Question

If each side of a triangle is doubled, then find percentage increase in its area.

Answer 1

Side of original triangle be a,b,c
$\therefore s=\frac{a+b+c}{2}$
and area $\left(\Delta \right)=\sqrt{s(s-a)(s-b)(s-c)}$
Let the sides of new triangle be 2a, 2b,2c, then
$s^{\prime }=\frac{2a+2b+2c}{c}=\frac{2(a+b+c)}{2}$
$\Rightarrow 2s^{\prime }=a+b+c$
Now area $\left({\Delta }^{\prime }s\right)=\sqrt{2s(2s-2a)(2s-2b)(2s-2c)}$
= $\sqrt{2\times 2\times 2\times 2\times s(s-a)(s-b)(s-c)}$
= $4\sqrt{s(s-a)(s-b)(s-c)}=4\Delta$
$\therefore {\Delta }^{\prime }=4\Delta$
Now increase in area of the triangle = ${\Delta }^{\prime }-\Delta 4\Delta -\Delta =3\Delta$
$\therefore Percentageincrease=\frac{3\Delta \times 100}{\Delta }=300$