Let ABC be the equilateral triangle whose side is a cm.
Let us draw altitude AD(h) such that AD ⊥ BC.
We know that altitude bisects the opposite side.
So, BD = DC = a cm.
In ADC, ∠ADC = 90°.
We know that the Pythagoras Theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
So, by applying Pythagoras Theorem,
AC2 = AD2 + DC2
(a cm)2 = AD2 + (a/2 cm)2
a2 cm2 = AD2 + a2/4 cm2
AD2 = 3a2/4 cm2
AD = √3 a/2 cm = h
We know that area of a triangle = 1/2 × base × height
Ar(ΔABC) = 1/2 × a cm × √3 a/2 cm
⇒ ar(ΔABC) = √3 a2/4 cm2
ar(ΔABC) = √3 a2/4 cm2
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