These NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Exercise 12.1

Question 1.

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

Solution:

In first case, length of each sides of an equilateral triangle is a

s = \(\frac{a+b+c}{2}=\frac{3 a}{2}\)

Therefore by Heron’s formula-

or, area of triangle = \(\frac{\sqrt{3} \cdot a^{2}}{4}\)

In case II,

Perimeter of an equilateral triange is 180 cm

∴ One side = \(\frac{180}{3}\) = 60 cm

s = \(\frac{60+60+60}{2}\) = 90 cm.

∴ Area of signal board = \(\sqrt{s(s-a)(s-b)(s-c)}\)

= \(\sqrt{90(90-60)(90-60)(90-60)}\)

= \(\sqrt{90 \times 30 \times 30 \times 30}\)

= 900√3 cm^{2}

or, Area of signal board = 900√3 cm^{2}.

Question 2.

The triangular side walls of a flyover have been used for advertisements. The sides of the wall are 122 m, 22 m, and 120 m (see fig. 12.9). The advertisements yield an earning of Rs. 5000 per m^{2} per year. A company hired one of its walls for 3 months. How much rent did it pay?

Solution:

The sides of the triangular wall are 122 m, 22 m, and 120 m.

∴ s = \(\frac{120+22+120}{2}\) = 132 m.

Therefore, by Heron’s Formula.

Area of triangular wall

= \(\sqrt{132(132-122)(132-122)(132-122)}\)

= \(\sqrt{132 \times 10 \times 110 \times 12}\)

= 1320 m^{2}

∴ Rent = 5000 × 1320 × \(\frac{3}{12}\) = 16,50,000

Therefore, rent of this wall for 3 months is Rs. 16,50,000

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. 12.10 ). If the sides of the wall are 15m, 11m, and 6m, find the area painted in colour.

Solution:

The sides of the wall are 15 m, 11 m, and 6 m.

∴ s = \(\frac{15+11+6}{2}\) = 16 m

∴ By Heron’s formula, we know that

Area of the wall

Therefore, the Area of the wall is 20√2 m^{2}.

Question 4.

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

Solution:

Let third side of triangle is x cm.

∴ Perimeter = 18 + 10 + x

⇒ 42 = 18 + 10 + x

⇒ x = 14 cm

s = \(\frac{18+10+14}{2}=\frac{42}{2}\) = 21

By Heron’s formula we know that

Area of triangle

Therefore, the area of the triangle = 2√11 cm^{2}.

Question 5.

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

Solution:

Let the first side of triangle = 12x

Second side of triangle = 17x

and third side of triangle = 25x

Therefore,

perimeter of triangle = 12x + 17x + 25x

⇒ 540 = 54x

⇒ x = 10

First side of triangle = 12x = 12 × 10 = 120 cm

Second side of triangle = 17x = 17 × 10 = 170 cm

and Third side of triangle = 25x = 25 × 10 = 250 cm

∴ s = \(\frac{120+170+250}{2}=\frac{540}{2}\) = 270

∴ By Heron’s formula we know that

Area of triangle

Therefore, area of triangle = 9000 cm^{2}

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:

Let the third side of an isosceles triangle is x.

∴ Perimeter of triangle = 12 + 12 + x

or, 30 = 24 + x

or, x = 6x m.

The third side of an isosceles triangle is 6 cm.

∴ s = \(\frac{12+12+6}{2}\) = 15

By Heron’s formula, we know that

Area of triangle

Therefore, the area of an isosceles triangle is 9√15 cm^{2}.