Question

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

Answer 1

Let a, b, c be the sides of the old triangle and s be its semi-perimeter. Then,
$s=\frac{1}{2}(a+b+c)$
The sides of the new triangle are $2a,2b$ and $2c$
.Let s' be its semi-perimeter. Then,
$s^{\prime }=\frac{1}{2}\times (2a+2b+2c)$
$=a+b+c=2s$
Let $\Delta$ and ${\Delta }^{\prime }$ be the areas of the old and new traiangles respectively. Then,
$\Delta =\sqrt{s(s-a)(s-b)(s-c)}$ and
${\Delta }^{\prime }=\sqrt{s^{\prime }(s^{\prime }-2a)(s^{\prime }-2b)(s^{\prime }-2c)}$
$\Rightarrow {\Delta }^{\prime }=\sqrt{2s(2s-2a)(2s-2b)(2s-2c)}$ $[\because s^{\prime }=2s]$
${\Delta }^{\prime }=4\sqrt{s(s-a)(s-)(s-c)}=4\Delta$
$\therefore$ Increase in the are of the triangle
$={\Delta }^{\prime }-\Delta =4\Delta -\Delta =3\Delta$
Hence, percentage increase in area
$=(\frac{3\Delta }{\Delta }\times 100)=300\%$