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 cm, what will be the area of the signal board?

Answer 1

Hint: If we are given a triangle with sides a, b and c and the semi - perimeter (obtained by dividing the perimeter by 2) s, then the area of this triangle can be found out by heron’s formula i.e. $\Delta =\sqrt{\left( s \right)\left( s-a \right)\left( s-b \right)\left( s-c \right)}$. Using this formula, we can solve this question.

Complete step by step solution:
Before proceeding with the question, we must know the formula that will be required to solve this question.
If we have a triangle with sides a, b, c and the semi - perimeter of this triangle is s, then by heron’s formula, the area of this triangle is given by the formula,
$\Delta =\sqrt{\left( s \right)\left( s-a \right)\left( s-b \right)\left( s-c \right)}$ . . . . . . . . . . . . . . . (1)
In this question, we have a traffic signal board which is an equilateral triangle with all sides equal to ‘a’. The perimeter of this board is a + a + a = 3a. So, the semi - perimeter will be $s=\dfrac{3a}{2}$. Using heron’s formula, the area of this triangle will be,

& \Delta =\sqrt{\left( \dfrac{3a}{2} \right)\left( \dfrac{3a}{2}-a \right)\left( \dfrac{3a}{2}-a \right)\left( \dfrac{3a}{2}-a \right)} \\\ & \Rightarrow \Delta =\sqrt{\left( \dfrac{3a}{2} \right)\left( \dfrac{a}{2} \right)\left( \dfrac{a}{2} \right)\left( \dfrac{a}{2} \right)} \\\ & \Rightarrow \Delta =\sqrt{\left( \dfrac{3{{a}^{4}}}{{{2}^{4}}} \right)} \\\ \end{aligned}$$ $$\Rightarrow \Delta =\dfrac{\sqrt{3}{{a}^{2}}}{4}$$ . . . . . . . . . . . . (2) It is given that the perimeter of this signal board is 180 cm. So, the semi – perimeter will be $\dfrac{180}{2}=90$ cm. Also, the semi – perimeter of the equilateral triangle $s=\dfrac{3a}{2}$. So, we have, $\begin{aligned} & \dfrac{3a}{2}=90 \\\ & \Rightarrow a=60 \\\ \end{aligned}$ Substituting this value of a in equation (2), we get, $$\begin{aligned} & \Delta =\dfrac{\sqrt{3}{{\left( 60 \right)}^{2}}}{4} \\\ & \Rightarrow \Delta =\dfrac{\sqrt{3}\times 3600}{4} \\\ & \Rightarrow \Delta =900\sqrt{3}c{{m}^{2}} \\\ \end{aligned}$$ Hence, the area of the signal board is $$900\sqrt{3}c{{m}^{2}}$$. Note: There is a possibility that one may commit a mistake while using heron’s formula. It is a very common mistake that one does not convert the perimeter into the semi – perimeter which leads us to an incorrect answer.