- 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:
Each side of a triangle = a
Perimeter of a triangle = 3a
s = 3a/2
Area of the signal traingel signal board = √[s(s-a)(s-b)(s-c)] [sign board is an equilateral triangle a = b = c ]
= √[s(s-a)(s-a)(s-a)
= (s – a) √[s(s-a)]
= (3a/2 – a) √[3a/2(3a/2 – a)]
= a/2 √(3a^2/4)
=a/2 . a/2 .√3
= a^2/4 . √3
Thus, the area of the signal sign board = a^2/4. √3 sq. units
Therefore, perimeter = 180cm (given)
⇒each side of the triangle = 180/3 = 60 cm
⇒ area of the triangle = (60)^2/4 x √3 cm2 = 900√3 cm2
- 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:
Given sides of the triangular walls are 122m, 22m , 120m i.e., a = 122m , b = 22m , c = 120m
Therefore, s = 3a/2 = 122+22+120/2 = 132 m
Area of the triangular side wall = √[s(s-a)(s-b)(s-c)
= √[132(132 – 122)(132 – 22)(132 – 120)] m2
= √[132(10)(110)(12) m2
= 1320 m2
Given: The advertisements yield an earning of Rs 5000 per m2 per year. i.e., the advertisement earning of 1m2 of the wall for one year is Rs. 5000
⸫ advertisement earning of 1m2 wall for 1month = Rs. 5000/12
⸫ advertisement earning of 1m2 of the wall for 3 months = Rs. 5000/12 x 3
⸫ advertisement earning of complete wall for 3 months = Rs. 5000/12 x 3 x 1320 = Rs. 15, 50, 000
- 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. Fig. 12.10
Solution:
The sides of the walls are 15m , 11m and 6m
Therefore, s = 15+11+6/2 = 32/2 = 16m
Area of the triangular walls = √[s(s-a)(s-b)(s-c)] [a= 15m, b = 11m, c = 6m]
=√[16(16-15)(16-11)(16-6)]
=√[16x1x5x10]
=4×5√2
= 20√2
- Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.
Solution:
Perimeter = 42m
Two sides of the triangle are 18cm and 10 cm , i.e., a = 18cm, b = 10cm
Perimeter = a + b + c
42 = 18 + 10 + c
c = 42 – 18 – 10 = 14m
Therefore, s = a+b+c/2 = 18+10+14/2 = 42/2 = 21cm
Area of the triangle = √[s(s-a)(s-b)(s-c)
= √[21(21-18)(21-10)(21-14)] cm2
=√[21x3x11x7] cm2
=√[7x3x3x11x7] cm2
= 21√11cm2
- Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find it’s area.
Solution:
Sides of the triangle are in the ratio 12:17:25 i.e., a = 12x , b = 17x, c = 25x
Perimeter = 540cm = 12x + 17x + 25x
54x = 540
x = 540/54 = 10cm
Thus, a = 120 , b = 170 and c = 250
s =a+b+c/2 = 540/2 = 270 cm
Area of the triangle = √[s(s-a)(s-b)(s-c)
= √[270(270-120)(270-170)(270-250)
=√[270x150x100x20]
= √30x9x30x5x10x10x5x4]
=30x3x5x10x2
= 9000cm2
- An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Solution:
Perimeter of an isosceles triangle = 30 cm
Perimeter = a+b+c = 30cm
Each of the equal side is 12cm i.e., a = b = 12 cm
12+12+c = 30
12+12+c = 60
24 + c = 30
c = 30 – 24 = 6 cm
s = 30/2 = 15 cm
Area of the triangle = √[s(s-a)(s-b)(s-c)
=√[15(15-12)(15-12)(15-6)]
=√[15x3x3x9]
=√[3x5x9x9]
=9√15 cm2
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