Solutions of class 9 maths exercise 12.1 of Chapter 12-Heron's Formula - Future Study Point

Solutions of class 9 maths exercise 12.1 of Chapter 12-Heron’s Formula

Exercise 12.1 herons formula

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

Exercise 12.1 herons 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.

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NCERT Solutions of class 9 maths

Chapter 1- Number SystemChapter 9-Areas of parallelogram and triangles
Chapter 2-PolynomialChapter 10-Circles
Chapter 3- Coordinate GeometryChapter 11-Construction
Chapter 4- Linear equations in two variablesChapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s GeometryChapter 13-Surface Areas and Volumes
Chapter 6-Lines and AnglesChapter 14-Statistics
Chapter 7-TrianglesChapter 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’

SCHOOL AHEAD

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

If the perimeter of the triangle is 180 cm

a +a +a = 180⇒3a = 180⇒a = 60 m

Area of the triangle

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

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

=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.

 

keep the park green and clean

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

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

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

= 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

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

Where

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

ABCD

 

 

 

 

 

 

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

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