We know that,

Each side of the equilateral triangle is equal.

Given,

Length of each side of an equilateral triangle = a cm.

Perimeter of traffic signal board (equilateral triangle) = sum of all the sides = a + a + a = 3a cm.

By formula,

Semi perimeter (s) = $\dfrac{\text{Perimeter of triangle}}{2} = \dfrac{3a}{2}$ cm.

By Heron's formula,

Area of triangle (A) = $\sqrt{s(s - a)(s - b)(s - c)}$ sq.units, where a, b and c are sides of triangle.

Substituting values we get :

$A = \sqrt{\dfrac{3a}{2} \times \Big(\dfrac{3a}{2} - a\Big) \times \Big(\dfrac{3a}{2} - a\Big) \times \Big(\dfrac{3a}{2} - a\Big)} \\[1em] = \sqrt{\dfrac{3a}{2} \times \Big(\dfrac{3a - 2a}{2}\Big) \times \Big(\dfrac{3a - 2a}{2}\Big) \times \Big(\dfrac{3a - 2a}{2}\Big)} \\[1em] = \sqrt{\dfrac{3a}{2} \times \dfrac{a}{2} \times \dfrac{a}{2} \times \dfrac{a}{2}} \\[1em] = \sqrt{\dfrac{3a^4}{16}} \\[1em] = \dfrac{\sqrt{3}}{4}a^2.$

Given,

Perimeter = 180 cm

∴ 3a = 180

⇒ a = $\dfrac{180}{3}$ = 60 cm.

Substituting value of a, we get :

$\text{Area of triangle (A)} = \dfrac{\sqrt{3}}{4}a^2 \\[1em] = \dfrac{\sqrt{3}}{4} \times 60^2 \\[1em] = \dfrac{\sqrt{3}}{4} \times 3600 \\[1em] = 900\sqrt{3}\text{ cm}^2.$

Hence, the area of the signal board is $900\sqrt{3}$ cm².

BRIGHT TUTORIALS

CBSE Class IX | Academic Year 2026-2027

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