NCERT Solutions Class 7 Maths Chapter 6

 NCERT Solutions for Class 7 Mathematics Chapter 6 The Triangle and its Properties

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties

You may have learnt about triangles in your earlier classes while learning about the different types of geometric shapes. In Chapter 6 Maths Class 7 you’ll be able to develop a better understanding of its core concept:  A triangle, its- types,  angles , sides and its properties in more detail. NCERT Solutions of Class 7th Mathematics Chapter 6 offers you solutions to the textbook questions that will act as a reference guide while you solve the exercise questions of the chapters.

NCERT Solutions for Class 7 Mathematics Chapter 6 The Triangle and Its Properties – 

(include NCERT Solutions for Class 7 Mathematics Chapter 6 The Triangle and Its Properties)

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Access NCERT Solutions for Class 7 Mathematics Chapter 6– The Triangle and its Properties

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NCERT Solutions for Class 7 Mathematics Chapter 6 

Go through the following concepts before referring to the solutions of NCERT Class 7 Mathematics Chapter 6. It will help you in understanding the solutions better: 

  1. A median of a triangle is a line segment drawn from a vertex of a triangle to the midpoint of the opposite side. A triangle can, thus, have three medians in all.
  2. An altitude of a triangle is a perpendicular line drawn from a vertex of a triangle to the opposite side. In simpler words, a triangle can have three altitudes in total, just like the median.
  3. The sum of the interior angles of a triangle is always equal to its opposite exterior angle.  
  4. Rule of angle sum property of a triangle: The sum of all the angles of a triangle is always equal to 180°.

NCERT Solutions for Class 7 Mathematics Chapter 6

Chapter 6 of Mathematics NCERT Class 7 covers the following topics:

UNIT TOPIC
6.1 Introduction
6.2 Medians of a Triangle
6.3 Altitudes of a Triangle 
6.4 Exterior Angle of a Triangle and Its Properties
6.5 Angle Sum Property of a Triangle
6.6 Two Special Triangles – Equilateral and Isosceles 
6.7 Sum of the Lengths of Two Sides of a Triangle
6.8 Right Angle Triangle and Pythagoras Property 

6.1 Introduction

A triangle is a closed curve formed by the intersection of three line segments. The line segments are called sides, and the points of intersections are called the vertices while the angles formed at the vertices are known as the angles of the triangle. So, in a triangle, we can say that there are three sides, vertices and angles. . 

Based on sides, the classification of triangles is of the following types: 

  1.  Equilateral triangle- A triangle in which all three sides are equal. 
  2. Isosceles triangle- A triangle in which only  two sides are equal.
  3. Scalene triangle- A triangle in which no two sides are equal.

Based on the angles, the classification of triangles is of the following types: 

  1. Acute angled triangle- Any triangle in which all three interior angles of the triangle are less than 90°. 
  2. Obtuse angled triangle- Any triangle with one of the interior angles of the triangle is more than 90°. 
  3. Right-angled triangle- A triangle in which one of the interior angles of the triangle is 90°. 

6.2 Medians of Triangle

This section deals with a line segment joining a vertex of a triangle to the midpoint of its opposite side, which is called a median. It bisects the opposite side of the triangle into two equal parts and further divides the triangle into two, the two of them having the same area.

6.3 Altitudes of a Triangle

This section deals with an altitude of a triangle which is a perpendicular line segment drawn from the vertex of the triangle to its opposite side. When meeting the opposite side, it makes an angle of 90°. A triangle has three altitudes and the point of intersection of all the three is known as the orthocenter.

6.4 Exterior Angle of a Triangle and its Property

 According to this law the exterior angle of a triangle is always equal to the sum of its opposite interior angles as this is a property of every exterior angle concerning a triangle. 

6.5 Angle Sum Property of a Triangle

In this section, students need to learn yet another important property of a triangle called the angle sum property. It states that the sum of all three angles is always equal to 180°. To prove this, the students can refer to the NCERT Solutions Class 7 Mathematics Chapter 6 and understand how it  can  be proved by using the exterior angle property rule. 

6.6 Two Special Triangles- Equilateral and Isosceles

This section helps the students get a detailed understanding of the different types of triangles based on sides, with a prime focus on Equilateral (all sides equal implying all angles of the triangle are also equal) and Isosceles triangle (any two sides equal implying two angles of the triangle also equal).

6.7 Sum of the Lengths of Two Sides of Triangle

This section explains  how the length of any two sides of  a triangle is always greater  than the length of its third side. To understand the concept clearer, students can refer to the NCERT Solutions Class 7 Mathematics Chapter 6 where this rule is proven and solved through many illustrations. 

6.8 Right Angle Triangle and Pythagoras Property

A right-angle triangle is the one in which at least one angle of the triangle is 90°. Such triangles have a special property which is called the Pythagoras property or Pythagoras theorem. As per this property, the area of a triangle whose one side is a slanting line, called hypotenuse (it is the side opposite to the right angle) is always equal to the squares of the sum of the other two sides. From the solutions, students will be able to get a better understanding of this property. 

Key Features of NCERT Solutions for Class 7 Mathematics Chapter 6

The key points regarding the benefits that the students will receive through the solutions provided on this official website of Extramarks are:

  • Our NCERT Solutions will provide you with a better understanding of all the topics in this chapter. 
  • It will enrich the young learners  as the solutions are easily available  and can be accessed anywhere, anytime as per your schedule and convenience. 
  • It will explain everything in a simple language,  so that the maximum number of students can benefit from it.  

NCERT Solutions for Class 7 Mathematics

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NCERT Solutions for Class 7

 The NCERT Solutions helps students in developing a better understanding of the complicated topics . It provides you with in-depth comprehension of the course and encourages you to develop deductive reasoning skills to solve the questions. It encourages the students to master the topic and increase their confidence in achieving a higher grade.

The NCERT Solutions of Extramarks for Class 7 cover all the subjects and concepts in a detailed manner. Students are assisted with  up-to-date solutions and studying Mathematics  couldn’t have  been easier.  So in case you are interested in looking for the solutions for all the subjects of Class 7, you can click on the link given below. 

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Q.1

In Δ PQR, Disthemidpoint of QR.¯

PM ¯ is _________________. PD ¯ is _________________. Is QM=MR? MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqdaaqaaiaabcfacaqGnbaaaiaabcca caqGPbGaae4CaiaabccacaGGFbGaai4xaiaac+facaGGFbGaai4xai aac+facaGGFbGaai4xaiaac+facaGGFbGaai4xaiaac+facaGGFbGa ai4xaiaac+facaGGFbGaai4xaiaac6caaeaadaqdaaqaaiaabcfaca qGebaaaiaabccacaqGPbGaae4CaiaabccacaGGFbGaai4xaiaac+fa caGGFbGaai4xaiaac+facaGGFbGaai4xaiaac+facaGGFbGaai4xai aac+facaGGFbGaai4xaiaac+facaGGFbGaai4xaiaac6caaeaacaqG jbGaae4CaiaabccacaqGrbGaaeytaiabg2da9iaab2eacaqGsbGaai 4paaaaaa@6DF7@

Ans

PM ¯ is Altitude _ . PD ¯ is Median _ . No. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqdaaqaaiaabcfacaqGnbaaaiaabcca caqGPbGaae4CaiaabccadaqjEaqaamaamaaabaGaaeyqaiaabYgaca qG0bGaaeyAaiaabshacaqG1bGaaeizaiaabwgaaaaaaiaac6caaeaa daqdaaqaaiaabcfacaqGebaaaiaabccacaqGPbGaae4Caiaabccada qjEaqaamaamaaabaGaaeytaiaabwgacaqGKbGaaeyAaiaabggacaqG UbaaaaaacaGGUaaabaGaaeOtaiaab+gacaqGUaaaaaa@5830@

Q.2

Draw rough sketches for the following:a In ΔABC, BE is a median.b In Δ PQR, PQ and PR are a ltitudes of the triangle.cIn Δ XYZ, YL is an altitude in the exterior of the triangle.

Ans

(a)

(b)

(c)

For ΔXYZ, YL is an altitude drawn exterior to side XZ which is extended up to point L.

Q.3

Verify by drawing a diagram if the median and altitude of anisosceles triangle can be same.

Ans

Draw a line segment AD perpendicular to BC. It is an altitude for the triangle.
The length of BD and DC are same
.

Therefore, AD is also the median of this triangle
.

Q.4

Find the value of the unknown exterior angle x in thefollowing diagrams:

Ans

(i) x=70°+50°( exterior angle theorem ) = 120° (ii) x=65°+45°( exterior angle theorem ) = 110° (iii) x=40°+30°( exterior angle theorem ) = 70° (iv) x=60°+60°( exterior angle theorem ) = 120° (v) x=50°+50°( exterior angle theorem ) = 100° (vi) x=60°+30°( exterior angle theorem ) = 90°

Q.5

Find the value of the unknown x in the following diagrams:

Ans

Since, the sum of all interior angles of a triangle is 180°. So, (i) x+50°+60°=180° x+110°=180° x=180°110°= 70° (ii) x+90°+30°=180° x+120°=180° x=180°120°= 60° (iii)x+110°+30°=180° x+140°=180° x=180°140°= 40° (iv)x+x+50°=180° 2x+50°=180° 2x=180°50° =130° x= 130° 2 = 65° (v)x+x+x=180° 3x=180° x= 180° 3 = 60° (vi)x+2x+90°=180° 3x+90°=180° x=180°90° 3x=90° x= 90° 3 = 30° MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGtbGaaeyAaiaab6gacaqGJbGaaeyz aiaabYcacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabohacaqG1b GaaeyBaiaabccacaqGVbGaaeOzaiaabccacaqGHbGaaeiBaiaabYga caqGGaGaaeyAaiaab6gacaqG0bGaaeyzaiaabkhacaqGPbGaae4Bai aabkhacaqGGaGaaeyyaiaab6gacaqGNbGaaeiBaiaabwgacaqGZbGa aeiiaiaab+gacaqGMbGaaeiiaiaabggacaqGGaGaaeiDaiaabkhaca qGPbGaaeyyaiaab6gacaqGNbGaaeiBaiaabwgacaqGGaGaaeyAaiaa bohacaqGGaGaaeymaiaabIdacaqGWaGaeyiSaaRaaeOlaiaabccaae aacaqGtbGaae4BaiaabYcaaeaacaqGOaGaaeyAaiaabMcacaqGGaGa amiEaiabgUcaRiaaiwdacaaIWaGaeyiSaaRaey4kaSIaaGOnaiaaic dacqGHWcaScqGH9aqpcaaIXaGaaGioaiaaicdacqGHWcaSaeaacaWG 4bGaey4kaSIaaGymaiaaigdacaaIWaGaeyiSaaRaeyypa0JaaGymai aaiIdacaaIWaGaeyiSaalabaGaamiEaiabg2da9iaaigdacaaI4aGa aGimaiabgclaWkabgkHiTiaaigdacaaIXaGaaGimaiabgclaWkabg2 da9maaL4babaGaaG4naiaaicdacqGHWcaSaaaabaGaaeikaiaabMga caqGPbGaaeykaiaabccacaWG4bGaey4kaSIaaGyoaiaaicdacqGHWc aScqGHRaWkcaaIZaGaaGimaiabgclaWkabg2da9iaaigdacaaI4aGa aGimaiabgclaWcqaaiaadIhacqGHRaWkcaaIXaGaaGOmaiaaicdacq GHWcaScqGH9aqpcaaIXaGaaGioaiaaicdacqGHWcaSaeaacaWG4bGa eyypa0JaaGymaiaaiIdacaaIWaGaeyiSaaRaeyOeI0IaaGymaiaaik dacaaIWaGaeyiSaaRaeyypa0ZaauIhaeaacaaI2aGaaGimaiabgcla WcaaaeaacaGGOaGaamyAaiaadMgacaWGPbGaaiykaiaaysW7caWG4b Gaey4kaSIaaGymaiaaigdacaaIWaGaeyiSaaRaey4kaSIaaG4maiaa icdacqGHWcaScqGH9aqpcaaIXaGaaGioaiaaicdacqGHWcaSaeaaca WG4bGaey4kaSIaaGymaiaaisdacaaIWaGaeyiSaaRaeyypa0JaaGym aiaaiIdacaaIWaGaeyiSaalabaGaamiEaiabg2da9iaaigdacaaI4a GaaGimaiabgclaWkabgkHiTiaaigdacaaI0aGaaGimaiabgclaWkab g2da9maaL4babaGaaGinaiaaicdacqGHWcaSaaaabaGaaiikaiaadM gacaWG2bGaaiykaiaaysW7caWG4bGaey4kaSIaamiEaiabgUcaRiaa iwdacaaIWaGaeyiSaaRaeyypa0JaaGymaiaaiIdacaaIWaGaeyiSaa labaGaaGOmaiaadIhacqGHRaWkcaaI1aGaaGimaiabgclaWkabg2da 9iaaigdacaaI4aGaaGimaiabgclaWcqaaiaaikdacaWG4bGaeyypa0 JaaGymaiaaiIdacaaIWaGaeyiSaaRaeyOeI0IaaGynaiaaicdacqGH WcaSaeaacqGH9aqpcaaIXaGaaG4maiaaicdacqGHWcaSaeaacaWG4b Gaeyypa0ZaaSaaaeaacaaIXaGaaG4maiaaicdacqGHWcaSaeaacaaI Yaaaaiabg2da9maaL4babaGaaGOnaiaaiwdacqGHWcaSaaaabaGaai ikaiaadAhacaGGPaGaaGjbVlaadIhacqGHRaWkcaWG4bGaey4kaSIa amiEaiabg2da9iaaigdacaaI4aGaaGimaiabgclaWcqaaiaaiodaca WG4bGaeyypa0JaaGymaiaaiIdacaaIWaGaeyiSaalabaGaamiEaiab g2da9maalaaabaGaaGymaiaaiIdacaaIWaGaeyiSaalabaGaaG4maa aacqGH9aqpdaqjEaqaaiaaiAdacaaIWaGaeyiSaalaaaqaaiaacIca caWG2bGaamyAaiaacMcacaaMe8UaamiEaiabgUcaRiaaikdacaWG4b Gaey4kaSIaaGyoaiaaicdacqGHWcaScqGH9aqpcaaIXaGaaGioaiaa icdacqGHWcaSaeaacaaIZaGaamiEaiabgUcaRiaaiMdacaaIWaGaey iSaaRaeyypa0JaaGymaiaaiIdacaaIWaGaeyiSaalabaGaamiEaiab g2da9iaaigdacaaI4aGaaGimaiabgclaWkabgkHiTiaaiMdacaaIWa GaeyiSaalabaGaaG4maiaadIhacqGH9aqpcaaI5aGaaGimaiabgcla WcqaaiaadIhacqGH9aqpdaWcaaqaaiaaiMdacaaIWaGaeyiSaalaba GaaG4maaaacqGH9aqpdaqjEaqaaiaaiodacaaIWaGaeyiSaalaaaaa aa@9045@

Q.6

Find the values of the unknowns x and y in the followingdiagrams:

Ans

(i)y+120°=180° (Linear pair) y=180°120°=60° Now, x+y+50°=180° (angle sum property) x+60°+50°=180° x=180°110°= 70° (ii) y=80° (vertically opposite angle) y+x+50°=180° 80°+x+50°=180° x+130°=180° x=180°130°= 50° (iii) y+50°+60°=180° (angle sum property) y+110°=180° y=180°110°= 70° (iv) x=60° (vertically opposite angle) 30°+x+y=180° 30°+60°+y=180° y=180°90°= 90° (v) y=90° (vertically opposite angle) x+x+y=180° 2x+90°=180° 2x=180°90°=90° 2x=90° x= 90° 2 = 45° (vi) y=x (vertically opposite angle) a=x (vertically opposite angle) b=x (vertically opposite angle) a+b+y=180° x+x+x=180° 3x=180° x= 180° 3 =60° So, we get x=y=60° MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGOaGaaeyAaiaabMcacaaMe8UaaeyE 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Q.7

Is it possible to have a triangle with the following sides?i 2cm,3cm,5cmii 3cm,6cm,7cmiii 6cm,3cm,2cm

Ans

For a triangle, the sum of lengths of either two sides is always greater than the third side. (i) Given sides of the triangle are 2 cm, 3 cm and 5 cm. 2 cm + 3 cm = 5 cm So, the sum of two side is not greater than the third side. Therefore, this triangle is not possible. (ii) Given sides of the triangle are 3 cm, 6 cm and 7 cm. 3 cm + 6 cm = 9 cm So, the sum of two side is greater than the third side. Therefore, this triangle is possible. (iii) Given sides of the triangle are 6 cm, 3 cm and 2 cm. 2 cm + 3 cm = 5 cm So, the sum of two side is not greater than the third side. Therefore, this triangle is not possible.

Q.8 Take any point O in the interior of a triangle PQR. Is
(i) OP + OQ > PQ
?
(II) OQ + OR > QR?
(
iii )
OR + OP > RP?

Ans

If O is a point in the interior of a given triangle, then three triangles ΔOPQ,ΔOQR,andΔORP can be constructed. In a triangle, the sum of the lengths of either two sides is always greater than the third side. (i) Yes, as ΔOPQ is a triangle with side OP, OQ and PQ. So, OP + OQ > PQ (ii) Yes, as ΔOQR is a triangle with side OP, OQ and PQ. So, OQ + OR > QR (iii)Yes, as ΔORP is a triangle with side OP, OQ and PQ. So, OP + OR > PR MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGjbGaaeOzaiaabccacaqGpbGaaeii aiaabMgacaqGZbGaaeiiaiaabggacaqGGaGaaeiCaiaab+gacaqGPb GaaeOBaiaabshacaqGGaGaaeyAaiaab6gacaqGGaGaaeiDaiaabIga caqGLbGaaeiiaiaabMgacaqGUbGaaeiDaiaabwgacaqGYbGaaeyAai aab+gacaqGYbGaaeiiaiaab+gacaqGMbGaaeiiaiaabggacaqGGaGa ae4zaiaabMgacaqG2bGaaeyzaiaab6gacaqGGaGaaeiDaiaabkhaca qGPbGaaeyyaiaab6gacaqGNbGaaeiBaiaabwgacaqGSaGaaeiiaiaa bshacaqGObGaaeyzaiaab6gacaqGGaGaaeiDaiaabIgacaqGYbGaae yzaiaabwgaaeaacaqG0bGaaeOCaiaabMgacaqGHbGaaeOBaiaabEga caqGSbGaaeyzaiaabohacaqGGaGaeyiLdqKaae4taiaabcfacaqGrb GaaeilaiaaysW7cqGHuoarcaqGpbGaaeyuaiaabkfacaqGSaGaaGjb VlaabggacaqGUbGaaeizaiaaysW7cqGHuoarcaqGpbGaaeOuaiaabc facaqGGaGaae4yaiaabggacaqGUbGaaeiiaiaabkgacaqGLbGaaeii aiaabogacaqGVbGaaeOBaiaabohacaqG0bGaaeOCaiaabwhacaqGJb GaaeiDaiaabwgacaqGKbGaaeOlaaqaaiaabMeacaqGUbGaaeiiaiaa bggacaqGGaGaaeiDaiaabkhacaqGPbGaaeyyaiaab6gacaqGNbGaae iBaiaabwgacaqGSaGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqG ZbGaaeyDaiaab2gacaqGGaGaae4BaiaabAgacaqGGaGaaeiDaiaabI gacaqGLbGaaeiiaiaabYgacaqGLbGaaeOBaiaabEgacaqG0bGaaeiA aiaabohacaqGGaGaae4BaiaabAgacaqGGaGaaeyzaiaabMgacaqG0b GaaeiAaiaabwgacaqGYbGaaeiiaiaabshacaqG3bGaae4Baiaabcca caqGZbGaaeyAaiaabsgacaqGLbGaae4CaiaabccaaeaacaqGPbGaae 4CaiaabccacaqGHbGaaeiBaiaabEhacaqGHbGaaeyEaiaabohacaqG GaGaae4zaiaabkhacaqGLbGaaeyyaiaabshacaqGLbGaaeOCaiaabc cacaqG0bGaaeiAaiaabggacaqGUbGaaeiiaiaabshacaqGObGaaeyz aiaabccacaqG0bGaaeiAaiaabMgacaqGYbGaaeizaiaabccacaqGZb GaaeyAaiaabsgacaqGLbGaaeOlaaqaaiaabIcacaqGPbGaaeykaiaa bccacaqGzbGaaeyzaiaabohacaqGSaGaaeiiaiaabggacaqGZbGaae iiaiabgs5aejaab+eacaqGqbGaaeyuaiaabccacaqGPbGaae4Caiaa bccacaqGHbGaaeiiaiaabshacaqGYbGaaeyAaiaabggacaqGUbGaae 4zaiaabYgacaqGLbGaaeiiaiaabEhacaqGPbGaaeiDaiaabIgacaqG GaGaae4CaiaabMgacaqGKbGaaeyzaiaabccacaqGpbGaaeiuaiaabY cacaqGGaGaae4taiaabgfacaqGGaGaaeyyaiaab6gacaqGKbGaaeii aiaabcfacaqGrbGaaeOlaaqaaiaabofacaqGVbGaaeilaiaabccaca qGpbGaaeiuaiaabccacaqGRaGaaeiiaiaab+eacaqGrbGaaeiiaiaa b6dacaqGGaGaaeiuaiaabgfaaeaacaqGOaGaaeyAaiaabMgacaqGPa GaaeiiaiaabMfacaqGLbGaae4CaiaabYcacaqGGaGaaeyyaiaaboha caqGGaGaeyiLdqKaae4taiaabgfacaqGsbGaaeiiaiaabMgacaqGZb GaaeiiaiaabggacaqGGaGaaeiDaiaabkhacaqGPbGaaeyyaiaab6ga caqGNbGaaeiBaiaabwgacaqGGaGaae4DaiaabMgacaqG0bGaaeiAai aabccacaqGZbGaaeyAaiaabsgacaqGLbGaaeiiaiaab+eacaqGqbGa aeilaiaabccacaqGpbGaaeyuaiaabccacaqGHbGaaeOBaiaabsgaca qGGaGaaeiuaiaabgfacaqGUaaabaGaae4uaiaab+gacaqGSaGaaeii aiaab+eacaqGrbGaaeiiaiaabUcacaqGGaGaae4taiaabkfacaqGGa GaaeOpaiaabccacaqGrbGaaeOuaaqaaiaabIcacaqGPbGaaeyAaiaa bMgacaqGPaGaaGjbVlaabMfacaqGLbGaae4CaiaabYcacaqGGaGaae yyaiaabohacaqGGaGaeyiLdqKaae4taiaabkfacaqGqbGaaeiiaiaa bMgacaqGZbGaaeiiaiaabggacaqGGaGaaeiDaiaabkhacaqGPbGaae yyaiaab6gacaqGNbGaaeiBaiaabwgacaqGGaGaae4DaiaabMgacaqG 0bGaaeiAaiaabccacaqGZbGaaeyAaiaabsgacaqGLbGaaeiiaiaab+ eacaqGqbGaaeilaiaabccacaqGpbGaaeyuaiaabccacaqGHbGaaeOB aiaabsgacaqGGaGaaeiuaiaabgfacaqGUaaabaGaae4uaiaab+gaca qGSaGaaeiiaiaab+eacaqGqbGaaeiiaiaabUcacaqGGaGaae4taiaa bkfacaqGGaGaaeOpaiaabccacaqGqbGaaeOuaaaaaa@A737@

Q.9

AM is a median of a triangle ABC.Is AB + BC + CA > 2 AM?

(Consider the sides of triangles ΔABM and ΔAMC.)

Ans

Since, in a triangle, the sum of either two sides is always greater than the third side. So, in ΔABM AB+BM>AM ( i ) Similary in ΔACM, we get AC+CM>AM ( ii ) Adding (i) and (ii) to get AB+BM+AC+CM>2AM AB+CA+(BM+CM)>2AM AB+AC+BC>2AM AB+BC+CA>2AM Therefore, the given expression is true.

Q.10 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.11

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.12

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.13

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

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.14

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.15

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.16

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.17

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.18

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.19

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.20

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@

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FAQs (Frequently Asked Questions)
1. How to overcome the fear of Mathematics with NCERT Solutions provided by Extramarks?

Mathematics is a subject that is always feared by a lot of students. This could be because a lot of them are unable to develop  interest in Mathematics due to the lack of proper guidance. Moreover, it’s a vast subject and  new topics are added every year.  It requires conceptual clarity and regular practice. Being unable to understand its core concepts properly,  students dislike or avoid Mathematics. Thus, the best way to overcome this fear is to identify your problem, take the required assistance from Extramarks solution to overcome it. Clarifying your doubts and practising in- text and end-text exercises thoroughly will surely encourage you to be better at Mathematics. 

Our NCERT Solutions of Mathematics for all classes are one of the most essential study guides. Extramarks subject matter experts  have framed it with the utmost care to make sure your exam preparation isn’t difficult anymore, in fact it’s fun and easy.

2. List out some of the important points to remember in Chapter 6 of Class 7 Mathematics.

Following are some key points to remember while solving exercise questions:

  • Sum of all interior angles of a triangle is 180°.
  • Kinds of angles including alternate, vertically opposite, adjacent and corresponding angles.
  • Sum of two angles lying on the same plane.
  • Angles based on degrees are of three kinds- acute, obtuse and right-angle.
  • Parallel lines and transversal in a triangle.
  • Properties of interior and exterior angles of a triangle.
  • Sum of vertically opposite angles in a triangle.

3. How to score full marks in Class 7 Mathematics Chapter 6?

To score full marks in any topic of Mathematics, revision and practice is the key. Practice will not only help in understanding all the concepts of Chapter 6  at a deeper level, but it will also help you avoid silly mistakes while solving related problems. The more you practice , the easier it gets. Practice day in and day out, to experience exceptional results over time.

4. What is the difference between a triangle and a triangular region?

A triangle is a closed curve made by the intersection of three line segments. The line segments are called sides, and the points of intersections are called the vertices. The angles formed at the vertices are called the angles of the triangle. Whereas a triangular region includes the interior of the triangle along with the triangle itself.

5. Should the examples of Chapter 6 also be solved after solving the unsolved questions?

There is no hard and fast rule. However, it’s good to make a habit of first going through all the examples in the chapter and then solve the unsolved questions. The more you practice, it will be easier for you to grasp all the concepts of this chapter. And in case you fail to understand any of the concepts or problems, the solutions offered by Extramarks for NCERT solutions Class 7 Mathematics will come handy. You  won’t require any other assistance once you begin using solutions and it will save time for other subjects as well.

6. Explain types of triangles based on sides.

Based on sides, the classification of triangles is of the following types: 

  •  Equilateral triangle- A triangle in which all three sides are equal. 
  • Isosceles triangle- A triangle in which  any two sides are equal.
  • Scalene triangle- A triangle in which no two sides are equal.

7. Explain types of triangles based on angles.

Based on the angles, the classification of triangles is of the following types: 

  • Acute angled triangle- Any triangle in which all three interior angles of the triangle are less than 90°. 
  • Obtuse angled triangle- Any triangle with one of the interior angles of the triangle is more than 90°. 
  • Right-angled triangle- A triangle in which one of the interior angles of the triangle is 90°.