Heron’s Formula Class 9 – NCERT Solutions Chapter 10

NCERT Solutions for Class 9 Maths Chapter 10 Heron’s Formula can be accessed here in a few scrolls. Mathematics focuses on various concepts, theories and formulas, and Heron’s Formula is among those. The formula used in various advanced chapters will be prescribed in senior classes. You must understand the concept well to score well in your examinations. The solutions are designed by keeping the guidelines of the updated syllabus. The answers are provided in a structurally step-wise manner to ensure clarity.

Overview of the Exercises of NCERT Solutions for Class 9 Maths Chapter 10 Heron’s Formula

Exercise 10.1: The first exercise of the chapter focuses on the basics concepts of Heron’s Formula. Students are asked to find the area, missing sides and height of a triangle, when certain specifics are given.

Exercise 10.2: The second exercise consists of a total of 9 questions and is the continuation of the first exercise.

NCERT Solutions for Class 9 Maths Chapter 10 Heron’s Formula Exercise 10.1

Question 1:

Find the area of a triangle with a perimeter of 180 cm using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?

Solution:

Side length of the triangle:
The perimeter of the equilateral triangle is given by:
Perimeter=3a=180 cm\text{Perimeter} = 3a = 180 \, \text{cm}Perimeter=3a=180cm
Therefore, the side length aaa is:
a=1803=60 cma = \frac{180}{3} = 60 \, \text{cm}a=3180​=60cm

Calculate the semi-perimeter:
The semi-perimeter sss of the triangle is:
s=3a2=3×602=90 cms = \frac{3a}{2} = \frac{3 \times 60}{2} = 90 \, \text{cm}s=23a​=23×60​=90cm

Apply Heron’s formula:
Heron’s formula for the area of a triangle is:
Area=s(s−a)(s−b)(s−c)\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}Area=s(s−a)(s−b)(s−c)​
For an equilateral triangle, all sides are equal (a=b=c)(a = b = c)(a=b=c). Thus, the formula simplifies to:
Area=s(s−a)(s−a)(s−a)=s(s−a)3\text{Area} = \sqrt{s(s-a)(s-a)(s-a)} = \sqrt{s(s-a)^3}Area=s(s−a)(s−a)(s−a)​=s(s−a)3​
Substituting the values:
Area=90(90−60)3=90×30×30×30=90×30=2700 cm2\text{Area} = \sqrt{90(90-60)^3} = \sqrt{90 \times 30 \times 30 \times 30} = 90 \times 30 = 2700 \, \text{cm}^2Area=90(90−60)3​=90×30×30×30​=90×30=2700cm2
Thus, the area of the signal board is approximately 2700 cm22700 \, \text{cm}^22700cm2.

Question 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. The advertisements yield an earning of Rs. 5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?

Solution:

Given:

a=122 m, b=22 m, c=120 ma = 122 \, \text{m}, \, b = 22 \, \text{m}, \, c = 120 \, \text{m}a=122m,b=22m,c=120m

Calculate the semi-perimeter:
s=a+b+c2=122+22+1202=132 ms = \frac{a+b+c}{2} = \frac{122 + 22 + 120}{2} = 132 \, \text{m}s=2a+b+c​=2122+22+120​=132m

Calculate the area using Heron’s formula:
Area=s(s−a)(s−b)(s−c)=132(132−122)(132−22)(132−120)=132×10×110×12=1320 m2\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{132(132-122)(132-22)(132-120)} = \sqrt{132 \times 10 \times 110 \times 12} = 1320 \, \text{m}^2Area=s(s−a)(s−b)(s−c)​=132(132−122)(132−22)(132−120)​=132×10×110×12​=1320m2

Calculate the rent:
Rent per month per m2=500012 Rs\text{Rent per month per m}^2 = \frac{5000}{12} \, \text{Rs}Rent per month per m2=125000​Rs Rent for 3 months for 1320 m2=3×500012×1320=Rs. 16,50,000\text{Rent for 3 months for 1320 m}^2 = 3 \times \frac{5000}{12} \times 1320 = \text{Rs. 16,50,000}Rent for 3 months for 1320 m2=3×125000​×1320=Rs. 16,50,000

Question 3:

There is a slide in a park. One of its side walls has been painted with a message. If the sides of the wall are 15 m, 11 m, and 6 m, find the area painted in color.

Solution:

Given:

a=15 m, b=11 m, c=6 ma = 15 \, \text{m}, \, b = 11 \, \text{m}, \, c = 6 \, \text{m}a=15m,b=11m,c=6m

Calculate the semi-perimeter:
s=15+11+62=16 ms = \frac{15 + 11 + 6}{2} = 16 \, \text{m}s=215+11+6​=16m

Calculate the area using Heron’s formula:
Area=s(s−a)(s−b)(s−c)=16(16−15)(16−11)(16−6)=202 m2\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{16(16-15)(16-11)(16-6)} = 20\sqrt{2} \, \text{m}^2Area=s(s−a)(s−b)(s−c)​=16(16−15)(16−11)(16−6)​=202​m2
Hence, the area painted in color is 202 m220\sqrt{2} \, \text{m}^2202​m2.

Question 4:

Find the area of a triangle with sides 18 cm, 10 cm, and 14 cm, and a perimeter of 42 cm.

Solution:

Calculate the semi-perimeter:
s=42 cm2=21 cms = \frac{42 \, \text{cm}}{2} = 21 \, \text{cm}s=242cm​=21cm

Calculate the area using Heron’s formula:
Area=21(21−18)(21−10)(21−14)=21×3×11×7=2111 cm2\text{Area} = \sqrt{21(21-18)(21-10)(21-14)} = \sqrt{21 \times 3 \times 11 \times 7} = 21\sqrt{11} \, \text{cm}^2Area=21(21−18)(21−10)(21−14)​=21×3×11×7​=2111​cm2
Therefore, the area of the triangle is 2111 cm221\sqrt{11} \, \text{cm}^22111​cm2.

Question 5:

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

Solution:

Find the sides of the triangle:
a=120 cm, b=170 cm, c=250 cma = 120 \, \text{cm}, \, b = 170 \, \text{cm}, \, c = 250 \, \text{cm}a=120cm,b=170cm,c=250cm

Calculate the semi-perimeter:
s=540 cm2=270 cms = \frac{540 \, \text{cm}}{2} = 270 \, \text{cm}s=2540cm​=270cm

Calculate the area using Heron’s formula:
Area=270(270−120)(270−170)(270−250)=270×150×100×20=9000 sq. cm\text{Area} = \sqrt{270(270-120)(270-170)(270-250)} = \sqrt{270 \times 150 \times 100 \times 20} = 9000 \, \text{sq. cm}Area=270(270−120)(270−170)(270−250)​=270×150×100×20​=9000sq. cm
Hence, the area is 9000 sq. cm9000 \, \text{sq. cm}9000sq. cm.

Question 6:

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

Solution:

Find the third side of the triangle:
x=6 cmx = 6 \, \text{cm}x=6cm

Calculate the semi-perimeter:
s=30 cm2=15 cms = \frac{30 \, \text{cm}}{2} = 15 \, \text{cm}s=230cm​=15cm

Calculate the area using Heron’s formula:
Area=15(15−12)(15−12)(15−6)=915 cm2\text{Area} = \sqrt{15(15-12)(15-12)(15-6)} = 9\sqrt{15} \, \text{cm}^2Area=15(15−12)(15−12)(15−6)​=915​cm2
Hence, the area is 915 cm29\sqrt{15} \, \text{cm}^2915​cm2.