Question

The percentage increase in the area of a triangle if its each side is quadrupled is equal to

• A. $1500$%
• B. $1200$%
• C. $900$%
• D. $800$%

Answer 1

The correct option is A $1500$%

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

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

$2s=a+b+c$ .......$\left(1\right)$

The sides of a new $\Delta$ are $4a,4b,4c$ [ given: Sides are quadrupled]

Let $s^{\prime }$ be the new semi perimeter.

$s^{\prime }=\frac{1}{2}(4a+4b+4c)=2a+2b+2c$

$\therefore s^{\prime }=4s$ ......$\left(2\right)$ from $\left(1\right)$

Let $\Delta =$ area of original triangle

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

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

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

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

$=\sqrt{256s(s-a)(s-b)(s-c)}$

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

$\therefore {\Delta }^{\prime }=16\Delta$ from $\left(3\right)$

Increase in the area of the triangle$={\Delta }^{\prime }-\Delta =16\Delta –\Delta =15\Delta$

$\%$ increase in area$=\frac{increaseintheareaofthetriangle}{originalareaofthetriangle}\times 100$

$\%$ increase in area$=\frac{15\Delta }{\Delta }\times 100$

$\%$ increase in area$=15\times 100=1500\%$

Hence, the percentage increase in the area of a triangle is $1500\%$