HBSE 9th Class Maths Solutions Chapter 12 Heron’s Formula Ex 12.1

Haryana State Board HBSE 9th Class Maths Solutions Chapter 12 Heron’s Formula Ex 12.1 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions 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 Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?
Solution:
Let ABC be the signal board is shape of an equilateral triangle with side a.
s = \(\frac{a+a+a}{2}\)
⇒ s = \(\frac{3 a}{2}\)
By Heron’s formula, we have
Area of triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{\frac{3 a}{2}\left(\frac{3 a}{2}-a\right)\left(\frac{3 a}{2}-a\right)\left(\frac{3 a}{2}-a\right)}\)
= \(\sqrt{\frac{3 a}{2} \times \frac{a}{2} \times \frac{a}{2} \times \frac{a}{2}}\)
= \(\frac{a}{2} \times \frac{a}{2} \times \sqrt{3}\)
= \(\frac{a^2}{4} \times \sqrt{3}\)
= \(\frac{\sqrt{3}}{4}\) a² square units

Perimeter = 180 cm
⇒ 3a = 180
⇒ a = \(\frac{180}{3}\) = 60
∴ Area of triangle = \(\frac{\sqrt{3}}{4}\) × (60)²
= 900\(\sqrt{3}\) cm²
Hence, area of an equilateral triangle = \(\frac{\sqrt{3}}{4}\) a² square units and 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. 12.9). 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:
Sides of a triangular wall of a flyover are 122 m, 22 m and 120 m. Here,
a = 122 m, b = 22 m and c = 120 m
s = \(\frac{122+22+120}{2}\)
⇒ s = \(\frac{264}{2}\) = 132 m
By Heron’s formula, we have
Area of side walls of flyover
= \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{132(132-122)(132-22)(132-120)}\)
= \(\sqrt{132 \times 10 \times 110 \times 12}\)
= \(\sqrt{11 \times 12 \times 10 \times 10 \times 11 \times 12}\)
= 10 × 11 × 12
= 1320 m²
∵ Rent paid by a company in 12 months for 1 m² = Rs. 5000
∴ Rent paid by a company in 1 month for 1 m² = Rs. \(\frac{5000}{12}\)
∴ Rent paid by company in 3 months for 1320 m² = Rs. \(\frac{5000 \times 3 \times 1320}{12}\)
= Rs. 1250 × 1320
= Rs. 1650000
Hence, rent paid by company = Rs. 1650000.

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 15 m, 11 m and 6 m, find the area painted in colour.

Solution:
Sides of the triangular park are 15 m, 11 m and 6 m.
Here, a = 15 m, b = 11 m and c = 6 m
s = \(\frac{15+11+6}{2}\)
s = \(\frac{32}{2}\)
s = 16 m
By Heron’s formula, we have
Area of triangular park = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{16(16-15)(16-11)(16-6)}\)
= \(\sqrt{16 \times 1 \times 5 \times 10}\)
= \(\sqrt{4 \times 4 \times 1 \times 5 \times 5 \times 2}\)
= 4 × 5\(\sqrt{2}\)
= 20\(\sqrt{2}\) m²
Hence, area painted in colour of triangular park = 20\(\sqrt{2}\) m².

Question 4.
Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Solution:
We have,
Two sides of a triangle are a = 18 cm and b = 10 cm
and perimeter of triangle (2s) = 42 cm
We know that,
Perimeter (2s) = a + b + c
⇒ 42 = 18 + 10 + c
⇒ 42 = 28 + c
⇒ c = 42 – 28 = 14 cm
Now 2s = 42
s = \(\frac{42}{2}\) = 21 cm
By Heron’s formula, we have
Area of a triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{21(21-18)(21-10)(21-14)}\)
= \(\sqrt{21 \times 3 \times 11 \times 7}\)
= \(\sqrt{3 \times 7 \times 3 \times 11 \times 7}\)
= 3 × 7\(\sqrt{11}\)
= 21\(\sqrt{11}\) cm²
Hence,area of a triangle = 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.
Solution:
Ratio of sides of a triangle = 12 : 17 : 25 and perimeter is 540 cm.
Let sides of a triangle be 12x cm, 17x cm and 25x cm.
Perimeter = Sum of the sides of a triangle
⇒ 540 = 12x + 17x + 25x
⇒ 540 = 54x
x = \(\frac{540}{51}\) = 10
So,sides of the triangle = 12 × 10, 17 × 10, 25 × 10 = 120 cm, 170 cm, 250 cm
Here, a = 120 cm, b = 170 cm and c = 250 cm
s = \(\frac{120+170+250}{2}\)
⇒ s = \(\frac{540}{2}\)
⇒ s = 270 cm
By Heron’s formula, we have
Area of a triangler = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{270(270-120)(270-170)(270-250)}\)
= \(\sqrt{270 \times 150 \times 100 \times 20}\)
= \(\sqrt{3 \times 3 \times 30 \times 30 \times 5 \times 5 \times 20 \times 20}\)
= 3 × 5 × 20 × 30
= 9000 cm²
Hence,area of a triangle = 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.
Solution:
Two equal sides of an isosceles triangle are 12 cm and 12 cm and perimeter = 30 cm
Perimeter of a triangle = Sum of sides of a triangle
⇒ 30 = 12 + 12 + IIIrd side
⇒ 30 = 24 + IIIrd side
⇒ IIIrd side = 30 – 24 = 6 cm
Here, a = 12 cm, b = 12 cm and c = 6 cm
s = \(\frac{12+12+6}{2}\)
⇒ s = \(\frac{30}{2}\)
= 15 cm
By Heron’s formula, we have
Area of a triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{15(15-12)(15-12)(15-6)}\)
= \(\sqrt{15 \times 3 \times 3 \times 9}\)
= \(\sqrt{15 \times 3 \times 3 \times 3 \times 3}\)
= 3 × 3\(\sqrt{15}\)
= 9\(\sqrt{15}\) cm²
Hence, area of an isosceles triangle = 9\(\sqrt{15}\) cm².