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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