Question

The perimeter of a right angled triangle is $30cm.$ If the hypotenuse is $13cm,$find its area using Heron's formula.

Answer 1

Step 1: Find the sides of the triangle

Let the other two sides i.e. base and perpendicular of a right angled triangle be $p$and $q$ respectively.

Given, Perimeter of a right angled triangle $=30cm.$

$$\begin{gathered} \Rightarrow p+q+13=30 \\ \Rightarrow p+q=30-13 \\ \Rightarrow p+q=17 \\ \Rightarrow q=17-p \end{gathered}$$

We know that, In a right angled triangle ,

$$\begin{gathered} {\left(\mathrm{Hypotenuse}\right)}^2={\left(\mathrm{Base}\right)}^2+{\left(\mathrm{Perpendicular}\right)}^2\left(\mathrm{Pythagoras}\mathrm{Theorm}\right) \\ {\left(13\right)}^2={\left(p\right)}^2+{\left(17-p\right)}^2 \\ 169=p^2+289+p^2-34p \\ 2p^2-34p+120=0 \\ {p}^2-17p+60=0 \\ {p}^2-12p-5p+60=0 \\ p\left(p-12\right)-5\left(p-12\right)=0 \\ \left(p-5\right)\left(p-12\right)=0 \\ p-5=0,p-12=0 \\ p=5,p=12 \end{gathered}$$

When $p=12,q=5$

and $p=5,q=12$

So , the other two sides are $5cm$and $12cm.$

Step 2: Find the area of the triangle

Area of triangle using Heron's Formula = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ (Where $s$ is the semi perimeter and $a,b,c$ are sides of triangle)

Here , $a=13cm,b=12cm,c=5cm.$

$$\begin{gathered} s=\frac{a+b+c}{2} \\ =\frac{13+12+5}{2} \\ =\frac{30}{2} \\ =15cm \end{gathered}$$

Substituting the values in the given formula ,

Area of the triangle

$$\begin{gathered} =\sqrt{15\times \left(15-13\right)\times \left(15-12\right)\times \left(15-5\right)} \\ =\sqrt{15\times 2\times 3\times 10} \\ =\sqrt{900} \\ =\sqrt{30\times 30} \\ =30c{m}^2 \end{gathered}$$

Hence, the area of a right angled triangle using heron's formula is $30c{m}^2$.