# Ex.12.1 Q5 Heron’s Formula Solution - NCERT Maths Class 9

## Question

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

## Text Solution

What is known?

Ratio of sides of the triangle and its perimeter.

What is unknown?

Area of the triangle.

Reasoning:

By using Heron’s formula we can calculate the area of triangle.

The formula given by Heron about the area of a triangle

$$=\sqrt{s(s-a)(s-b)(s-c)}$$

Where $$a, b$$ and $$c$$ are the sides of the triangle, and

\begin{align}s &= \text{Semi-perimeter}\\& = \begin{Bmatrix} \text{Half the Perimeter } \\ \text{ of the triangle} \;\end{Bmatrix} \\&=\frac{(a+b+c)}{2}\end{align}

Steps:

Suppose the sides are $$12x\;\rm {cm}$$, $$17x\;\rm {cm}$$ and $$25x\rm\; {cm}$$.

Perimeter of the triangle $$=$$ $$540\; \rm{cm}$$

\begin{align}{12 x+17 x+25 x=540} \\ {54 x=540} \\ {x=\frac{540}{54}} \\ {x=10 \mathrm{cm}}\end{align}

Therefore sides of triangle:

\begin{align}&12 x=12 \times 10=120 \;\mathrm{cm},\\ & 17 x=17 \times 10=170 \;\mathrm{cm}, \\ & 25 x=25 \times 10=250 \;\mathrm{cm}\end{align}

$$a = 120\; {\rm{cm}} ,\\ b = 170\;{\rm{cm}},\\ c = 250\;{\rm{cm}}$$

\begin{align}s &= \text{Half the Perimeter}\\ s&=\frac{540}{2}\\&=270 \;\mathrm{cm}\end{align}

By using Heron’s formula,

Area of a triangle

\begin{align}&={\sqrt{s(s-a)(s-b)(s-c)}} \\&=\!\sqrt{270(270\!-\!120)\!(270\!-\!170)\!(270\!-\!250)}\\ &=\sqrt{270 \times 150 \times 100 \times 20} \\ &=\sqrt{81000000} \\ &=9000 \mathrm{cm}^{2}\\\end{align}

Area of a triangle $$=$$$$9000 \;\rm{cm^2}.$$

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