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

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#### Solution

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

𝐴𝐵^{2} = (2𝑥)^{2} − (𝑥)^{2}

⇒ 𝐴𝐷^{2} = 4𝑥^{2} − 𝑥^{2} = 3𝑥^{2}

⇒ 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`

Concept: Areas of Similar Triangles

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