If each side of a triangle is doubled, the find percentage increase in its area.
Solution
The area of a triangle having sides a, b, c and s as semi-perimeter is given by,
`A = sqrt(s(s-a)(s-b)(s-c))`
Where,
`s = (a+b+c)/2`
`2s = a+b+c`
We take the sides of a new triangle as 2a, 2b, 2c that is twice the sides of previous one
Now, the area of a triangle having sides 2a, 2b, and 2c and s1 as semi-perimeter is given by,
`A_1= sqrt(s_1(s_1-2a)(s_1-2b)(s_1-2c))`
Where,
`s_1 = (2a+2b+2c)/2`
`s_1 = (2(a+b+c))/2`
s1 = a+ b+ c
s1 = 2s
Now,
`A_1 = sqrt(2s (2s-2a)(2s-2b)(2s-2c))`
`A_1 = sqrt(2s xx 2 (s-a) xx 2 (s-b) xx 2 (s-c))`
`A_1 = 4 sqrt(s(s-a)(s-b)(s-c))`
`A_1 = 4A`
Therefore, increase in the area of the triangle
=A1 -A
=4A-A
=3A
Percentage increase in area
`=(3A)/A xx 100 `
= 300%