Question

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

• A. $200\%$
• B. $300\%$
• C. $400\%$
• D. $500\%$

Answer 1

The correct option is B $300\%$

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

S, $a+b+c/2$

$2s.a+b+c$ ……………..$\left(1\right)$

The sides of a new triangle are $2a,2b,2c$

[Given: side is doubled]

Let's be the new semi perimeter

$s^{\prime }=(2a+2b+2c)/2$

$s^{\prime }=2(a+b+c)/2$

$s^{\prime }=a+b+c$

$s^{\prime }=2S$( from equation $\left(1\right)$) …………..$\left(2\right)$

Let $\Delta =$ area of original triangle

$\Delta =\sqrt{s(s-a)(s-b)(s-c)}$ …………$\left(3\right)$

${\Delta }^{\prime }=$area of new triangle

${\Delta }^{\prime }=\sqrt{s^{\prime }(s^{\prime }-2a)(s^{\prime }-2b)(s^{\prime }-2c)}$

${\Delta }^{\prime }=\sqrt{2s(2s-2a)(2s-2b)(2s-2c)}$

From equation- $\left(2\right)$

${\Delta }^{\prime }=\sqrt{2s.2(s-a)2.(s-b)2(s-c)}$

${\Delta }^{\prime }=\sqrt{16s(s-a)(s-b)(s-c)}$

${\Delta }^{\prime }=4\sqrt{s(s-a)(s-b)(s-c)}$

${\Delta }^{\prime }=4\Delta$

Increase in the area of the triangle $={\Delta }^{\prime }-\Delta$

$=4\Delta -1\Delta$

$=3\Delta$.

$\%$ increase in are $=$ (increase in the area of the triangle/ original area of the triangle)$\times 100$

$\%$ incomes in area $=3\Delta /\Delta \times 100$

$\%$ increase in area $=3\times 100$

$\%$ increase in area $=300\%$

Hence the percentage increase in the area of a triangle is $300\%$.

$\therefore$ So, the answer is B. $300\%$.