Question

In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitude.

Answer 1

Given that ABC is an equilateral triangle. Therefore, its sides and angles are equal.

Hence, AB = BC = AC and \(\angle\)ABC = \(\angle\)ACB = \(\angle\)BAC = 60°.

Draw an altitude CD from C to the side AB.

Now, in right angle triangle ∆BDC,

⇒ 3AB² = 4CD² which implies that three times the square of one side is equal to four times the square of one of its altitude.