Question

An isosceles triangle has a perimeter $30\mathrm{cm}$ and each of the equal sides is $12\mathrm{cm}$. Find the area of the triangle.

Answer 1

Given, the length of two equal sides of an isosceles triangle is $12\mathrm{cm}$.

Perimeter is $30\mathrm{cm}$.

Step 1: Find the third side of the triangle

Let $\mathrm{a},\mathrm{b},\mathrm{c}$ are sides of the triangle and $\mathrm{a}=\mathrm{b}=12\mathrm{cm}$

We know that the perimeter of a triangle is equal to the sum of three sides of the triangle.

$$\begin{gathered} \mathrm{perimeter}=\mathrm{a}+\mathrm{b}+\mathrm{c} \\ \Rightarrow 30=12+12+\mathrm{c} \\ \Rightarrow 30=24+\mathrm{c} \\ \Rightarrow \mathrm{c}=30-24 \\ \Rightarrow \mathrm{c}=6 \end{gathered}$$

Thus the length of the third side of the triangle is $6\mathrm{cm}$.

Step 2: Find the area of the triangle

By using Heron’s formula,

The area of a triangle $=\sqrt{\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)}$

where $\mathrm{a},\mathrm{b},\mathrm{c}$ are the sides of the triangle and $\mathrm{s}$ is the semi perimeter of the triangle.

$\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}=\frac{12+12+6}{2}=\frac{30}{2}=15$

$\mathrm{s}=15\mathrm{cm}$

$\therefore \mathrm{area}=\sqrt{15\left(15-12\right)\left(15-12\right)\left(15-6\right)}$

$$\begin{gathered} =\sqrt{15\times 3\times 3\times 9} \\ =\sqrt{15\times 81} \\ =9\sqrt{15}{\mathrm{cm}}^2 \end{gathered}$$

Therefore, the area of the triangle is $9\sqrt{15}{\mathrm{cm}}^2$.