Question

AD is an altitude of an equilateral triangle ABC. On AD as base, another equilateral triangle ADE is constructed. Prove that Area (ΔADE) : Area (ΔABC) = 3 : 4.

Answer 1

Area of the two triangles are in the ratio 3:4 proved.

Step-by-step explanation:

Given: ΔABC and ΔADE are equilateral.

Let's take AB = BC = CA = a units.

We know that altitude of an equilateral triangle is √3/2 times its side.

So AD = √3a/2

ΔABC ΔADE ~ (AAA similarity)

So area of ΔABC/ area of ΔADE = AD² / BC²

area of ΔABC/ area of ΔADE = (√3a/2)² / a² = 3/4

So area of the two triangles are in the ratio 3:4. Hence proved.