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

We have,

ΔABC is an equilateral triangle

Then, AB = BC = AC

Let, AB = BC = AC = 2x

Since, AD ⊥ BC then BD = DC = x

In ΔADB, by Pythagoras theorem

AB² = (2x)² − (x)²

⇒ AD² = 4x² − x² = 3x²

⇒ AD = `sqrt3`x cm

Since, ΔABC and ΔADE both are equilateral triangles then they are equiangular

∴ ΔABC ~ ΔADE [By AA similarity]

By area of similar triangle theorem

`("area"(triangleADE))/("area"(triangleABC))="AD"^2/"AB"^2`

`=(sqrt3x)^2/(2x)^2`

`=(3x^2)/(4x^2)`

`=3/4`