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 the perimeter is 180 cm, what will be the area of the signal board?
Solution: Let 2s be the perimeter of the signal board, where 2s = 180 , then
We know,
2s = a + b + c [ ∵ a = b = c]
or, 2s = a + a +a = 3a ……(1)
or, s = (3a)/2
By using,Heron’s formula
area of triangle, Δ is,
Now, from (1)
2s = 3a
or, 180 = 3a
or, a = 180/3 = 60 cm
Hence, Δ =( √3/4) × a² = (√3/4) × 60²
∴ area of the triangle is 900√3 cm²
2). The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122m, 22m and 120m in figure below. The advertisements yield an earning of Rs. 5000 per m² per year. A company hired both walls for 3 months. How much rent did it pay?
Solution: Given in figure,
Lengths of the sides of the walls are,
a = 122m, b = 120m and c = 22m
We have,
122² = 120² + 22²
So, walls are in the form of right triangles.
∴ Area of two walls = 2 × [ 1/2 × Base × Height]
∵ Base = 120m, Height = 22m
or, Area = 2 × 1/2 × 120 × 22 = 2640 m²
Now,
Yearly rent = Rs 5000 per m²
∴ Monthly rent = 5000/12 Rs.
Hence, rent paid by the company for 3 months = 3 × 5000/12 × 2640
= Rs 33 00000 Ans.
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” in figure below. If the sides of the wall are 15m, 11m and 6m, find the area painted in colour.
Solution: From figure, given
sides are a = 15m, b = 11m , c = 6m
We know,
Perimeter, 2s = a + b + c = 15 + 11 + 6 = 32
or, s = 32/2 = 16
Now,
Area to be painted in colour = Area of the side wall
= √[ s ( s - a)( s - b)(s - c)]
= √[ 16( 16 - 15)(16 - 11)(16 - 6)]
= √ [ 16 × 1 × 5 ×10]
= √ [ 2⁴ × 1 × 5 × 2 × 5] = 2² × 5 × √2
= 20√2 m²
Hence, the area to be painted in colour = 20√2 m²
4). Find the area of a triangle, two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Solution : Let a, b.c be the sides of the given triangle and 2s be its perimeter.
where, a = 18cm, b = 10cm and 2s = 42cm or s = 42/2 = 21 cm.
Now, a + b + c = 2s [ perimeter of Δ]
∴ 18 + 10 + c = 42
or, 28 + c = 42
or, c = 42 - 28 = 14 cm
Hence, Area of Δ
= √[ s ( s - a)( s - b)( s - c)]
= √[ 21( 21 - 18)(21 - 10) ( 21 - 14)]
= √ [ 21 × 3 × 11 × 7]
= √ [ 7 × 3 × 11 × 3 × 7]
= √[ 7² × 3² ×11] = 7 × 3 × √11 = 21√11
Hence, the area of the triangle is
21√11 cm². Ans
5). Sides of a triangle are in the ratio of 12:17:25 and its perimeter is 540 cm . find its area.
Solution :
Given, Sides = a : b : c = 12 : 17 : 25, perimeter, 2s = 540 , s = 540/2 = 270 cm
Now,
a = 12 k, b = 17 k, c = 25 k
We know,
perimeter, 2s = a + b + c ………(1)
Substitute, a = 12k, b = 17k, c = 25k and 2s = 540 in (1)
540 = 12k + 17k + 25k
or, 54k = 540
or, k = 540/54 = 10
∴ a = 12k = 12 × 10 = 120 cm
b = 17k = 17 × 10 = 170 cm
c = 25k = 25 × 10 = 250 cm
∵ Area of triangle ,
= [ s( s - a) ( s - b )( s - c)] ^ (1/2) ….(2)
Substitute, a, b,c and s in (2)
Area of triangle, A
A = [ 270 ( 270 - 120)( 270 - 170)(270 -250)] ^(1/2)
= [ 270 × 150 × 100 × 20] ^(1/2)
= [ 30 × 3 × 3 × 5 × 30 × 5 × 20 × 20 ] ^(1/2)
= 30 × 3 × 5 × 20 = 9000 cm²
Hence, the area of the triangle is 9000 cm².
6). An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Soluion: Let the length of the unequal sides be x cm.
Given, two equal sides are, 12cm and 12 cm, perimeter, 2s = 30 cm
∴ s = 30/2 = 15 cm
We know, perimeter = sum of all sides
or, 2s = x + 12 + 12
or, 30 = x + 12 + 12
or, x = 30 - 24 = 6 cm
Now, a = 12 cm, b = 12 cm, c = 6cm, and s = 15 cm .
By using, Heron's formula
Area of triangle is,
√ [ s( s - a)(s - b) ( s - c)]
= √ [ 15( 15 - 12) ( 15 - 12)( 15 -6)]
= √ [ 15 × 3 × 3 × 9]
= √ [ 5 × 3 × 3 × 3 × 3 × 3]
= √ [ 15 × 3² × 3² ] = 9 √15 cm²
Hence, the area of the triangle is 9 √15 cm²
