NCERT Solutions for Class 9 Mathematics Chapter 12 – Heron’s Formula

Mathematics requires a lot of practice. One can excel in Mathematics only if he practices as much numericals as he can. After understanding the concepts and developing the habit of practising, you will see yourself passing Mathematics exam with flying colours as it is rightly said; ‘Practice makes a man perfect.

Extramarks is a well-known learning platform trusted for providing high-quality study materials for Class 11 to Class 12. Our team of academic experts have created study materials based on the latest CBSE syllabus and guidelines. Students can confidently rely on our comprehensive study solutions.

The topics like introduction to triangles, area of equilateral triangles, area of isosceles triangles, and finding the area of any polygon using Heron’s formula are covered in Class 9 Chapter 12 Mathematics. Students are advised to complete the chapter triangles given in NCERT Class 9 Mathematics textbook and clear all the basics related to it.

You will find information related to the properties of equilateral and isosceles triangles, how to find their areas, who was a heron, what role he played in Mathematics and how to find the area of any polygon using Heron’s formula in NCERT Solutions for Class 9 Mathematics Chapter 12. You can also explore solutions for other classes such as  NCERT Class 9 Mathematics Chapter 7 and NCERT Class 9 Mathematics Chapter 12 in our NCERT Solutions for Class 9 Mathematics Chapter 12.

Extramarks is always a step ahead in providing quality resources. You can find NCERT Exemplars , NCERT related study material, NCERT revision notes, NCERT practice questions, NCERT Solutions and a lot more on our Extramarks website.

Key Topics Covered In NCERT Solutions for Class 9 Mathematics Chapter 12

You can recall what a triangle is, its definitions, the theorems related to it and all its properties from NCERT Class 9 Mathematics Chapter 7. All the applications covered in that chapter will be of great use in this chapter too. Hence, students are advised to thoroughly revise it before beginning NCERT Solutions for Class 9 Mathematics Chapter 12.

You will learn more about the calculations of areas of triangles and other polygons in this chapter. One of the highlight topics of this chapter is Heron’s formula which helps in solving sums with a lot of ease and thereby bringing accuracy. You don’t have to jump from NCERT Mathematics Chapter 7 to NCERT Mathematics Chapter 12 again and again if you refer to Extramarks’ resources.

We have covered all the interrelated topics and sub-topics from both the chapters in the NCERT Solutions for Class 9 Mathematics Chapter 12 available on the Extramarks’ website.

NCERT Solutions Class 9 Mathematics Chapter 12 requires students to use their critical thinking ability and students apply a wide range of formulas they have learnt. 

Introduction

In the earlier chapters of Class 9 Mathematics, we have already learnt about the different shapes like triangles, parallelograms, rhombus, rectangles etc.

We have also read about how to calculate the areas of these shapes.

We will learn more about the calculation of areas of the different types of triangles like Equilateral triangles, isosceles triangles, and right-angle triangles.

  • Equilateral triangles: In this triangle, all sides are equal in length.
    • Isosceles triangles: This triangle has two equal sides and two equal angles.
  • Right-angle triangles: This triangle has one angle of 90 degrees.

 You will also get to know how to calculate the area of a polygon using Heron’s formula.

Area of a triangle – by Heron’s formula

Heron was a well-known mathematician from Egypt. Due to his varied contributions to Mathematics, he was also considered as an encyclopaedia  for major concepts of mathematics. He derived the famous formula for the area of a triangle in terms of its three sides. This formula was named in his honour as Heron’s formula.

Heron’s formula is:

Area of triangle =√s(s-a)(s-b)(s-c), 

where s = Perimeter/2 = (a + b + c)/2

This formula is helpful when it is not possible to find the height of the triangle easily.

To learn more about the Heron’s formula and practice a lot of questions related to the area of the triangle. You can visit the website of  Extramarks and explore our study material and refer to NCERT Solutions for Class 9 Mathematics Chapter 12.

Application of Heron’s formula in finding the area of quadrilateral

In this section, we will get to know about how to calculate the area of a quadrilateral by splitting it into triangles. 

By doing so, you can quickly calculate the area with a lot of accuracy.

You can find the steps to calculate the area of the quadrilateral in the NCERT Solutions for Class 9 Mathematics Chapter 12 available on the Extramarks’ website.

Summary

In this chapter, we have learnt about

  1. Different types of triangles
  2. Calculation of the area of equilateral triangles
  3. Calculation of the area of the isosceles triangles
  4. Calculation of the area of the right angle triangles
  5. Calculation of the area of the polygons by Heron’s formula
  6. Calculate the area of quadrilaterals by splitting them into triangles

NCERT Solutions for Class 9 Mathematics Chapter 12 Exercise & Solutions

Extramarks believes that chapter exercises play a key role in students’ preparation. Students can ace their examinations if they solve every exercise given in the NCERT textbook properly and completely. Hence, Extramarks NCERT Solutions for Class 9 Mathematics Chapter 12 has all the chapter related exercises and solutions covered in a detailed manner. You can also find some extra questions to practice in our NCERT study material available on the Extramarks’ website.

You can find for exercise specific questions and solutions for NCERT Solutions for Class 9 Mathematics Chapter 12 by referring to the following links:

  •  Chapter 12: Exercise 12.1 Question and answers    
  •  Chapter 12: Exercise 12.2 Question and answers

Along with NCERT Solutions for Class 9 Mathematics Chapter 12, students can explore NCERT Solutions on our Extramarks website for all primary and secondary classes.

  • NCERT Solutions Class 1
  • NCERT Solutions Class 2
  • NCERT Solutions Class 3
  • NCERT Solutions Class 4
  • NCERT Solutions Class 5 
  • NCERT Solutions Class 6
  • NCERT Solutions Class 7
  • NCERT Solutions Class 8
  • NCERT Solutions Class 9
  • NCERT solutions Class 10
  • NCERT solutions Class 11
  • NCERT solutions Class 12

NCERT Exemplar for Class 9 Mathematics 

NCERT Exemplar Class 9 Mathematics book helps students to develop interest in the subject. One can learn about the application of Mathematics in real life because of the well-formed questions in the book. The wide range of concepts covered in the book makes it beneficial for all the curriculum students.

It has a set of various questions required to be good performers in the subject of Mathematics. The different types and ranges of questions covered throughout the book will prove to be a milestone in every aspect of your study of Mathematics. As a result, students are advised by the experts to include these books in their study material.

Students can easily refer to a more advanced level of questions once they develop conceptual understanding using this book. This book is an encyclopaedia for NCERT related questions. Students develop more skills based thinking once they start studying from this book. Thus, NCERT Exemplar is the right resource for your preparation.

Key Features of NCERT Solutions for Class 9 Mathematics Chapter 12

An active mind always yields positive results. Hence, NCERT Solutions for Class 9 Mathematics Chapter 12 has been designed in such a way that keeps your mind active and engaged. You can find the following key features:

  • NCERT Solutions for Class 9 Mathematics Chapter 12 has a wide range of questions that gives a good exercise to your brain; thus making you feel energetic
  • The gradual increase in the difficulty of the questions keeps you motivated during your practice sessions.
  • After completing the NCERT Solutions for Class 9 Mathematics Chapter 12, you can ensure you have practised questions from all the corners of the chapter, and thus, no concept remains untouched.

Q.1 A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?

Ans.

Side of equlateral triangle=aArea of triangle Δ=s(sa)(sb)(sc)[Heron’s Formula]where, s=a+b+c2=a+a+a2=3a2Δ=3a2(3a2a)(3a2a)(3a2a)=3a2×a2×a2×a2=a2×a23=34a2 square units.If perimeter of triangle(3a) =180 cma=1803=60 cmThen, area of equilateral triangular signal board =34(60)2 square cm=9003cm2Thus, the area of the traffic signal board is 9003cm2.

Q.2 The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see following figure.). The advertisements yield an earning of ₹ 5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?

Ans.

The sides of triangular walls are 122 m, 22 m and 120 m.Area of triangle             Δ=s(sa)(sb)(sc)      [Heron’s Formula]where, s=a+b+c2 and a=122m,b=22m and c=120m.             s=122+22+1202               =2642               =132m            Δ=132(132122)(13222)(132120)               =132×10×110×12              =1320 m2Thus, the area of one wall is 1320 m2.Earning of 1m2 wall by the advertisement in 1 year               =5000Earning of 1m2 wall by the advertisement in 3 months               =500012×3               =1250Earning of 1320 m2 wall by the advertisement in 3 months               =1250×1320              =1650000Thus, the company paid the rent for one wall is 1650000.

Q.3 There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN”.
If the sides of the wall are 15m, 11 m and 6 m, find the area painted in colour.

Ans.

The sides of the wall are 15m, 11 m and 6 m.Heron’s Formula: Δ=s(sa)(sb)(sc)where, s=a+b+c2 and a=15m,b=11m and c=6m.s=15+11+62=322=16mΔ=16(1615)(1611)(166)=16×1×5×10=202 m2Thus, the painted area of wall is 202 m2.

Q.4 Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.

Ans.

Two sides of triangle are 18 cm and 10 cm.Perimeter of triangle=42 cm                a+b+c=42              18+10+c=42                        c=4228                          =14Area of triangle by Heron’s formula,                      Δ=s(sa)(sb)(sc)                       s=a+b+c2                         =422=21                      Δ=21(2118)(2110)(2114)                         =21×3×11×7                         =2111 cm2Thus, the area of triangle is 2111 cm2.

Q.5 Sides of a triangle are in the ratio of 12:17:25 and its perimeter is 540 cm. Find its area.

Ans.

Ratio of sides of triangle=12:17:25        Perimeter of triangle=540 cmLet a=12 k, b=17k and 25 kthen,   12k+17k+25k=540                           54 k=540                          k=10So,​ a=12×10=120 cm         b=14×10=170 cm         c=25×10=250 cmArea of triangle, by Heron’s Formula:       Δ=s(sa)(sb)(sc)       s=a+b+c2=5402=270      Δ=s(sa)(sb)(sc)      =270(270120)(270170)(270250)       =270(150)(100)(20)       =270(150)(100)(20)       =9000 cm2Thus, the area of triangle is 9000 cm2.

Q.6 An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

Ans.

Perimeter of isosceles triangle=30 cmEach of the equal side=12 cmThen, third side of triangle=301212=6 cmArea of triangle, Δ=s(sa)(sb)(sc)[Heron’s Formula]s=a+b+c2=302=15Δ=15(1512)(1512)(156)=15(3)(3)(9)=915 cm2Thus, the area of isosceles triangle is 915 cm2.

Q.7 A park, in the shape of a quadrilateral ABCD, has ∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

Ans.

In ΔBCD, BCD=90°So, Pythagoras theoremBD=BC2+CD2=(12)2+(5)2=144+25=169=13cmar(ΔBCD)=12×BC×DC=12×12×5=30cm2Area ΔABD, by Heron’s formulaΔ=s(sa)(sb)(sc)s=a+b+c2=9+8+132=302=15cmΔ=15(159)(158)(1513)=15×6×7×2=635cm=35.5cm2(Approx.)Area(ABCD)=ar(ΔBCD)+ar(ΔABD)=30+35.5=65.5cm2Thus, area of ABCD is 65.5 cm2 approx.

Q.8 Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

Ans.

InΔABC,3,4 and 5 are Pythagorian triplet.So, area(ΔABC)=12×BC×AB                     =12×4×3                     =6cm2Area of ΔACD,by Pythagoras Theorem,                  Δ=s(sa)(sb)(sc)                   s=5+5+42                      =142=7cm                  Δ=7(75)(75)(74)                    =7(2)(2)(3)                     =9.2(Approx.)Area(ABCD)=area(ΔABC)+area(ΔACD)                   =6+9.5                   =15.2cm2      (Approx)

Q.9 Radha made a picture of an aeroplane with coloured paper as shown in the following figure. Find the total area of the paper used.

Ans.

For I triangle:sides are 5 cm, 1 cm and 5 cm.By Heron’s formula:Area of triangle(Δ)=s(sa)(sb)(sc)where,​              s=a+b+c2                       =5+1+52=5.5cm                     Δ=5.5(5.55)(5.51)(5.55)                       =5.5(0.5)(4.5)(0.5)                        =2.488 cm2Area of II figure:   Area of rectangle=Length×Breadth                        =6.5×1                        =6.5 cm2Area of III figure:

Height of ΔBCF=12(0.5)2 =10.25 =0.75=0.866cmArea of trapezium ABCD=12(sum of parallel sides)×height =12(1+2)×0.866=1.299cm2Area of IV figure:Area of right triangle =12×1.5×6=4.5Area of V figure:Area of right triangle =12×1.5×6 =4.5Total area of used paper=2.488+6.5+1.299+4.5+4.5 =19.287cm2=19.3cm2(Approx.)

Q.10 A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.

Ans.

Sides of triangle are 26 cm, 28 cm and 30 cm.Area of triangle by Heron’s fromula,                Δ=s(sa)(sb)(sc)                 s=26+28+302=42                Δ=42(4226)(4228)(4230)                   =42(16)(14)(12)                   =336cm2Let height of parallelogram be h cm.Since, area of parallelogram=Area of triangle                           base×h=336cm2                             28×h=336cm2                                  h=33628=12cmThus, height of parallelogram is 12 cm.

Q.11 A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?

Ans.

Sides of triangle are 30 m, 30 m and 48 m. By Heron’s formula: Area ( ABC )= s( sa )( sb )( sc ) s= a+b+c 2 = 30+30+48 2 =54 Area ( ABC )= 54( 5430 )( 5430 )( 5448 ) = 54( 24 )( 24 )( 6 ) =432 cm 2 Area of rhombus ABCD =2×Area( ΔABC ) =2×432 =864 m 2 Number of cows=18 Area grazed by each cow= 864 18 =48 m 2 Thus, the area of grass grazed by each cow is 48 m 2 of the field. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqGabeqaaa9cbaGaam4uaiaadMgacaWGKbGaamyzaiaadohacaqGGaGaae4BaiaabAgacaqGGaGaaeiDaiaabkhacaqGPbGaaeyyaiaab6gacaqGNbGaaeiBaiaabwgacaqGGaGaaeyyaiaabkhacaqGLbGaaeiiaiaabodacaqGWaGaaeiiaiaab2gacaqGSaGaaeiiaiaabodacaqGWaGaaeiiaiaab2gacaqGGaGaaeyyaiaab6gacaqGKbGaaeiiaiaabsdacaqG4aGaaeiiaiaab2gacaqGUaaabaGaamOqaiaadMhacaqGGaGaaeisaiaabwgacaqGYbGaae4Baiaab6gacaqGNaGaae4CaiaabccacaqGMbGaae4BaiaabkhacaqGTbGaaeyDaiaabYgacaqGHbGaaeOoaaqaaiaadgeacaWGYbGaamyzaiaadggacaqGGaWaaeWaaeaacaWGbbGaamOqaiaadoeaaiaawIcacaGLPaaacqGH9aqpdaGcaaqaaiaadohadaqadaqaaiaadohacqGHsislcaWGHbaacaGLOaGaayzkaaWaaeWaaeaacaWGZbGaeyOeI0IaamOyaaGaayjkaiaawMcaamaabmaabaGaam4CaiabgkHiTiaadogaaiaawIcacaGLPaaaaSqabaaakeaacaaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caWGZbGaeyypa0ZaaSaaaeaacaWGHbGaey4kaSIaamOyaiabgUcaRiaadogaaeaacaaIYaaaaaqaaiaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8Uaeyypa0ZaaSaaaeaacaaIZaGaaGimaiabgUcaRiaaiodacaaIWaGaey4kaSIaaGinaiaaiIdaaeaacaaIYaaaaiabg2da9iaaiwdacaaI0aaabaGaamyqaiaadkhacaWGLbGaamyyaiaabccadaqadaqaaiaadgeacaWGcbGaam4qaaGaayjkaiaawMcaaiabg2da9maakaaabaGaaGynaiaaisdadaqadaqaaiaaiwdacaaI0aGaeyOeI0IaaG4maiaaicdaaiaawIcacaGLPaaadaqadaqaaiaaiwdacaaI0aGaeyOeI0IaaG4maiaaicdaaiaawIcacaGLPaaadaqadaqaaiaaiwdacaaI0aGaeyOeI0IaaGinaiaaiIdaaiaawIcacaGLPaaaaSqabaaakeaacaaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlabg2da9maakaaabaGaaGynaiaaisdadaqadaqaaiaaikdacaaI0aaacaGLOaGaayzkaaWaaeWaaeaacaaIYaGaaGinaaGaayjkaiaawMcaamaabmaabaGaaGOnaaGaayjkaiaawMcaaaWcbeaaaOqaaiaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8Uaeyypa0JaaGinaiaaiodacaaIYaGaaeiiaiaabogacaqGTbWaaWbaaSqabeaacaqGYaaaaaGcbaGaamyqaiaadkhacaWGLbGaamyyaiaabccacaqGVbGaaeOzaiaabccacaqGYbGaaeiAaiaab+gacaqGTbGaaeOyaiaabwhacaqGZbGaaeiiaiaabgeacaqGcbGaae4qaiaabseaaeaacaaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGPaVlabg2da9iaaikdacqGHxdaTcaWGbbGaamOCaiaadwgacaWGHbWaaeWaaeaacqGHuoarcaWGbbGaamOqaiaadoeaaiaawIcacaGLPaaaaeaacaaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGPaVlabg2da9iaaikdacqGHxdaTcaaI0aGaaG4maiaaikdaaeaacaaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGPaVlabg2da9iaaiIdacaaI2aGaaGinaiaaykW7caWGTbWaaWbaaSqabeaacaaIYaaaaaGcbaGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caWGobGaamyDaiaad2gacaWGIbGaamyzaiaadkhacaqGGaGaae4BaiaabAgacaqGGaGaae4yaiaab+gacaqG3bGaae4Caiabg2da9iaabgdacaqG4aaabaGaaeyqaiaabkhacaqGLbGaaeyyaiaabccacaqGNbGaaeOCaiaabggacaqG6bGaaeyzaiaabsgacaqGGaGaaeOyaiaabMhacaqGGaGaaeyzaiaabggacaqGJbGaaeiAaiaabccacaqGJbGaae4BaiaabEhacqGH9aqpdaWcaaqaaiaaiIdacaaI2aGaaGinaaqaaiaaigdacaaI4aaaaaqaaiaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGPaVlabg2da9iaaisdacaaI4aGaaGPaVlaad2gadaahaaWcbeqaaiaaikdaaaaakeaacaWGubGaamiAaiaadwhacaWGZbGaaiilaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeyyaiaabkhacaqGLbGaaeyyaiaabccacaqGVbGaaeOzaiaabccacaqGNbGaaeOCaiaabggacaqGZbGaae4CaiaabccacaqGNbGaaeOCaiaabggacaqG6bGaaeyzaiaabsgacaqGGaGaaeOyaiaabMhacaqGGaGaaeyzaiaabggacaqGJbGaaeiAaiaabccacaqGJbGaae4BaiaabEhacaqGGaGaaeyAaiaabohacaqGGaGaaeinaiaabIdacaaMc8UaaeyBamaaCaaaleqabaGaaeOmaaaakiaabccacaqGVbGaaeOzaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeOzaiaabMgacaqGLbGaaeiBaiaabsgacaqGUaaaaaa@26D9@

Q.12 An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig. 12.16), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

Ans.

         Number of triangular pieces in umbrella=10Number of triangular pieces of each coulour=5       Sides of each triangle=20 cm, 50 cm, 50 cmarea of triangle, by Heron’s fromula                             Δ=s(sa)(sb)(sc)where,         s=a+b+c2=20+50+502=60                             Δ=60(6020)(6050)(6050)                             Δ=60(40)(10)(10)=2006cm2Area of required cloth for each colour in umbrella                               =5×2006                               =10006  cm2Thus, the cloth required for one colour is 10006 cm2 and clothrequired for other colour is 10006 cm2.

Q.13 A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Figure. How much paper of each shade has been used in it?

Ans.

Area of square =12×Product of diagonals                    =12×32×32                    =512cm2Since, paper required for area of shade I                     =paper required for area of shade II                    =5122cm2                    =216cm2Thus, the required paper of each part I and II is 216 cm2.Sides of triangle are 6cm, 6 cm and 8 cm.Area of triangle=s(sa)(sb)(sc)       [By Heron’s formula]                    =10(106)(106)(108)                                    [s=6+6+82=10]Area of triangle=10(4)(4)(2)                    =85                    =8×2.24Area of triangle=17.92cm2Thus, the area of required paper for I and II each part is 216 cm2 and that of for III shaded part is 17.92 cm2.

Q.14 A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm.
Find the cost of polishing the tiles at the rate of 50p per cm2.

Ans.

Sides of triangle used in floral design are 9 cm,28 cm and 35 cm.So, area of a triangle=s(sa)(sb)(sc)                           =36(369)(3628)(3635)                                   [s=9+28+352=36cm]                           =36(27)(8)(1)                           =366cm2                           =36×2.45cm2                           =88.2cm2    Area of 16 triangles=16×88.2                           =1411.2cm2Cost of polishing1 cm2 area of floor                           =50100                           =0.5Cost of polishing 1411.2 cm2 area of floor                           =0.5×1411.2                           =705.6Thus, the cost of polishing the tiles at the rate of 50p per cm2 is₹705.6.

Q.15 A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

Ans.

Let ABCD be trapezium and parallel sides AB=10 m and CD = 25 m.
Non-parallel sides are AD = 13 m and BC = 14 m.
Let height of ABCD be BE. Draw BF parallel to AD.
So, BF = 13 cm because opposite sides of parallelogram are equal.
FC = 25 – DF
= 25 – 10 = 15 m

Sides of ΔBFC are 13 m, 15 m and 14 m.Area of ΔBFC=s(sa)(sb)(sc)          [By Heron’s formula]                  =21(2113)(2115)(2114)                                [s=13+15+142=21m]                 =21(8)(6)(7)                  =7056                 =84​ m2Area of ΔBFC=12×FC×BE                 84​ m2=12×15×BEheight,  BE=2×8415=11.2marea(ABCD)=12(AB+CD)×BE         [area of trapezium=12(sumof parallel sides)×h]area(ABCD)=12(10+25)×11.2                 =12(35)×11.2                =196m2

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FAQs (Frequently Asked Questions)
1. How many exercises and solutions are included in NCERT Solutions for Class 9 Mathematics Chapter 12?

There are a total of 2 exercises and a total of 15 questions included in the NCERT Solutions for Class 9 Mathematics Chapter 12. You can find a detailed solution to each exercise and every question and test knowledge and step up your preparation.

You can avail of NCERT Solutions from the Extramarks website anytime to leverage your performance and excel in your examinations.

2. Why has NCERT Chapter 12 been included in the Class 9 Mathematics syllabus?

You have already been learning about the basics of triangles right from your lower classes. By now, you might have understood the importance of triangles in the Mathematics syllabus. As a result, NCERT Class 9 Mathematics textbook itself contains two chapters based on triangles.

It is necessary that students know all the applications of triangles and the role of Heron’s formula in calculating area in a stipulated time. The concepts covered in this chapter will largely be used in Class 11 and Class 12 Mathematics. Hence, this chapter carries a lot of importance and has been included in NCERT Class 9 Mathematics syllabus.