# Ex.12.1 Q4 Heron’s Formula Solution - NCERT Maths Class 9

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## Question

Find the area of a triangle two sides of which are $$18\;\rm{cm}$$ and $$10\;\rm{cm}$$ and the perimeter is $$42\;\rm{cm}$$.

## Text Solution

What is known?

Two sides of the triangle and its perimeter.

What is unknown?

Area of the triangle and one of its side.

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:

The sides of triangle given: $$a =18 \, {\rm{cm}} , b = 10 \, {\rm{cm}}$$

Perimeter of the triangle

\begin{align} &= a + b + c\\ {42}&={18+10+c} \\ {42}&={28+c} \\ {c}&={42-28} \\ {c}&={14}\end{align}

Semi Perimeter

\begin{align}s & =\frac{(a+b+c)}{2}\\ &=\frac{42}{2} \\ &=21 \mathrm{cm}\end{align}

By using Heron’s formula,

Area of a triangle

\begin{align}&={\sqrt{s(s-a)(s-b)(s-c)}} \\ &=\sqrt{ 21(21-18) (21-10) (21-14) } \\ &=\sqrt{21 \times 3 \times 11 \times 7} \\ &=21 \sqrt{11} \mathrm{cm}^{2}\end{align}

Area of a triangle $$=21\sqrt{11}\mathrm{cm}^{2}$$

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