Question

Sides of a triangle are in the ratio of $12:17:25$ and its perimeter is$540cm$. Find its area.

Answer 1

Area of the triangle :

The ratio of the sides of the triangle is given as$12:17:25$

Perimeter is$540\mathrm{cm}$

By using Heron’s formula,

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

Let’s assume the common ratio between the sides of the triangle be$\mathrm{x}$

Then, the sides are$12\mathrm{x},17\mathrm{x}$ and $25\mathrm{x}$

The perimeter of the triangle is $540\mathrm{cm}$

$$\begin{gathered} 12\mathrm{x}+17\mathrm{x}+25\mathrm{x}=540\mathrm{cm} \\ \Rightarrow 54\mathrm{x}=540\mathrm{cm} \\ \Rightarrow \mathrm{x}=10 \end{gathered}$$

Hence, the sides of a triangle are $120\mathrm{cm},170\mathrm{cm},250\mathrm{cm}.$

$$\begin{gathered} \mathrm{a}=250\mathrm{cm} \\ \mathrm{b}=120\mathrm{cm} \\ \mathrm{c}=170\mathrm{cm} \end{gathered}$$

Semi-perimeter of the triangle:

$$\begin{gathered} \mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2} \\ \mathrm{s}=\frac{250+120+170}{2} \\ \mathrm{s}=270\mathrm{cm} \end{gathered}$$

Area of the triangle:

$\begin{array}{rcl}\mathrm{ar}(△\mathrm{ABC})& =& \sqrt{270(270-250)(270-120)(270-170)}\\ & =& \sqrt{270\times \left(20\right)\times \left(150\right)\times \left(100\right)}\\ & =& 9000{\mathrm{cm}}^2\end{array}$

Hence, the triangle's area is determined as $9000{\mathrm{cm}}^2$