Question

Prove that the area of the equilateral triangle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the equilateral triangles drawn on the other two sides of the triangle.

Answer 1

Let a right triangle BAC in which ∠A is right angle and AC = y, AB = x.

Three equilateral triangles ΔAEC, ΔAFB and ΔCBD are drawn on the three sides of ΔABC.

Again, let area of triangles made on AC, AB and BC are A₁, A₂ and A₃, respectively.

To prove: A₃ = A₁ + A₂ 

Proof: In ΔCAB,

By pythagoras theorem,

BC² = AC² + AB²

⇒ BC² = y² + x²

⇒ BC = `sqrt(y^2 + x^2)`

We know that,

Area of an equilateral triangle = `sqrt(3)/4 ("Side")^2`

∴ Area of equilateral ΔAEC,

A₁ = `sqrt(3)/4 ("AC")^2`

⇒ A₁ = `sqrt(3)/4 y^2`   ...(i)

And area of equilateral ΔAFB,

A₂ = `sqrt(3)/4 ("AB")^2`

= `(sqrt(3)x^2)/4`   ...(ii)

Area of equilateral ΔCBD,

A₃ = `sqrt(3)/4 ("CB")^2`

= `sqrt(3)/4 (y^2 + x^2)`

= `sqrt(3)/4 y^2 + sqrt(3)/4 x^2`

= A₁ + A₂   ...[From equations (i) and (ii)]

⇒ A₃ = A₁ + A₂

Hence proved.