# Ex.12.1 Q1 Heron’s Formula Solution - NCERT Maths Class 9

## Question

A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘$$a$$’. Find the area of the signal board, using Heron’s formula. If its perimeter is $$180 \;\rm{cm}$$, what will be the area of the signal board?

## Text Solution

What is known?

Dimensions of the traffic signal board (triangle) and its perimeter.

#### What is Unknown?

Area of the signal board.

Reasoning:

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

The formula given by Heron about the area of a triangle

\begin{align}=\sqrt{s(s-a)(s-b)(s-c)}\end{align}

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:

Each side of traffic signal board (triangle)$$=‘a’ \;\rm{cm}$$

Perimeter of traffic signal board (triangle) $$=$$ sum of all the sides

\begin{align} &=a+a+a \\ &=3 a \end{align}

Semi Perimeter

\begin{align} s&=\frac{(a+b+c)}{2}\\&=\frac{a+a+a}{2}\\&=\frac{3 a}{2} \end{align}

By using Heron’s formula,

Area of a triangle

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

Area of a (triangle) traffic signal board

\begin{align}&=\sqrt{s(s-a)(s-b)(s-c)} \\& =\sqrt{s(s-a)(s-a)(s-a)} \\& =(s-a) \sqrt{s(s-a)}\quad....(1)\end{align}

We know \begin{align}s=\frac{3 a}{2}\end{align} so substituting in equation (1)

\begin{align}&{=\left(\frac{3 a}{2}-a\right) \sqrt{\frac{3 a}{2}\left(\frac{3 a}{2}-a\right)}} \\ &{=\left(\frac{a}{2}\right) \sqrt{\frac{3 a}{2}\left(\frac{a}{2}\right)}} \\ &{=\frac{a}{2} \times \frac{a}{2} \sqrt{3}} \\ &{=\frac{a^{2}}{4} \sqrt{3}}\end{align}

Area of the signal board

\begin{align}=\frac{a^{2}}{4} \sqrt{3} \;\; \text{sq. units}\end{align}

Now given perimeter $$=$$ $$180 \;\rm{cm}$$

Each side of triangle

\begin{align}&=\frac{{180}}{3}\,\mathrm{cm} \\ a &={60 \,\mathrm{cm}}\end{align}

Area of the signal board

\begin{align}&={\frac{a^{2}}{4} \sqrt{3}} \\ &={\frac{60^{2}}{4} \sqrt{3}}\end{align}

Area of the signal board

$$=900 \sqrt{3} \;\rm{cm^2}$$

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