzcqqGJbWycqqG0baDcqqGGaaicqqGVbWBcqqGWbaCcqqG0baDcqqGPbqAcqqGVbWBcqqGUbGBcqqGUaGlaaa@825B@ Consider net (iii) It can be folded as: MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGdbWqcqqGVbWBcqqGUbGBcqqGZbWCcqqGPbqAcqqGKbazcqqGLbqzcqqGYbGCcqqGGaaicqqGUbGBcqqGLbqzcqqG0baDcqqGGaaicqqGOaakcqqGPbqAcqqGPbqAcqqGPbqAcqqGPaqkcqqGGaaicqqGjbqscqqG0baDcqqGGaaicqqGJbWycqqGHbqycqqGUbGBcqqGGaaicqqGIbGycqqGLbqzcqqGGaaicqqGMbGzcqqGVbWBcqqGSbaBcqqGKbazcqqGLbqzcqqGKbazcqqGGaaicqqGHbqycqqGZbWCcqqG6aGoaaa@70F6@

It is a net of cylinder. Hence, (b) is the correct option. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@87EF@ Consider net (iv) It can be folded as: MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGdbWqcqqGVbWBcqqGUbGBcqqGZbWCcqqGPbqAcqqGKbazcqqGLbqzcqqGYbGCcqqGGaaicqqGUbGBcqqGLbqzcqqG0baDcqqGGaaicqqGOaakcqqGPbqAcqqG2bGDcqqGPaqkcqqGGaaicqqGjbqscqqG0baDcqqGGaaicqqGJbWycqqGHbqycqqGUbGBcqqGGaaicqqGIbGycqqGLbqzcqqGGaaicqqGMbGzcqqGVbWBcqqGSbaBcqqGKbazcqqGLbqzcqqGKbazcqqGGaaicqqGHbqycqqGZbWCcqqG6aGoaaa@6FB7@

It is a net of cone. Hence, (c) is the correct option. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGjbqscqqG0baDcqqGGaaicqqGPbqAcqqGZbWCcqqGGaaicqqGHbqycqqGGaaicqqGUbGBcqqGLbqzcqqG0baDcqqGGaaicqqGVbWBcqqGMbGzcqqGGaaicqqGJbWycqqGVbWBcqqGUbGBcqqGLbqzcqqGUaGlcqqGGaaicqqGibascqqGLbqzcqqGUbGBcqqGJbWycqqGLbqzcqqGSaalcqqGGaaicqqGOaakcqqGJbWycqqGPaqkcqqGGaaicqqGPbqAcqqGZbWCcqqGGaaicqqG0baDcqqGObaAcqqGLbqzcqqGGaaicqqGJbWycqqGVbWBcqqGYbGCcqqGYbGCcqqGLbqzcqqGJbWycqqG0baDcqqGGaaicqqGVbWBcqqGWbaCcqqG0baDcqqGPbqAcqqGVbWBcqqGUbGBcqqGUaGlaaa@826B@

Q.71

Use isometric dot paper and make an isometric sketch for each one of the given shapes:

Ans

Q.72

The dimensions of a cuboid are 5 cm, 3 cm and 2 cm. Draw three different isometric sketches of this cuboid.

Ans

Q.73

Three cubes each with 2 cm edge are placed side by sideto form a cuboid. Sketch an oblique or isometric sketchof this cuboid.

Ans

Q.74

Make an oblique sketch for each one of the given isometric shapes:

Ans

Q.75

Give i an oblique sketch and ii an isometric sketch foreach of the following:a A cuboid of dimensions 5 cm, 3 cm and 2 cm.Is your sketch unique?b A cube with an edge 4 cm long.An isometric sheet is attached at the end of the book.You could try to make on it somecubes or cuboids ofdimensions specified by your friend.

Ans

(i) oblique sketch: MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqGPaqkcqqGGaaicqqGVbWBcqqGIbGycqqGSbaBcqqGPbqAcqqGXbqCcqqG1bqDcqqGLbqzcqqGGaaicqqGZbWCcqqGRbWAcqqGLbqzcqqG0baDcqqGJbWycqqGObaAcqqG6aGoaaa@5980@

(ii) Isometric sketch: MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqGPbqAcqqGPaqkcqqGGaaicqqGjbqscqqGZbWCcqqGVbWBcqqGTbqBcqqGLbqzcqqG0baDcqqGYbGCcqqGPbqAcqqGJbWycqqGGaaicqqGZbWCcqqGRbWAcqqGLbqzcqqG0baDcqqGJbWycqqGObaAcqqG6aGoaaa@5D63@

Q.76

What cross-sections do you get when you give ai vertical cut ii horizontal cut to the following solids?a A brick b Around applec A die d A circular pipee A nice cream cone

Ans

(a) A brickWe can give a vertical cut to a brick in the following way:

(b) A round apple We can give a vertical cut to a roudn apple in the following way:

We can give a horizontal cut to a round apple in the following way:

(c) A die

We can give a vertical cut to a die in the following ways:

We can give a horizontal cut to a die in the following way: MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@8B3E@

(d) A circular pipe We can give a vertical cut to a circular pipe in the following way:

We can give a horizontal cut to a circular pipe in the following way: MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@9866@

(e) An ice cream cone We can give a vertical cut to an ice cream cone in the following way:

We can give a horizontal cut to an ice cream cone in the following way:

Q.77

A bulb is kept burning just right above the following solids.Name the shape of the shadows obtained in each case.Attempt to give a rough sketch of the shadow.(You maytry to experiment first and then answer these questions).

Ans

The shapes of the shadows of these figures will be as follows: (i) A ball MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@9AEC@

The shape of the shadow of a ball will be a circle. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGubavcqqGObaAcqqGLbqzcqqGGaaicqqGZbWCcqqGObaAcqqGHbqycqqGWbaCcqqGLbqzcqqGGaaicqqGVbWBcqqGMbGzcqqGGaaicqqG0baDcqqGObaAcqqGLbqzcqqGGaaicqqGZbWCcqqGObaAcqqGHbqycqqGKbazcqqGVbWBcqqG3bWDcqqGGaaicqqGVbWBcqqGMbGzcqqGGaaicqqGHbqycqqGGaaicqqGIbGycqqGHbqycqqGSbaBcqqGSbaBcqqGGaaicqqG3bWDcqqGPbqAcqqGSbaBcqqGSbaBcqqGGaaicqqGIbGycqqGLbqzcqqGGaaicqqGHbqycqqGGaaicqqGJbWycqqGPbqAcqqGYbGCcqqGJbWycqqGSbaBcqqGLbqzcqqGUaGlaaa@8010@ (ii) A cylindrical pipe MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqGPbqAcqqGPaqkcqqGGaaicqqGbbqqcqqGGaaicqqGJbWycqqG5bqEcqqGSbaBcqqGPbqAcqqGUbGBcqqGKbazcqqGYbGCcqqGPbqAcqqGJbWycqqGHbqycqqGSbaBcqqGGaaicqqGWbaCcqqGPbqAcqqGWbaCcqqGLbqzaaa@5E4E@

The shape of the shadow of a cylindrical pipe will be a rectangle . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGubavcqqGObaAcqqGLbqzcqqGGaaicqqGZbWCcqqGObaAcqqGHbqycqqGWbaCcqqGLbqzcqqGGaaicqqGVbWBcqqGMbGzcqqGGaaicqqG0baDcqqGObaAcqqGLbqzcqqGGaaicqqGZbWCcqqGObaAcqqGHbqycqqGKbazcqqGVbWBcqqG3bWDcqqGGaaicqqGVbWBcqqGMbGzcqqGGaaicqqGHbqycqqGGaaicqqGJbWycqqG5bqEcqqGSbaBcqqGPbqAcqqGUbGBcqqGKbazcqqGYbGCcqqGPbqAcqqGJbWycqqGHbqycqqGSbaBcqqGGaaicqqGWbaCcqqGPbqAcqqGWbaCcqqGLbqzcqqGGaaicqqG3bWDcqqGPbqAcqqGSbaBcqqGSbaBcqqGGaaicqqGIbGycqqGLbqzcqqGGaaicqqGHbqycqqGGaaicqqGYbGCcqqGLbqzcqqGJbWycqqG0baDcqqGHbqycqqGUbGBcqqGNbWzcqqGSbaBcqqGLbqzcqqGUaGlaaa@9401@ (iii) A book

athType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGOaakcqqGPbqAcqqGPbqAcqqGPbqAcqqGPaqkcqqGGaaicqqGbbqqcqqGGaaicqqGIbGycqqGVbWBcqqGVbWBcqqGRbWAaaa@4FF1@ The shape of the shadow of a book will be a rectangle.MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGubavcqqGObaAcqqGLbqzcqqGGaaicqqGZbWCcqqGObaAcqqGHbqycqqGWbaCcqqGLbqzcqqGGaaicqqGVbWBcqqGMbGzcqqGGaaicqqG0baDcqqGObaAcqqGLbqzcqqGGaaicqqGZbWCcqqGObaAcqqGHbqycqqGKbazcqqGVbWBcqqG3bWDcqqGGaaicqqGVbWBcqqGMbGzcqqGGaaicqqGHbqycqqGGaaicqqGIbGycqqGVbWBcqqGVbWBcqqGRbWAcqqGGaaicqqG3bWDcqqGPbqAcqqGSbaBcqqGSbaBcqqGGaaicqqGIbGycqqGLbqzcqqGGaaicqqGHbqycqqGGaaicqqGYbGCcqqGLbqzcqqGJbWycqqG0baDcqqGHbqycqqGUbGBcqqGNbWzcqqGSbaBcqqGLbqzcqqGUaGlaaa@844B@

Q.78

Here are the shadows of some 3-D objects, when seenunder the lamp of an over head projector. Identify thesolids that match each shadow. (There may bemultipleanswers for these!).

Ans

The given shadows can be obtained in case of the following objects. (i) Compact disk (Sphere) (ii) A ​dice (Cube) (iii) Triangular pyramid (Cone) (iv) Note Book (Cuboid) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbsfaujabbIgaOjabbwgaLjabbccaGiabbEgaNjabbMgaPjabbAha2jabbwgaLjabb6gaUjabbccaGiabbohaZjabbIgaOjabbggaHjabbsgaKjabb+gaVjabbEha3jabbohaZjabbccaGiabbogaJjabbggaHjabb6gaUjabbccaGiabbkgaIjabbwgaLjabbccaGiabb+gaVjabbkgaIjabbsha0jabbggaHjabbMgaPjabb6gaUjabbwgaLjabbsgaKjabbccaGiabbMgaPjabb6gaUjabbccaGiabbogaJjabbggaHjabbohaZjabbwgaLjabbccaGiabb+gaVjabbAgaMjabbccaGiabbsha0jabbIgaOjabbwgaLjabbccaGiabbAgaMjabb+gaVjabbYgaSjabbYgaSjabb+gaVjabbEha3jabbMgaPjabb6gaUjabbEgaNbqaaiabb+gaVjabbkgaIjabbQgaQjabbwgaLjabbogaJjabbsha0jabbohaZjabb6caUaqaaiabbIcaOiabbMgaPjabbMcaPiabbccaGiabboeadjabb+gaVjabb2gaTjabbchaWjabbggaHjabbogaJjabbsha0jabbccaGiabbsgaKjabbMgaPjabbohaZjabbUgaRjabbccaGiabbIcaOiabbofatjabbchaWjabbIgaOjabbwgaLjabbkhaYjabbwgaLjabbMcaPaqaaiabbIcaOiabbMgaPjabbMgaPjabbMcaPiabbccaGiabbgeabjabbccaGiaaygW7cqqGKbazcqqGPbqAcqqGJbWycqqGLbqzcqqGGaaicqqGOaakcqqGdbWqcqqG1bqDcqqGIbGycqqGLbqzcqqGPaqkaeaacqqGOaakcqqGPbqAcqqGPbqAcqqGPbqAcqqGPaqkcqqGGaaicqqGubavcqqGYbGCcqqGPbqAcqqGHbqycqqGUbGBcqqGNbWzcqqG1bqDcqqGSbaBcqqGHbqycqqGYbGCcqqGGaaicqqGWbaCcqqG5bqEcqqGYbGCcqqGHbqycqqGTbqBcqqGPbqAcqqGKbazcqqGGaaicqqGOaakcqqGdbWqcqqGVbWBcqqGUbGBcqqGLbqzcqqGPaqkaeaacqqGOaakcqqGPbqAcqqG2bGDcqqGPaqkcqqGGaaicqqGobGtcqqGVbWBcqqG0baDcqqGLbqzcqqGGaaicqqGcbGqcqqGVbWBcqqGVbWBcqqGRbWAcqqGGaaicqqGOaakcqqGdbWqcqqG1bqDcqqGIbGycqqGVbWBcqqGPbqAcqqGKbazcqqGPaqkaaaa@092B@

Q.79

Examine if the following are true statements:i The cube can cast a shadow in the shape of a rectangle.iiThe cube can cast a shadow in the shape of a hexagon.

Ans

A cube can cast shadow only in the shape of a square. Therefore, any other shapes are not possible. So, (i) False (ii) False MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@D573@

Q.80 Find the area of a square park whose perimeter is 320 m.

Ans

Let the side of the square park be ‘a. Since perimeter of a square=4a So, we get 4a=320 a= 320 4 =80m Therefore, the area of the square park= a 2 = ( 80m ) 2 =6400 m 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@1E02@

Q.81 Find the breadth of a rectangular plot of land, if its area is 440 m2 and the length is 22 m. Also find its perimeter.

Ans

Let the breadth of the rectangular plot be x. Area of the rectangular plot=length×breadth SO, we get 440m 2 =22m×x x= 440 m 2 22m =20m Perimeter of the rectangular plot=2( length+breadth ) =2( 20m+22m ) =2( 42m ) =84mMathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@5ED5@

Q.82 The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area.

Ans

Let the breadth of the rectangular sheet be x. Perimeter of the rectangular sheet=2( length+breadth ) 100 cm=2( 35cm+x ) 100 2 cm=( 35 cm+x ) 50cm=35 cm+x x=50 cm35 cm =15 cm Area of rectangular sheet=length×breadth =35 cm×15 cm = 525 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@9138@

Q.83

The area of a square park is the same as of a rectangularpark. If the side of the square park is 60 m and the lengthof the rectangular park is 90 m, find the breadth of therectangular park.

Ans

Let the breadth of rectangular park be x. Since, area of a square park is the same as of a rectangular park. Area of square=Area of rectangular park ( 60m ) 2 =90m×x 3600 m 2 =90m×x x= 3600 m 2 90m =40m MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@44FC@

Q.84

A wire is in the shape of a rectangle. Its length is 40 cmand breadth is 22 cm. If the same wire is rebent in theshape of a square, what will be the measure of each side.Also find which shape encloses more area?

Ans

Perimeter of a rectangle=Perimeter of square 2( length+breadth )=4×side 2( 40m+22m )=4×side side= 124 4 =31cm So, Area of rectangle=length×breadth =40cm×22cm = 880cm 2 Area of square= ( side ) 2 = ( 31cm ) 2 =961 cm 2 Therefore, the square-shaped wire encloses more area. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@AE17@

Q.85

The perimeter of a rectangle is 130 cm. If the breadthof the rectangle is 30 cm, find its length. Also find thearea of the rectangle.

Ans

Let the length of the rectangle be x cm. Perimeter of rectangle = 2 (length+breadth) 130cm=2(30 cm+x) 130 2 cm=30 cm+x x=65 cm30 cm =35 cm Now, area of rectangle=length×breadth =30 cm×35 cm = 1050 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbYeamjabbwgaLjabbsha0jabbccaGiabbsha0jabbIgaOjabbwgaLjabbccaGiabbYgaSjabbwgaLjabb6gaUjabbEgaNjabbsha0jabbIgaOjabbccaGiabb+gaVjabbAgaMjabbccaGiabbsha0jabbIgaOjabbwgaLjabbccaGiabbkhaYjabbwgaLjabbogaJjabbsha0jabbggaHjabb6gaUjabbEgaNjabbYgaSjabbwgaLjabbccaGiabbkgaIjabbwgaLjabbccaGiabdIha4jabbccaGiabbogaJjabb2gaTjabb6caUaqaaiabbcfaqjabbwgaLjabbkhaYjabbMgaPjabb2gaTjabbwgaLjabbsha0jabbwgaLjabbkhaYjabbccaGiabb+gaVjabbAgaMjabbccaGiabbkhaYjabbwgaLjabbogaJjabbsha0jabbggaHjabb6gaUjabbEgaNjabbYgaSjabbwgaLjabbccaGiabg2da9iabbccaGiabbkdaYiabbccaGiabbIcaOiabbYgaSjabbwgaLjabb6gaUjabbEgaNjabbsha0jabbIgaOjabgUcaRiabbkgaIjabbkhaYjabbwgaLjabbggaHjabbsgaKjabbsha0jabbIgaOjabbMcaPaqaaiaaxMaacaWLjaGaaCzcaiaaysW7caaMe8UaaGjbVlaaysW7cqaIXaqmcqaIZaWmcqaIWaamcaaMe8Uaee4yamMaeeyBa0Maeyypa0JaeeOmaiJaeeikaGIaee4mamJaeeimaaJaeeiiaaIaee4yamMaeeyBa0Maey4kaSIaemiEaGNaeeykaKcabaGaaCzcaiaaxMaacaWLjaGaaGjbVlaaysW7caaMe8+aaSaaaeaacqaIXaqmcqaIZaWmcqaIWaamaeaacqaIYaGmaaGaeeiiaaIaee4yamMaeeyBa0Maeyypa0Jaee4mamJaeeimaaJaeeiiaaIaee4yamMaeeyBa0Maey4kaSIaemiEaGhabaGaaCzcaiaaxMaacaWLjaGaaCzcaiaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlabdIha4jabg2da9iabbAda2iabbwda1iabbccaGiabbogaJjabb2gaTjabgkHiTiabbodaZiabbcdaWiabbccaGiabbogaJjabb2gaTbqaaiaaxMaacaWLjaGaaCzcaiaaxMaacaWLjaGaaGjbVlaaysW7cqGH9aqpcqqGZaWmcqqG1aqncqqGGaaicqqGJbWycqqGTbqBaeaacqqGobGtcqqGVbWBcqqG3bWDcqqGSaalcqqGGaaiaeaacqqGHbqycqqGYbGCcqqGLbqzcqqGHbqycqqGGaaicqqGVbWBcqqGMbGzcqqGGaaicqqGYbGCcqqGLbqzcqqGJbWycqqG0baDcqqGHbqycqqGUbGBcqqGNbWzcqqGSbaBcqqGLbqzcqGH9aqpcqqGSbaBcqqGLbqzcqqGUbGBcqqGNbWzcqqG0baDcqqGObaAcqGHxdaTcqqGIbGycqqGYbGCcqqGLbqzcqqGHbqycqqGKbazcqqG0baDcqqGObaAaeaacaWLjaGaaCzcaiaaxMaacaaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMc8Uaeyypa0Jaee4mamJaeeimaaJaeeiiaaIaee4yamMaeeyBa0Maey41aqRaee4mamJaeeynauJaeeiiaaIaee4yamMaeeyBa0gabaGaaCzcaiaaxMaacaWLjaGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGPaVlabg2da9iabbgdaXiabbcdaWiabbwda1iabbcdaWiabbccaGiabbogaJjabb2gaTnaaCaaaleqabaGaeGOmaidaaaaaaa@601F@

Q.86 Find the area of each of the following parallelogram :

Ans

(a) Area of Parallelogram=Base×Height Height=4 cm Base=7 cm So, area of parallelogram=7×4 = 28cm 2 (b) Area of Parallelogram=Base×Height Height=3 cm Base=5 cm So, area of parallelogram=5×3 = 15 cm 2 (c) Area of Parallelogram=Base×Height Height=3.5 cm Base=2.5 cm So, area of parallelogram=2.5×3.5 =8 .75 cm 2 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(d) Area of Parallelogram=Base×Height Height=4.8 cm Base=5 cm So, area of parallelogram=5×4.8 = 24 cm 2 (e) Area of Parallelogram=Base×Height Height=4.4 cm Base=2 cm So, area of parallelogram=2×4.4 =8 .8 cm 2 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Q.87 Find the area of each of the following triangles :


Ans

(a) Area of a triangle= 1 2 ×Base×Height = 1 2 ×4×3 = 1 2 ×12=6 cm 2 (b) Area of a triangle= 1 2 ×Base×Height = 1 2 ×5×3.2 = 1 2 ×16=8 cm 2 (c) Area of a triangle= 1 2 ×Base×Height = 1 2 ×3×4 = 1 2 ×12=6 cm 2 (d) Area of a triangle= 1 2 ×Base×Height = 1 2 ×3×2 = 1 2 ×6=3 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@6A5E@

Q.88 Find the missing values :

S. NO. Base Height Area of the Parallelogram
a. 20 cm 246 cm2
b. 15 cm 154.5 cm2
c. 8.4 cm 78.72 cm2
d. 15.6 cm 16.38 cm2

Ans

Let the height be h and base be b. (a) Area of parallelogram=Base×Height So,we get 246 cm 2 =20 cm×h h= 246 cm 2 20cm =12.3cm (b) Area of parallelogram=Base×Height So,we get 154 .5 cm 2 =b×15 cm b= 154.5 cm 2 15cm =10.3cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@3796@ (c) Area of parallelogram=Base×Height So,we get 78 .72 cm 2 =b×8.4 cm b= 48.72 cm 2 8.4cm =5.8cm (d) Area of parallelogram=Base×Height So,we get 16 .38 cm 2 =15.6 cm×h h= 16.38 cm 2 15.6cm =1.05cm So, we get S.No. Base Height Area of the Parallelogram a. 20 cm 12.3cm 246 cm 2 b. 10.3cm 15 cm 154.5 cm 2 c. 5.8cm 8.4 cm 78.72 cm 2 d. 15.6 cm 1.05cm 16.38 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@BA07@

Q.89 Find the missing values :

Base Height Area of Triangle
15 cm …………….. 84 cm2
…………….. 31.4 mm 1256 mm2
22 cm …………….. 170.5 cm2

Ans

(a) Let the height be h and base be b. Area of triangle= 1 2 ×base×height 87cm 2 = 1 2 ×15cm×h h= 87 cm 2 ×2 15cm =11.6 cm (b) Let the height be h and base be b. Area of triangle= 1 2 ×base×height 1256 mm 2 = 1 2 ×b×31.4 mm b= 1256 mm 2 ×2 31.4mm =80 mm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@7243@ (c) Let the height be h and base be b. Area of triangle= 1 2 ×base×height 170 .5 cm 2 = 1 2 ×22cm×h h= 170.5 cm 2 ×2 22cm =15.5 cm So we get Base Height Area of Triangle 15 cm 11.6 cm 84 cm 2 80mm 31.4 mm 1256 mm 2 22 cm 15.5cm 170.5 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbccaGiabbIcaOiabbogaJjabbMcaPiabbccaGiabbYeamjabbwgaLjabbsha0jabbccaGiabbsha0jabbIgaOjabbwgaLjabbccaGiabbIgaOjabbwgaLjabbMgaPjabbEgaNjabbIgaOjabbsha0jabbccaGiabbkgaIjabbwgaLjabbccaGiabdIgaOjabbccaGiabbggaHjabb6gaUjabbsgaKjabbccaGiabbkgaIjabbggaHjabbohaZjabbwgaLjabbccaGiabbkgaIjabbwgaLjabbccaGiabdkgaIjabb6caUaqaaiabbgeabjabbkhaYjabbwgaLjabbggaHjabbccaGiabb+gaVjabbAgaMjabbccaGiabbsha0jabbkhaYjabbMgaPjabbggaHjabb6gaUjabbEgaNjabbYgaSjabbwgaLjabg2da9maalaaabaGaeGymaedabaGaeGOmaidaaiabgEna0kabbkgaIjabbggaHjabbohaZjabbwgaLjabgEna0kabbIgaOjabbwgaLjabbMgaPjabbEgaNjabbIgaOjabbsha0bqaaiaaxMaacaWLjaGaaGPaVlabbgdaXiabbEda3iabbcdaWiabb6caUiabbwda1iabbccaGiabbogaJjabb2gaTnaaCaaaleqabaGaeGOmaidaaOGaeyypa0ZaaSaaaeaacqaIXaqmaeaacqaIYaGmaaGaey41aqRaeGOmaiJaeGOmaiJaaGjbVlabbogaJjabb2gaTjabgEna0kabbIgaObqaaiaaxMaacaWLjaGaaCzcaiaaysW7caaMe8UaeeiAaGMaeyypa0ZaaSaaaeaacqaIXaqmcqaI3aWncqaIWaamcqGGUaGlcqaI1aqncqqGGaaicqqGJbWycqqGTbqBdaahaaWcbeqaaiabikdaYaaakiabgEna0kabikdaYaqaaiabikdaYiabikdaYiaaysW7cqqGJbWycqqGTbqBaaaabaGaaCzcaiaaxMaacaWLjaGaaGjbVlaaysW7caaMe8UaaGjbVlaaykW7cqGH9aqpcqaIXaqmcqaI1aqncqGGUaGlcqaI1aqncqqGGaaicqqGJbWycqqGTbqBaeaacqqGtbWucqqGVbWBcqqGGaaicqqG3bWDcqqGLbqzcqqGGaaicqqGNbWzcqqGLbqzcqqG0baDaeaafaqabiabduvbvfaacqqGcbGqcqqGHbqycqqGZbWCcqqGLbqzaeaacqqGibascqqGLbqzcqqGPbqAcqqGNbWzcqqGObaAcqqG0baDaeaacqqGbbqqcqqGYbGCcqqGLbqzcqqGHbqycqqGGaaicqqGVbWBcqqGMbGzcqqGGaaicqqGubavcqqGYbGCcqqGPbqAcqqGHbqycqqGUbGBcqqGNbWzcqqGSbaBcqqGLbqzaeaacqaIXaqmcqaI1aqncqqGGaaicqqGJbWycqqGTbqBaeaacqaIXaqmcqaIXaqmcqGGUaGlcqaI2aGncqqGGaaicqqGJbWycqqGTbqBaeaacqaI4aaocqaI0aancqqGGaaicqqGJbWycqqGTbqBdaahaaWcbeqaaiabikdaYaaaaOqaaiabiIda4iabicdaWiaaysW7cqqGTbqBcqqGTbqBaeaacqaIZaWmcqaIXaqmcqGGUaGlcqaI0aancqqGGaaicqqGTbqBcqqGTbqBaeaacqaIXaqmcqaIYaGmcqaI1aqncqaI2aGncaaMe8UaeeyBa0MaeeyBa02aaWbaaSqabeaacqaIYaGmaaaakeaacqaIYaGmcqaIYaGmcqqGGaaicqqGJbWycqqGTbqBaeaacqaIXaqmcqaI1aqncqGGUaGlcqaI1aqncaaMe8Uaee4yamMaeeyBa0gabaGaeGymaeJaeG4naCJaeGimaaJaeiOla4IaeGynauJaeeiiaaIaee4yamMaeeyBa02aaWbaaSqabeaacqaIYaGmaaaaaaaaaa@4B95@

Q.90

PQRS is a parallelogram.QM is the height from Q to SRand QN is the height from Q to PS. If SR =12 cm andQM = 7.6 cm.Find:a the area of the parallegram PQRSb QN, if PS = 8 cm

Ans

(a) Area of a parallelogram=Base×Height =SR×QM =7.6×12 =91.2 cm 2 (b) Area of a parallelogram=Base×Height =PS×QN =91.2 cm 2 QN×8=91.2 QN= 91.2 8 =11.4cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@246A@

Q.91

DL and BM are the heights on sides AB and AD respectivelyof parallelogram ABCD. If the area of the parallelogram is1470 cm2,AB = 35 cm and AD = 49 cm, find the length ofBMand DL.

Ans

Area of parallelogram=Base×Height =AB×DL 1470=35×DL DL= 1470 35 =42cm Also, 1470=AD×BM 1470=49×BM BM= 1470 49 =30cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@ECDA@

Q.92

â–³ABC is right angled at A .AD is perpendicular to BC.If AB=5 cm,BC=13 cm and AC=12 cm, Find the area ofâ–³ABC. Also find the length of AD.

Ans

Since, Area= 1 2 ×Base×Height = 1 2 ×5×12 = 30 cm 2 Also, area of triangle= 1 2 ×AD×BC 30= 1 2 ×AD×13 AD= 30×2 12 =4.6cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@DD81@

Q.93

â–³ABC is isosceles with AB=AC=7.5 cm and BC=9 cm. The height AD from A to BC, is 6 cm. Find the area of â–³ABC. What will be the height from C to AB i.e., CE?

Ans

Area of ΔABC= 1 2 ×Base×Height = 1 2 BC×AD= 1 2 ×9×6= 27cm 2 Also, area of ΔABC= 1 2 ×Base×Height = 1 2 AB×CE 27= 1 2 7.5×CE CE= 27×2 7.5 =7.2 cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@E877@

Q.94

Find the circumference of the circles with the followingradius: Take π=227a 14 cm b 28 mm c 21 cm

Ans

(a) Circumference of a circle is =2πr =2× 22 7 ×14 =2×22×2 =88cm (b) Circumference of a circle is =2πr =2× 22 7 ×28 =2×22×4 =176mm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@0482@ (c) Circumference of a circle is =2πr =2× 22 7 ×21 =2×22×3 =132cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A3AF@

Q.95

Find the area of the following circles, given that:a radius = 14 mm Take π=227b diameter = 49 mc radius = 5 cm

Ans

(a) Area of a circle=π r 2 = 22 7 ( 14 ) 2 =616 mm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@7CF8@ (b) Radius= Diameter 2 = 49 2 =24.5mm Area of a circle=π r 2 = 22 7 ( 24.5 ) 2 =1886.5 m 2 (c) Area of a circle=π r 2 = 22 7 ( 5 ) 2 = 550 7 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbIcaOiabbkgaIjabbMcaPaqaaiabbkfasjabbggaHjabbsgaKjabbMgaPjabbwha1jabbohaZjabg2da9maalaaabaGaeeiraqKaeeyAaKMaeeyyaeMaeeyBa0MaeeyzauMaeeiDaqNaeeyzauMaeeOCaihabaGaeGOmaidaaiabg2da9maalaaabaGaeGinaqJaeGyoaKdabaGaeGOmaidaaiabg2da9iabikdaYiabisda0iabc6caUiabiwda1iaaysW7cqqGTbqBcqqGTbqBaeaacqqGbbqqcqqGYbGCcqqGLbqzcqqGHbqycqqGGaaicqqGVbWBcqqGMbGzcqqGGaaicqqGHbqycqqGGaaicqqGJbWycqqGPbqAcqqGYbGCcqqGJbWycqqGSbaBcqqGLbqzcqGH9aqpcqaHapaCcqqGYbGCdaahaaWcbeqaaiabikdaYaaaaOqaaiaaxMaacaWLjaGaaCzcaiaaysW7caaMe8UaaGjbVlabg2da9maalaaabaGaeGOmaiJaeGOmaidabaGaeG4naCdaamaabmaabaGaeGOmaiJaeGinaqJaeiOla4IaeGynaudacaGLOaGaayzkaaWaaWbaaSqabeaacqaIYaGmaaaakeaacaWLjaGaaCzcaiaaxMaacaaMe8UaaGjbVlaaysW7cqGH9aqpcqaIXaqmcqaI4aaocqaI4aaocqaI2aGncqGGUaGlcqaI1aqncaaMe8UaeeyBa02aaWbaaSqabeaacqaIYaGmaaaakeaacqqGOaakcqqGJbWycqqGPaqkaeaacqqGbbqqcqqGYbGCcqqGLbqzcqqGHbqycqqGGaaicqqGVbWBcqqGMbGzcqqGGaaicqqGHbqycqqGGaaicqqGJbWycqqGPbqAcqqGYbGCcqqGJbWycqqGSbaBcqqGLbqzcqGH9aqpcqaHapaCcqqGYbGCdaahaaWcbeqaaiabikdaYaaaaOqaaiaaxMaacaWLjaGaaCzcaiaaysW7caaMe8UaaGjbVlabg2da9maalaaabaGaeGOmaiJaeGOmaidabaGaeG4naCdaamaabmaabaGaeGynaudacaGLOaGaayzkaaWaaWbaaSqabeaacqaIYaGmaaaakeaacaWLjaGaaCzcaiaaxMaacaaMe8UaaGjbVlaaysW7cqGH9aqpdaWcaaqaaiabiwda1iabiwda1iabicdaWaqaaiabiEda3aaacaaMe8Uaee4yamMaeeyBa02aaWbaaSqabeaacqaIYaGmaaaaaaa@DCD5@

Q.96 If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π =

227

)

Ans

Since, circumference=2πr So, we get MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbofatjabbMgaPjabb6gaUjabbogaJjabbwgaLjabbYcaSiabbccaGiabbogaJjabbMgaPjabbkhaYjabbogaJjabbwha1jabb2gaTjabbAgaMjabbwgaLjabbkhaYjabbwgaLjabb6gaUjabbogaJjabbwgaLjabg2da9iabbkdaYiabec8aWjabbkhaYbqaaiabbofatjabb+gaVjabbYcaSiabbccaGiabbEha3jabbwgaLjabbccaGiabbEgaNjabbwgaLjabbsha0baaaa@6CE5@ 154=2× 22 7 ×r r= 154×7 2×22 = 49 2 m=4.5m Now, Area=π r 2 = 22 7 × 49 2 × 49 2 =1886.5 m 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A695@

Q.97 From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)

Ans

Outer radius of circular sheet=4 cm Inner radius of circular sheet=3 cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@99EC@ Remaining area=( 3.14×3×3 ) =50.2428.26 =21.98 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@884A@

Q.98
< Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

Ans

Radius=5 cm Length of curved part=πr = 22 7 ×5 =15.71 cm Total permieter=Length of the curved part+Length of diameter =15.71+10 =25.71 cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbkfasjabbggaHjabbsgaKjabbMgaPjabbwha1jabbohaZjabg2da9iabbwda1iabbccaGiabbogaJjabb2gaTbqaaiabbYeamjabbwgaLjabb6gaUjabbEgaNjabbsha0jabbIgaOjabbccaGiabb+gaVjabbAgaMjabbccaGiabbogaJjabbwha1jabbkhaYjabbAha2jabbwgaLjabbsgaKjabbccaGiabbchaWjabbggaHjabbkhaYjabbsha0jabg2da9iabec8aWjabbkhaYbqaaiaaxMaacaWLjaGaaCzcaiaaxMaacaaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7cqGH9aqpdaWcaaqaaiabikdaYiabikdaYaqaaiabiEda3aaacqGHxdaTcqaI1aqnaeaacaWLjaGaaCzcaiaaxMaacaWLjaGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8Uaeyypa0JaeGymaeJaeGynauJaeiOla4IaeG4naCJaeGymaeJaeeiiaaIaee4yamMaeeyBa0gabaGaeeivaqLaee4Ba8MaeeiDaqNaeeyyaeMaeeiBaWMaeeiiaaIaeeiCaaNaeeyzauMaeeOCaiNaeeyBa0MaeeyAaKMaeeyzauMaeeiDaqNaeeyzauMaeeOCaiNaeyypa0JaeeitaWKaeeyzauMaeeOBa4Maee4zaCMaeeiDaqNaeeiAaGMaeeiiaaIaee4Ba8MaeeOzayMaeeiiaaIaeeiDaqNaeeiAaGMaeeyzauMaeeiiaaIaee4yamMaeeyDauNaeeOCaiNaeeODayNaeeyzauMaeeizaqMaeeiiaaIaeeiCaaNaeeyyaeMaeeOCaiNaeeiDaqNaee4kaSIaeeitaWKaeeyzauMaeeOBa4Maee4zaCMaeeiDaqNaeeiAaGMaeeiiaaIaee4Ba8MaeeOzayMaeeiiaaIaeeizaqMaeeyAaKMaeeyyaeMaeeyBa0MaeeyzauMaeeiDaqNaeeyzauMaeeOCaihabaGaaCzcaiaaxMaacaWLjaGaaGjbVlaaysW7caaMe8UaaGjbVlabg2da9iabbgdaXiabbwda1iabb6caUiabbEda3iabbgdaXiabbUcaRiabbgdaXiabbcdaWaqaaiaaxMaacaWLjaGaaCzcaiaaysW7caaMe8UaaGjbVlaaysW7cqGH9aqpcqqGYaGmcqqG1aqncqqGUaGlcqqG3aWncqqGXaqmcqqGGaaicqqGJbWycqqGTbqBaaaa@045B@

Q.99

Shazli took a wire of length 44 cm and bent it into theshape of a circle. Find the radius of that circle. Also findits area. If the same wire is bent into the shape of asquare, what will be the length of each of its sides?Which figure encloses more.

Ans

Circumference=2πr=44 cm 2× 22 7 ×r=44 r=7 cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabboeadjabbMgaPjabbkhaYjabbogaJjabbwha1jabb2gaTjabbAgaMjabbwgaLjabbkhaYjabbwgaLjabb6gaUjabbogaJjabbwgaLjabg2da9iabbkdaYiabec8aWjabbkhaYjabg2da9iabbsda0iabbsda0iabbccaGiabbogaJjabb2gaTbqaaiaaxMaacaaMe8UaaGjbVlabbkdaYiabgEna0oaalaaabaGaeGOmaiJaeGOmaidabaGaeG4naCdaaiabgEna0kabdkhaYjabg2da9iabisda0iabisda0aqaaiaaxMaacaWLjaGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaemOCaiNaeyypa0JaeG4naCJaeeiiaaIaee4yamMaeeyBa0gaaaa@8092@ Area=π r 2 = 22 7 ×7×7 =154 cm 2 If the wire is bent into a square, then the length of each side would be= 44 4 =11 cm Area of square= ( 11 ) 2 =121 cm 2 Therefore, circle encloses more area. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@1AB3@

Q.100

From a circular card sheet of radius 14 cm, two circles ofradius 3.5 cm and a rectangle of length 3 cm and breadth1 cm are removed. (as shown in the adjoiningfigure).Find the area of the remaining sheet. Takeπ=227

Ans

Area of bigger cirle= 22 7 ×14×14 =616 cm 2 Area of 2 small circles=2×π r 2 =2× 22 7 ×3.5×3.5 =77 cm 2 Area of rectangle=Length×Breadth =3×1= 3 cm 2 Remaining area of sheet= 616 cm 2 77 cm 2 3 cm 2 = 536 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbgeabjabbkhaYjabbwgaLjabbggaHjabbccaGiabb+gaVjabbAgaMjabbccaGiabbkgaIjabbMgaPjabbEgaNjabbEgaNjabbwgaLjabbkhaYjabbccaGiabbogaJjabbMgaPjabbkhaYjabbYgaSjabbwgaLjabg2da9maalaaabaGaeGOmaiJaeGOmaidabaGaeG4naCdaaiabgEna0kabigdaXiabisda0iabgEna0kabigdaXiabisda0aqaaiaaxMaacaWLjaGaaCzcaiaaxMaacaaMe8UaaGjbVlabg2da9iabiAda2iabigdaXiabiAda2iabbccaGiabbogaJjabb2gaTnaaCaaaleqabaGaeGOmaidaaaGcbaGaeeyqaeKaeeOCaiNaeeyzauMaeeyyaeMaeeiiaaIaee4Ba8MaeeOzayMaeeiiaaIaeeOmaiJaeeiiaaIaee4CamNaeeyBa0MaeeyyaeMaeeiBaWMaeeiBaWMaeeiiaaIaee4yamMaeeyAaKMaeeOCaiNaee4yamMaeeiBaWMaeeyzauMaee4CamNaeyypa0JaeeOmaiJaey41aqRaeqiWdaNaeeOCai3aaWbaaSqabeaacqaIYaGmaaaakeaacaWLjaGaaCzcaiaaxMaacaWLjaGaaCzcaiabg2da9iabikdaYiabgEna0oaalaaabaGaeGOmaiJaeGOmaidabaGaeG4naCdaaiabgEna0kabiodaZiabc6caUiabiwda1iabgEna0kabiodaZiabc6caUiabiwda1aqaaiaaxMaacaWLjaGaaCzcaiaaxMaacaWLjaGaeyypa0JaeG4naCJaeG4naCJaeeiiaaIaee4yamMaeeyBa02aaWbaaSqabeaacqaIYaGmaaaakeaacqqGbbqqcqqGYbGCcqqGLbqzcqqGHbqycqqGGaaicqqGVbWBcqqGMbGzcqqGGaaicqqGYbGCcqqGLbqzcqqGJbWycqqG0baDcqqGHbqycqqGUbGBcqqGNbWzcqqGSbaBcqqGLbqzcqGH9aqpcqqGmbatcqqGLbqzcqqGUbGBcqqGNbWzcqqG0baDcqqGObaAcqGHxdaTcqqGcbGqcqqGYbGCcqqGLbqzcqqGHbqycqqGKbazcqqG0baDcqqGObaAaeaacaWLjaGaaCzcaiaaxMaacaaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMc8Uaeyypa0Jaee4mamJaey41aqRaeeymaeJaeyypa0Jaee4mamJaeeiiaaIaee4yamMaeeyBa02aaWbaaSqabeaacqaIYaGmaaaakeaacqqGsbGucqqGLbqzcqqGTbqBcqqGHbqycqqGPbqAcqqGUbGBcqqGPbqAcqqGUbGBcqqGNbWzcqqGGaaicqqGHbqycqqGYbGCcqqGLbqzcqqGHbqycqqGGaaicqqGVbWBcqqGMbGzcqqGGaaicqqGZbWCcqqGObaAcqqGLbqzcqqGLbqzcqqG0baDcqGH9aqpcqqG2aGncqqGXaqmcqqG2aGncqqGGaaicqqGJbWycqqGTbqBdaahaaWcbeqaaiabikdaYaaakiabgkHiTiabbEda3iabbEda3iabbccaGiabbogaJjabb2gaTnaaCaaaleqabaGaeGOmaidaaOGaeyOeI0Iaee4mamJaeeiiaaIaee4yamMaeeyBa02aaWbaaSqabeaacqaIYaGmaaaakeaacaWLjaGaaCzcaiaaxMaacaWLjaGaaCzcaiaaysW7caaMe8UaaGjbVlaaysW7cqGH9aqpcqqG1aqncqqGZaWmcqqG2aGncqqGGaaicqqGJbWycqqGTbqBdaahaaWcbeqaaiabikdaYaaaaaaa@3FAA@

Q.101

A circle of radius 2 cm is cut out from a square piece of analuminium sheet of side 6 cm. What is the area of the leftover aluminium sheet?Take π= 3.14

Ans

Area of square-shaped sheet= ( side ) 2 = ( 6 cm 2 ) 2 =36 cm 2 Area of circle=3.14×2×2 =12.56 cm 2 Remaining area= 36 cm 2 12 .56 cm 2 =23 .44 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@0452@

Q.102 The circumference of a circle is 31.4 cm. Find the radius and the area of the circle ? (Take π = 3.14)

Ans

Circumference=2πr=31.4cm 2×3.14×r=31.4 r= 31.4 2×3.14 =5 cm So, Area=3.14×5×5 =78 .50 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@AB49@

Q.103 A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (π = 3.14)

Ans

Radius of the flower bed= 66 2 =33m So, radius of flower bed and path together=33+4 =37m Area of flower baed and path together=3.14×37×37 =4298.66 m 2 Area of flower bed =3.14×33×33 =3419 .46 m 2 Area of path =4298 .66 m 2 3419 .46m 2 =879 .20m 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@6874@

Q.104

A circular flower garden has an area of 314 m2. A sprinklerat the centre of the garden can cover an area that has aradius of 12 m. Will the sprinkler water the entire garden?Takeπ= 3.14

Ans

Area=π r 2 =314 m 2 3.14× r 2 =314 r 2 =100 r=10 m Yes, the sprinkle will water the whole garden. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@B781@

Q.105 Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)

Ans

Radius of outer circle=19 m Circumference=2πr =2×3.14×19 =119.32 m Radius of the inner circle=1910=9 m Circumference=2πr =2×3.14×9 =56.52 m MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbkfasjabbggaHjabbsgaKjabbMgaPjabbwha1jabbohaZjabbccaGiabb+gaVjabbAgaMjabbccaGiabb+gaVjabbwha1jabbsha0jabbwgaLjabbkhaYjabbccaGiabbogaJjabbMgaPjabbkhaYjabbogaJjabbYgaSjabbwgaLjabg2da9iabbgdaXiabbMda5iabbccaGiabb2gaTbqaaiaaxMaacaaMe8UaaGjbVlaaysW7cqqGdbWqcqqGPbqAcqqGYbGCcqqGJbWycqqG1bqDcqqGTbqBcqqGMbGzcqqGLbqzcqqGYbGCcqqGLbqzcqqGUbGBcqqGJbWycqqGLbqzcqGH9aqpcqqGYaGmcqaHapaCcqqGYbGCaeaacaWLjaGaaCzcaiaaxMaacaWLjaGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMc8Uaeyypa0JaeeOmaiJaey41aqRaee4mamJaeeOla4IaeeymaeJaeeinaqJaey41aqRaeeymaeJaeeyoaKdabaGaaCzcaiaaxMaacaWLjaGaaCzcaiaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGPaVlabg2da9iabbgdaXiabbgdaXiabbMda5iabb6caUiabbodaZiabbkdaYiabbccaGiabb2gaTbqaaiabbkfasjabbggaHjabbsgaKjabbMgaPjabbwha1jabbohaZjabbccaGiabb+gaVjabbAgaMjabbccaGiabbsha0jabbIgaOjabbwgaLjabbccaGiabbMgaPjabb6gaUjabb6gaUjabbwgaLjabbkhaYjabbccaGiabbogaJjabbMgaPjabbkhaYjabbogaJjabbYgaSjabbwgaLjabg2da9iabbgdaXiabbMda5iabgkHiTiabbgdaXiabbcdaWiabg2da9iabbMda5iabbccaGiabb2gaTbqaaiaaxMaacaWLjaGaaGjbVlaaysW7cqqGdbWqcqqGPbqAcqqGYbGCcqqGJbWycqqG1bqDcqqGTbqBcqqGMbGzcqqGLbqzcqqGYbGCcqqGLbqzcqqGUbGBcqqGJbWycqqGLbqzcqGH9aqpcqqGYaGmcqaHapaCcqqGYbGCaeaacaWLjaGaaCzcaiaaxMaacaWLjaGaaCzcaiaaysW7caaMe8UaaGjbVlaaysW7caaMe8Uaeyypa0JaeeOmaiJaey41aqRaee4mamJaeeOla4IaeeymaeJaeeinaqJaey41aqRaeeyoaKdabaGaaCzcaiaaxMaacaWLjaGaaCzcaiaaxMaacaaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlabg2da9iabbwda1iabbAda2iabb6caUiabbwda1iabbkdaYiabbccaGiabb2gaTbaaaa@18E1@

Q.106 How many times a wheel of radius 28 cm must rotate to go 352 m? (take π =

227

)

Ans

Radius, r=28 cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGsbGucqqGHbqycqqGKbazcqqGPbqAcqqG1bqDcqqGZbWCcqqGSaalcqqGGaaicqqGYbGCcqGH9aqpcqqGYaGmcqqG4aaocqqGGaaicqqGJbWycqqGTbqBaaa@5395@ Circumference=2πr =2× 22 7 ×28 =176 cm Number of rotatinos= Total distance to be covered Circumference of the wheel = 352 m 176cm = 35200 176 =200 Therefore, it will rotate 200 times. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@36A6@

Q.107 The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour. (Take π = 3.14)

Ans

Since, Distance travelled by the tip of minute hand =Circumference of the clock =2πr =2×3.14×15 =94.2 cm Therfore, minute hand move 94.2 cm in 1 hour. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@F536@

Q.108 A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectare.

Ans

Length (L) of garden=90 m Breadth (B) of garden=75 m Area of garden=L×B =90 m×75 m = 6750 m 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@C591@ From the figure, it can be observed that the new length and breadth of the garden,when path is also included are 100 m and 85 m respectively Area of the garden including path=100 m ×80 m = 8000 m 2 Area of the path =Area of the garden including pathArea of garden =8000 m 2 6750 m 2 =1750 m 2 1 hectare= 10000 m 2 Therefore, area of the garden in hectare= 6750 10000 =0.675 hectare MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@079B@

Q.109 A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65m. Find the area of the path.

Ans

Length (L) of park=125 m Breadth (B) of park=65 m Area of park=L×B =125 m×65 m = 8125 m 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@BF97@ From the figure, it can be observed that the new length and breadth of the park,when path is also included are 131 m and 71 m respectively Area of the park including path=131 m ×71 m = 9301 m 2 Area of the path =Area of the park including pathArea of park =9301 m 2 8125 m 2 =1176 m 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@8DBC@

Q.110 A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.

Ans

Length (L) of cardboard=125 m Breadth (B) of cardboard=65 m Area of cardboard including margin=L×B =8 cm×5 cm =40 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@E873@ From the figure, it can be observed that the new length and breadth of the cardboard,when margin is not included are 5 cm and 2 cm respectively Area of the cardboard not including margin=5 cm ×2 m = 10 cm 2 Area of the margin =Area of the cardboard including margin Area of cardboard not including margin =40 cm 2 10 cm 2 =30 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@D639@

Q.111

Two cross roads, each of width 10 m, cut at right anglesthrough the centre of a rectangular park of length 700 mand breadth 300 m and parallel to its sides. Find the areaof the roads. Also find the area of the park excluding crossroads. Give the answer in hectares.

Ans

Length (L) of the park=700 m Breadth (B) of the park=300 m Area of park=700×300 = 210000 m 2 Length of road PQRS=700 m Length of road ABCD=300 m Width of each road=10 m Area of the roads=area(PQRS)+area(ABCD)area(KLMN) =( 700×10 )+( 300×10 )( 10×10 ) =7000+3000100 =9900 m 2 =0.99 hectare MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@AE64@ Area of park excluding roads=210000-9900 = 200100 m 2 =20.01 hectare MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@9E5E@

Q.112 Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any left? (π = 3.14)

Ans

Perimeter of the square=4×side of the square =4×4 =16 cm Perimeter of circular pip=2πr =3×3.14×4 =25.12 cm Therefore, length of chord left with Pragya=25.12 cm16 cm =9.12 cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@2F91@

Q.113 The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find:
(i) the area of the whole land
(ii) the area of the flower bed
(iii) the area of the lawn excluding the area of the flower bed.

(iv) the circumference of the flower bed.

Ans

(i) Area of whole land=Length×Breadth =10×5 = 50 m 2 (ii) Area of flower bed=π r 2 =3.14×2×2 =12.56 m 2 (iii) Area of the lawn excluding the flower bed =Area of whole landArea of flower bed =50-12.56 =37 .44 m 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@5A62@

Q.114 In the following figures, find the area of the shaded portions:

Ans

(i) Area of EFDC=area(ABCD)area(BCE)area(AFE) =( 18×10 ) 1 2 ( 10×8 ) 1 2 ( 6×10 ) =1804030 =110 cm 2 (ii) area(QTU)=area(PQRS)area(TSU)area(RUQ)area(PQT) =( 20×20 ) 1 2 ( 10×10 ) 1 2 ( 20×10 ) 1 2 ( 20×10 ) =40050100100 =150 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@3542@

Q.115

Find the area of the quadrilateral ABCD.Here, AC=22 cm, BM=3 cm,DN=3 cm, and BMAC, DNAC.

Ans

area(ABCD)=area(ABC)+area(ADC) = 1 2 ( 3×22 ) 1 2 ( 3×22 ) =33+33 =66 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@9DB2@

Q.116

Get the algebraic expressions in the following cases usingvariables, constants and arithmetic operations.i Subtraction of z from y.ii Onehalf of the sum of numbers x and y.iii The number z multiplied by itself.iv Onefourth of the product of numbers p and q.v Numbersxandyboth squared and added.vi Number 5 added to three times the product ofnumbers m and n.vii Product of numbers y and z subtracted from 10.viii Sum of numbers a and b subtracted from theirproduct.

Ans

(i) yz (ii) 1 2 ( x+y ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbIcaOiabbMgaPjabbMcaPiabbccaGiabbMha5jabgkHiTiabbQha6bqaaiabbIcaOiabbMgaPjabbMgaPjabbMcaPiabbccaGmaalaaabaGaeGymaedabaGaeGOmaidaamaabmaabaGaemiEaGNaey4kaSIaemyEaKhacaGLOaGaayzkaaaaaaa@565F@ (iii) z 2 (iv) 1 4 ( pq ) (v) x 2 + y 2 (vi) 5+3( mn ) (vii) 10yz (viii) ab( a+b ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@8A1A@

Q.117

(i) Identify the terms and their factors in the followingexpressions.Show the terms and factors by tree diagrams.(a) x3 (b)1+x+x2 (c)yy3(d) 5xy2+7x2y (e)ab+2b23a2(ii) Identify terms and factors in the expressions givenbelow:(a) 4x+5 (b) 4x+5y (c) 5y+3y2(d) xy+2x2y2 (e) pq+q (f) 1.2ab2.4b+3.6a(g) 34x+14 (h) 0.1p2+0.2q2

Ans

(i)

a.

b.

c.

d.

e.

( ii ) Row Expression Terms Factors ( a ) 4x+5 4x 5 4,x 5 ( b ) 4x+5y 4x 5y 4,x 5,y ( c ) 5y+3 y 2 5y 3 y 2 5,y 3,y,y ( d ) xy+2 x 2 y 2 xy 2 x 2 y 2 x,y 2,x,x,y,y ( e ) pq+q pq q p,q q ( f ) 1.2ab2.4b+3.6a 1.2ab 2.4b 3.6a 1.2,a,b 2.4,b 3.6,a ( g ) 3 4 x+ 1 4 3 4 x 1 4 3 4 ,x 1 4 ( h ) 0.1 p 2 +0.2 q 2 0.1 p 2 0.2 q 2 0.1,p,p 0.2,q,q MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@5021@

Q.118

Identify the numerical coefficients of terms(other than constants) in the following expressions:(i) 53t2 (ii) 1+t+t2+t3 (iii) x+2xy+3y(iv) 100m+1000n (v) p2q2+7pq (vi) 1.2a+0.8b(vii) 3.14r2 (viii) 2(l+b) (ix) 0.1y+0.01y2

Ans

Row Expression Terms Coefficients ( i ) 53 t 2 3 t 2 3 ( ii ) 1+t+ t 2 + t 3 t t 2 t 3 1 1 1 ( iii ) x+2xy+3y x 2xy 3y 1 2 3 ( iv ) 100m+100n 100m 100n 100 100 ( v ) p 2 q 2 +7pq p 2 q 2 7pq 1 7 ( vi ) 1.2a+0.8b 1.2a 0.8b 1.2 0.8 ( vii ) 3.14 r 2 3.14 r 2 3.14 ( viii ) 2( l+b ) 2l 2b 2 2 ( ix ) 0.1y+0.01 y 2 0.1y 0.01 y 2 0.1 0.01 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@484B@

Q.119

Identify the numerical coefficients of terms(other than constants) in the following expressions:(i) 53t2 (ii) 1+t+t2+t3 (iii) x+2xy+3y(iv) 100m+1000n (v)p2q2+7pq (vi)1.2a+0.8b(vii) 3.14r2 (viii) 2(l+b) (ix) 0.1y+0.01y2

Ans

Row Expression Terms Coefficients ( i ) 53 t 2 3 t 2 3 ( ii ) 1+t+ t 2 + t 3 t t 2 t 3 1 1 1 ( iii ) x+2xy+3y x 2xy 3y 1 2 3 ( iv ) 100m+100n 100m 100n 100 100 ( v ) p 2 q 2 +7pq p 2 q 2 7pq 1 7 ( vi ) 1.2a+0.8b 1.2a 0.8b 1.2 0.8 ( vii ) 3.14 r 2 3.14 r 2 3.14 ( viii ) 2( l+b ) 2l 2b 2 2 ( ix ) 0.1y+0.01 y 2 0.1y 0.01 y 2 0.1 0.01 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@484B@

Q.120

(a) Identify terms which contain x and give thecoefficient of x.(i) y2x+y (ii) 13y28yx (iii)x+y+2(iv) 5+z+zx (v)1+x+xy (vi)12xy2+25(vii) 7x+xy2(b)Identify terms which contain y2 and give thecoefficient of y2.(i) 8xy2 (ii) 5y2+7x (iii) 2x2y15xy2+7y2

Ans

( a ) Row Expression Terms with x Cofficient of x ( i ) y 2 x+y y 2 x y 2 ( ii ) 13 y 2 8yx 8y 8 ( iii ) x+y+2 x 1 ( iv ) 5+z+zx zx z ( v ) 1+x+xy x xy 1 y ( vi ) 12x y 2 +25 12x y 2 12 y 2 ( vii ) 7+x y 2 x y 2 y 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@F5FA@ ( b ) Row Expression Terms with y 2 Cofficient of y 2 ( i ) 8x y 2 x y 2 x ( ii ) 5 y 2 +7x 5 y 2 5 ( iii ) 2 x 2 y+7 y 2 15x y 2 7 y 2 15x y 2 7 15x MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaamaabmaabaGaeeOyaigacaGLOaGaayzkaaaabaqbaeqacqabvvvvbeaacqqGsbGucqqGVbWBcqqG3bWDaeaacqqGfbqrcqqG4baEcqqGWbaCcqqGYbGCcqqGLbqzcqqGZbWCcqqGZbWCcqqGPbqAcqqGVbWBcqqGUbGBaeaacqqGubavcqqGLbqzcqqGYbGCcqqGTbqBcqqGZbWCcqqGGaaicqqG3bWDcqqGPbqAcqqG0baDcqqGObaAcqqGGaaicqqG5bqEdaahaaWcbeqaaiabikdaYaaaaOqaaiabboeadjabb+gaVjabbAgaMjabbAgaMjabbMgaPjabbogaJjabbMgaPjabbwgaLjabb6gaUjabbsha0jabbccaGiabb+gaVjabbAgaMjabbccaGiabbMha5naaCaaaleqabaGaeGOmaidaaaGcbaWaiaiMbmaabGaGykadaIzGPbqAaiacaIPLOaGaiaiMwMcaaaqaaiabiIda4iabgkHiTiabdIha4jabdMha5naaCaaaleqabaGaeGOmaidaaaGcbaGaeyOeI0IaemiEaGNaemyEaK3aaWbaaSqabeaacqaIYaGmaaaakeaacqGHsislcqWG4baEaeaadaqadaqaaiabbMgaPjabbMgaPbGaayjkaiaawMcaaaqaaiabiwda1iabdMha5naaCaaaleqabaGaeGOmaidaaOGaey4kaSIaeG4naCJaemiEaGhabaGaeGynauJaemyEaK3aaWbaaSqabeaacqaIYaGmaaaakeaacqaI1aqnaeaadaqadaqaaiabbMgaPjabbMgaPjabbMgaPbGaayjkaiaawMcaaaqaaiabikdaYiabdIha4naaCaaaleqabaGaeGOmaidaaOGaemyEaKNaey4kaSIaeG4naCJaemyEaK3aaWbaaSqabeaacqaIYaGmaaGccqGHsislcqaIXaqmcqaI1aqncqWG4baEcqWG5bqEdaahaaWcbeqaaiabikdaYaaaaOabaeqabaGaeG4naCJaemyEaK3aaWbaaSqabeaacqaIYaGmaaaakeaacqGHsislcqaIXaqmcqaI1aqncqWG4baEcqWG5bqEdaahaaWcbeqaaiabikdaYaaaaaGceaqabeaacqaI3aWnaeaacqGHsislcqaIXaqmcqaI1aqncqWG4baEaaaaaaaa@C830@

Q.121

Classify into monomials, binomials and trinomials.(i) 4y7