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## 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? It is given that each side of the equilateral triangle is a.
Therefore, its perimeter
= a + a + a = 180cm
= 3a = 180cm
= a = 60cm

Therefore the semi-perimeter of the equilateral triangle,
s = (a + a + a)/2
s = 3a/2

Area of triangle
= √[s (s - a) (s - b) (s - c)]
= √[3/2a (3a/2 - a) (3a/2 - a) (3/2a - a)]
= √[3/2a x a/2 x a/2 x a/2]
= √[3/4 a²]

On substituting a = 60cm, the area of triangle
= √3/4 (60)²
= √3/4 3600
= 900√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 ₹5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay? The length of sides of the triangular side wall of a flyover are 122m, 22m and 120m.
Therefore the semi-perimeter,
s = a + b + c/2
= s = 122 + 22 + 120/2
= s = 264/2
= s = 132 m

The area of the triangular side wall
= √[s (s - a) (s - b) (s - c)]
= √[132 (132 - 122) (132 - 22) (132 - 120)]
= √[132 x 10 x 110 x 12]
= √[17,42,400]
= 1320

Rent for 1 year (i.e. 12 months) per m² = ₹5000
∴ Rent for 3 months per m² =  ₹5000 x 312
= Rent for 3 months for 1320 m² = ₹5000 x 312 x 1320
= Rent for 3 months for 1320 m² = ₹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 15 m, 11 m and 6 m, find the area painted in colour. Let the sides of the wall be a = 15 m, b = 11 m, c = 6 m
Therefore semi-perimeter, s = a + b + c/2
s = 15 + 11 + 6/2
s = 32/2
s = 16 m

Area that needs to be painted
= √[s (s - a) (s - b) (s - c)]
= √[16 (16 - 15) (16 - 11) (16 - 6)]
= √[16 x 1 x 5 x 10]
= √800
= 20√2 m²

Thus, the required area painted in colour is 20√2 m².

Question 4: Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.

Let the sides of the triangle be a = 18 cm, b = 10 cm and c = x cm.
Since, perimeter of the triangle = 42 cm
(Therefore), 18 cm + 10 cm + x = 42
x = 42 - (18 + 10)
x = 14 cm

Now, semi-perimeter, s = a + b+ c/2
s = 18 + 10 + 14/2
s = 42/2
s = 21cm

Area of triangle
= √[s (s - a) (s - b) (s - c)]
= √[21 (21 - 18) (21 - 10) (21 - 14)]
= √[21 x 3 x 11 x 7]
= √4851
= 21√11 cm²

Thus, the area of the triangle is  21√11 cm².

Question 5: Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.

Let the sides of the triangle be x. Therefore, the sides of triangle are
a = 12x cm, b = 17x cm, c = 25x cm

Since, perimeter of the triangle = 540 cm
(therefore) 12x + 17x + 25x = 540
54x = 540
x = 10

∴ a = (12 x 10) cm = 120cm,
b = (17 x 10) cm = 170 cm
and c = (25 x 10) cm = 250 cm

Now the semi-perimeter, s = a + b +c/2
s = 120 + 170 +250/2
s = 540/2
s = 270 cm

Area of triangle
= √[s (s - a) (s - b) (s - c)]
= √[270 (270 - 120) (270 - 170) (270 - 250)]
= √[270 x 150 x 100 x 20]
= √81000000
= 9000 cm²

Thus, the area of the triangle is  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.

Let the sides of an isosceles triangle be a = 12cm, b = 12cm, c = x cm
Since, the perimeter of the triangle = 30 cm
(therefore) 12 cm + 12 cm + x cm = 30 cm
x + 24 = 30
x = 30 - 24
x = 6

Now the semi-perimeter, s = a + b + c/2
s = 12 + 12 + 6/2
s = 30/2
s = 15 cm

Area of triangle
= √[s (s - a) (s - b) (s - c)]
= √[15 (15 - 12) (15 - 12) (15 - 6)]
= √[15 x 3 x 3 x 9]
= √1215
= 9√15 cm²

Thus, the area of the triangle = 9√15 cm².