Question

The percentage increase in the area of a triangle if each of its side is doubled is

(a) 200% 

(b) 50% 

(c) 300% 

(d) 100%

Answer 1

(c) 300%.

Let a, b, c be the sides of the original triangle and s be its semi-perimeter.

Then, s = \(\frac{1}{2}\)(a+b+c)

Let s₁ be the semi-perimeter of the new triangle. Then,

s₁ = \(\frac{1}{2}\) (2a + 2b + 2c) = (a + b + c) = 2s

∴ Area A of original Δ = \(\sqrt{s(s-a)(s-b)(s-c)}\) and 

Area A₁ of new Δ = \(\sqrt{s_1(s_1-2a)(s_1-2b)(s_1-2c)}\)

= \(\sqrt{2s(2s-2a)(2s-2b)(2s-2c)}\)

= 4\(\sqrt{s(s-a)(s-b)(s-c)}\) = 4A.

∴ Percentage increase in area = \(\bigg(\frac{A_1-A}{A}\times100\bigg)\)%

= \(\frac{(4A-A)}{A}\times100\)% = 300%.