Question

Find the area of the triangle two sides of which are $18cm$ and $10cm$ and the perimeter is $42cm$.

Answer 1

Solve for area of the given triangle:

Given, the two sides of the triangle are $18cm$ and $10cm$ and perimeter is $42cm$

Step 1: Find the length of the third side

Let us assume, $a=18$, $b=10$ and length of the third side be $c$.

We know that, perimeter of triangle$=a+b+c$

$$\begin{gathered} \therefore 42=18+10+c \\ \Rightarrow c=42-28 \\ =14 \end{gathered}$$

Hence, the length of the third side $c$ is $14cm$.

Step 2: Find area of the triangle:

By using Heron's formula, area of triangle$\mathbf{=}\sqrt{\mathbf{s}\left(s-a\right)\left(s-b\right)\left(s-c\right)}$, where $\mathit{s}\mathbf{=}\frac{\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{+}\mathbf{c}}{\mathbf{2}}$.

Hence, $s=\frac{42}{2}=21$ $\left[\because a+b+c=42cm\right]$

$\therefore$ Area of the given triangle $=\sqrt{21\left(21-18\right)\left(21-10\right)\left(21-14\right)}$

$=\sqrt{21\times 3\times 11\times 7}$

$\Rightarrow$Area of the given triangle $=21\sqrt{11}c{m}^2$

Hence, the area of the given triangle is $21\sqrt{11}c{m}^2$.