# JAC Class 9 Maths Solutions Chapter 12 Heron’s Formula Ex 12.1

Jharkhand Board JAC Class 9 Maths Solutions Chapter 12 Heron’s Formula Ex 12.1 Textbook Exercise Questions and Answers.

## JAC Board Class 9th Maths Solutions Chapter 12 Heron’s Formula Ex 12.1

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Question 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?
Length of the side of equilateral triangle = a
Perimeter of the signal board = 3a = 180 cm
∴ 3a = 180 cm ⇒ a = 60 cm
Semi perimeter of the signal board 3a
S = $$\frac{3a}{2}$$

Using Herons formula,
Area of the signal board
= $$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{\left(\frac{3 a}{2}\right)\left(\frac{3 a}{2}-a\right)\left(\frac{3 a}{2}-a\right)\left(\frac{3 a}{2}-a\right)}$$
= $$\sqrt{\frac{3 a}{2} \times \frac{a}{2} \times \frac{a}{2} \times \frac{a}{2}}$$
= $$\sqrt{\frac{3 a^4}{16}}$$
= $$\frac{\sqrt{3}}{4}$$ a2
= $$\frac{\sqrt{3}}{4}$$ × 60 × 60
= 900 $$\sqrt{3}$$ cm2

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 (see Fig.). 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?

The sides of the triangle are 122 m, 22 m and 120 m.
Perimeter of the triangle is 122 + 22 + 120 = 264 m
Semi perimeter of triangle (s) = $$\frac{264}{2}$$
= 132 m
Using Herons formula,
= $$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{132(132-122)(132-22)(132-100)}$$
= $$\sqrt{132×10×110×12}$$ m2
= 1320 m2
Rent of advertising per year = ₹ 5000 per m2
Rent of one wall for 3 months
= ₹ $$\frac{1320 \times 5000 \times 300}{12}$$
= ₹ 1650000

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Question 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” (see Fig.). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Sides of the triangular wall are 15 m, 11 m and 6 m.
Semi perimeter of triangular wall
S = $$\frac{15+11+16}{2}$$ m = 16 m
Using Herons formula,
Area of the wall = $$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{16(16-15)(16-11)(16-6)}$$ m2
= $$\sqrt{16×1×5×10}$$ m2
= $$\sqrt{800}$$ m2
=$$20 \sqrt{2}$$ m2

Question 4.
Find the area of a triangle two sides of which are 18 cm and 10 cm and the pe-rimeter is 42 cm.
Two sides of the triangle are 18 cm and 10 cm
Perimeter of the triangle = 42 cm
Third side of triangle = 42 – (18 + 10) = 14 cm
Semi perimeter of triangle = $$\frac{42}{2}$$ = 21 cm
Using heron’s formula,
Area of the triangle
$$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{21(21 -18)(21 -10)(21 – 14)}$$ cm2
= $$\sqrt{21×3×11×7}$$ cm2
=$$21 \sqrt{11}$$ cm2

Question 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 the sides of the triangle = 12 : 17 : 25
Let the sides be 12x, 17x and 25x
Perimeter of the triangle = 540 cm
12x + 17x + 25x = 540 cm
⇒ 54x = 540 cm
⇒ x= 10
Sides of triangle are,
12x = 12 × 10 = 120 cm
17x = 17 × 10 = 170 cm
25x = 25 × 10 = 250 cm

Semi perimeter of triangle, S = $$\frac{540}{2}$$cm = 270 cm
Using Herons formula,
Area of the triangle = $$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{270(270 – 120)(270 – 170)(270 – 250)}$$
= $$\sqrt{270×150×100×20}$$ cm2
= 9000 cm²

Question 6.
An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
S = $$\frac{30}{2}$$ cm = 15 cm
Area of the triangle= $$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{15(15 – 12) (15 – 12) (15 – 6)}$$ cm2
= $$\sqrt{15×3×3×9}$$ cm2
=$$9 \sqrt{15}$$ cm2