Question

Prove that the area of an equilateral triangle described on a side of a right-angled isosceles triangle is half the area of the equilateral triangle described on its hypotenuse.

Answer 1

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✒ GiVen :

• In △ABC, ∠ABC = 90°
• AB = AC
• △ABD and △ACE are equilateral triangle's.

✒ To Prove :

• Ar(△ABD) = 1/2 × Ar(△CAE)

✒ SoluTion :

• Let AB = AC = x units.

Also,

Hypotenuse = CA = $\sf \sqrt{ {x}^{2} + {x}^{2} }$ = $\sf x \sqrt{2}$ units.

We know that,

• △ABD and △CAE are equilateral triangle's, with angle = 60°.

$\sf\therefore \: \triangle \: ABD \sim \triangle \: CAE$

Also, the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

So,

$\sf \longrightarrow \: \dfrac{ar( \triangle \: ABD )}{ar( \triangle \: CAE)} \\ \\ \sf \longrightarrow \: \dfrac{ {AB}^{2} }{ {CA}^{2} } \\ \\ \sf \longrightarrow \: \dfrac{ {x}^{2} }{ ({x \sqrt{2}})^{2} } \\ \\ \sf \longrightarrow \: \dfrac{ {x}^{2} }{2 {x}^{2} } \\ \\ \sf \longrightarrow \: \dfrac{1}{2}$

Hence, Ar(△ABD) = 1/2 × Ar(△CAE).

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