Question

Find the area of the triangle whose perimeter is $22cm$. One side measures $9cm$ and the difference of other two sides is $3cm$.

Answer 1

Step 1: Solve for other two sides of triangle

Given,

1. One side of Triangle $c=9cm$
2. Perimeter of Triangle $a+b+c=22cm$
3. Difference of the other two sides of Triangle $a-b=3cm$

$$\begin{gathered} a+b+c=22 \\ \Rightarrow a+b=22-c \\ \Rightarrow a+b=13....\left[i\right] \\ a-b=3cm....\left[ii\right] \end{gathered}$$

Solving $\left[i\right]$ and $\left[ii\right]$,

$a=8cm$ and $b=5cm$

Step 2: Solve for area of triangle

Heron's formula: area of the triangle $=\sqrt{\mathbf{s}\mathbf{(}\mathbf{s}\mathbf{-}\mathbf{a}\mathbf{)}\mathbf{(}\mathbf{s}\mathbf{-}\mathbf{b}\mathbf{)}\mathbf{(}\mathbf{s}\mathbf{-}\mathbf{c}\mathbf{)}}\mathbf{}$, where $\mathit{s}\mathbf{=}\frac{\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{+}\mathbf{c}}{\mathbf{2}}$

Here, $s=\frac{8+5+9}{2}=11$

Therefore,

Area of the required triangle $=\sqrt{11\left(11-8\right)\left(11-5\right)\left(11-9\right)}$

$=\sqrt{11\times 3\times 6\times 2}=6\sqrt{11}\approx 19.90c{m}^2$

Hence, area of the triangle is $19.90c{m}^2$