Question

Find the percentage increase in the area of a triangle if its each side is doubled.
(A). 50%
(B). 100%
(C). 300%
(D). 150%

Answer 1

Hint: In the above question we will have to know about the semi-perimeter of the triangle which is half its perimeter. Also, we will use the Heron’s formula to calculate the area of a triangle. The formulae that we will use are as below:
$\begin{align}
  & s=\text{semi-perimeter}=\dfrac{a+b+c}{2} \\
 & A=\text{area=}\sqrt{s(s-a)(s-b)(s-c)} \\
\end{align}$


Complete step-by-step solution -

Here a, b ,c are the sides of a triangle, s is semi- semi-perimeter and A is the area of a triangle.

Let us consider the sides of the triangle are x, y, z.
\[\begin{align}
  & s=\dfrac{x+y+z}{2} \\
 & A=\sqrt{s(s-x)(s-y)(s-z)} \\
\end{align}\]
When each side is doubled, the new sides are 2x, 2y , 2z.
 \[\begin{align}
  & \text{Hence, new s }\!\!'\!\!\text{ =}\dfrac{2x+2y+2z}{2}=2\left( \dfrac{x+y+z}{2} \right)=2s \\
 & \text{New area A }\!\!'\!\!\text{ =}\sqrt{2s(2s-2x)(2s-2y)(2s-2z)} \\
 & \sqrt{2\times 2\times 2\times 2\times s(s-x)(s-y)(s-z)} \\
 & 4\sqrt{s(s-x)(s-y)(s-z)}=4A \\
\end{align}\]
 \[\begin{align}
  & \therefore \%\text{ change in area =}\dfrac{A'-A}{A}\times 100 \\
 & =\dfrac{4\sqrt{s(s-x)(s-y)(s-z)}-\sqrt{s(s-x)(s-y)(s-z)}}{\sqrt{s(s-x)(s-y)(s-z)}}\times 100 \\
 & =\dfrac{4-1}{1}\times 100 \\
 & =300\% \\
\end{align}\]
Therefore, the correct option of the above question is option C.

Note: Remember Heron's formula to calculate the area of a triangle when the sides of a triangle are given in the question. Unlike other area formulae of a triangle, there is no need to calculate angles or other distances in the triangle first. Also it can be applied to any shape of triangle, as long as we know its three side lengths.
Also remember the formula of the semi-perimeter of a triangle.
Be careful while doing calculation as there is a chance that you might make a mistake and you will get the incorrect answer.