Future Study Point

# Solutions of Class 9 Maths Chapter 12-Heron’s Formula

Solutions of class 9 maths exercise 12.1 of Chapter 12-Heron’s Formula are solved here for the purpose of clearing the doubts of students that are required to get excellent marks in the exam. All solutions of class 9 maths exercise 12.1 of Chapter 12-Heron’s Formula are prepared by an expert of maths who has huge experience in teaching maths subject, each unsolved question of exercise 12.1 is created by a step by step method and as per the CBSE norms therefore every student of 9 class can understand the solutions easily without taking any help from tutor or maths teacher.

Click for online shopping

Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc

### NCERT Solutions of class 9 maths

 Chapter 1- Number System Chapter 9-Areas of parallelogram and triangles Chapter 2-Polynomial Chapter 10-Circles Chapter 3- Coordinate Geometry Chapter 11-Construction Chapter 4- Linear equations in two variables Chapter 12-Heron’s Formula Chapter 5- Introduction to Euclid’s Geometry Chapter 13-Surface Areas and Volumes Chapter 6-Lines and Angles Chapter 14-Statistics Chapter 7-Triangles Chapter 15-Probability Chapter 8- Quadrilateral

Q1.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. The side of an equilateral triangle given to us is ‘a’

The sum of its side is = a + a +a = 3a

Semiperimeter of the triagle, s = 3a/2

Since the triangle is equilateral, therefore  b=c=a

Applying heron’s formula for calculating its area

$\fn_cm Area \: of\: \Delta =\sqrt{s\left ( s-a \right )\left ( s-b \right )\left ( s-c \right )}$

$\fn_cm Area \: of\: \Delta =\sqrt{3a/2\left ( 3a/2-a \right )\left ( 3a/2-a \right )\left ( 3a/2-a \right )}$

$\fn_cm =\sqrt{3a/2\left ( a/2 \right )\left ( a/2 \right )\left ( a/2 \right )}$

$\fn_cm =\frac{a^{2}\sqrt{3}}{4}$

If the perimeter of the triangle is 180 cm

a +a +a = 180⇒3a = 180⇒a = 60 m

Area of the triangle

$\fn_cm =\frac{60^{2}\sqrt{3}}{4}=\frac{3600\sqrt{3}}{4}=900\sqrt{3}$

Therefore the area of the triangle is 900√3 cm²

Q2.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 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 triangular sides of the wall are,a = 122  m,b=22m, c=120 m

Semiperimeter of the triangular walls,s = (a+b+c)/2 =(122+22+120)/2 =264/2=132

Applying heron’s formula for calculating its area

$\fn_cm Area \: of\: \Delta =\sqrt{s\left ( s-a \right )\left ( s-b \right )\left ( s-c \right )}$

Putting the value of a,b and c in the formula

$\fn_cm =\sqrt{132\left ( 132-122 \right )\left ( 132-22 \right )\left ( 132-120 \right )}$

$\fn_cm =\sqrt{132\times 10\times 110\times 12}$

$\fn_cm =\sqrt{11\times 2\times 2\times 3\times 2\times 5\times 11\times 2\times 5\times 2\times 3\times 2}$

=11×2×2×2×5×3 = 1320

Area of the triangular walls is = 1320 m²

The cost of advertisement for hiring 1 year(12 months) = Rs 5000/m²

The cost of advertisement for hiring 3 months = (5000/12)×3 = Rs 1250/m²

The company will pay the rent for 3 months in hiring one of the walls = Area of the triangular walls× 1250= 1320 ×1250 = Rs 1650000

Q3.There is a slide in a park. One of its side company hired one of its walls for 3 months. walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN” (see figure). If the sides of the wall are 15 m, 11 m, and 6m, find the area painted in colour.

Ans.The sides  of triangular walls are a = 15 m, b= 11 m and c = 6m

Semiperimeter of the triangular walls,s = (a+b+c)/2 =(15+6+11)/2 =32/2=16

Applying heron’s formula for calculating its area

$\fn_cm Area \: of\: \Delta =\sqrt{s\left ( s-a \right )\left ( s-b \right )\left ( s-c \right )}$

$\fn_cm =\sqrt{16\left ( 16-15 \right )\left ( 16-11 \right )\left ( 16-6 \right )}$

$\fn_cm =\sqrt{16\times 1\times 5\times 10}=4\times 5\sqrt{2}=20\sqrt{2}$

The area of the triangular wall is 20√2 m²

Q4.Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.

Ans. The given two sides of triangle are ,a = 18 cm and b= 10 cm

Perimeter of the triangle is 42 cm

Third side of the triangle,c = Perimeter -(sum of both sides) =42 -(18+10) =42-28 =14 cm

Semiperimeter,s = Perimeter/2 = 42/2 = 21 cm

Applying heron’s formula for calculating its area

$\fn_cm Area \: of\: \Delta =\sqrt{s\left ( s-a \right )\left ( s-b \right )\left ( s-c \right )}$

$\fn_cm =\sqrt{21\left ( 21-18 \right )\left ( 21-10 \right )\left ( 21-14 \right )}$

$\fn_cm =\sqrt{21\times 3\times 11\times 7}=\sqrt{21\times 21\times 11}=21\sqrt{11}$

Required area of the triangle is 21√11 cm²

Q5.Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.

Ans. Since the ratio between the sides of triangle given to us is 12:17:25,therefore assuming the sides of triangle 12x, 17x and 25 x

The perimeter given to us = 540 cm

∴ 12x +17x + 25x = 540

54x = 540 ⇒ x =10

Sides of the triangle are,a = 12×10 = 120 cm, b= 17×10=170 cm, c = 25×10=250 cm

Semeiperimeter ,s = Perimeter/2 = 540/2 =270

Applying heron’s formula for calculating the area of triangle

$\fn_cm Area \: of\: \Delta =\sqrt{s\left ( s-a \right )\left ( s-b \right )\left ( s-c \right )}$

$\fn_cm =\sqrt{270\left ( 270-120 \right ) \left ( 270-170 \right )\left ( 270-250 \right )}$

$\fn_cm =\sqrt{270\times 150\times 100\times 20}$

$\fn_cm =\sqrt{100\times 100\times 10\times 27\times 15\times 2}$

$\fn_cm =\sqrt{100\times 100\times 10\times 9\times 3\times 3\times 10}$

= 100 × 10 ×3×3 = 9000

Therefore required area of the given triangle is 9000 cm²

Q6.An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle

Ans. Each of the equal sides of the isoscell triangle given to us is ,a = b= 12 cm,let the third side of the triangle is c

The perimeter given to us is = 30 cm

c = 30 -(12 + 12) = 30 -24 = 6 cm

Semiperimeter of the triangle is,s = Perimeter/2 =30/2 = 15 cm

Applying heron’s formula for calculating the area of triangle

$\fn_cm Area \: of\: \Delta =\sqrt{s\left ( s-a \right )\left ( s-b \right )\left ( s-c \right )}$

$\fn_cm =\sqrt{15\left ( 15-12 \right )\left ( 15-12 \right )\left ( 15-6 \right )}$

$\fn_cm =\sqrt{5\times 3\times 3\times 3\times 9}=9\sqrt{15}$

Therefore the required area of the triangle is 9√15 cm²

## Summary of the exercise 12.1 -Heron’s formula

According to Heron’s formula, if the sides of the triangle are a, b .c and the semi perimeter is S , then the area of the triangle is given as follows

$\fn_cm Area\: of \: the\: \Delta =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$

Where

$\fn_cm S=\frac{a+b+c}{2}$

The area of a quadrilateral whose sides and one diagonal is given then it area can also be calculated as follows

Area of the quadrilateral ABCD is = ar ABD + ar DBC

### NCERT Solutions of class 9 science

CBSE Class 9-Question paper of science 2020 with solutions

CBSE Class 9-Sample paper of science

CBSE Class 9-Unsolved question paper of science 2019

### NCERT Solutions of class 10 maths

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

### Solutions of class 10 last years Science question papers

CBSE Class 10 – Question paper of science 2020 with solutions

CBSE class 10 -Latest sample paper of science

### NCERT solutions of class 11 maths

 Chapter 1-Sets Chapter 9-Sequences and Series Chapter 2- Relations and functions Chapter 10- Straight Lines Chapter 3- Trigonometry Chapter 11-Conic Sections Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry Chapter 5-Complex numbers Chapter 13- Limits and Derivatives Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning Chapter 7- Permutations and Combinations Chapter 15- Statistics Chapter 8- Binomial Theorem Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

### NCERT solutions of class 12 maths

 Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals

50 important science questions for cbse 10 class

Chemical properties of acid and base

Anatomy of human brain

Physics viva voce questions

Chemistry viva voce questions & answers

Chemistry practical based questions

### Solutions of previous years science & maths question papers

Solutions of class 10 Science question paper 2019 CBSE

Solutions of class 10 CBSE maths question paper 2020

CBSE Class 10 science question paper 2020 SET -3 solutions

Download e-book of Science NCERT 634 questions with solutions for Polytechnic and NDA entrance exams