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

#### Solution

Let ABC be a right triangle, right-angled at B and AB = y, BC = x.

Three semi-circles are drawn on the sides AB, BC and AC, respectively with diameters AB, BC and AC, respectively.

Again, let area of circles with diameters AB, BC and AC are respectively A_{1}, A_{2} and A_{3}.**To prove:** `A_3 = A_1 + A_2`

**Proof:** In ΔABC, by Pythagoras theorem,

`AC^2 = AB^2 + BC^2`

⇒ `AC^2 = y^2 + x^2`

⇒ `AC = sqrt(y^2 + x^2)`

We know that, area of a semi-circle with radius, `r = (pir^2)/2`

∴ Area of semi-circle drawn on AC, `A_3 = pi/2((AC)/2)^2 = pi/2(sqrt(y^2 + x^2)/2)^2`

⇒ `A_3 = (pi(y^2 + x^2))/8` .......(i)

Now, area of semi-circle drawn on `AB, A_1 = pi/2 ((AB)/2)^2`

⇒ `A_1 = pi/2(y/2)^2`

⇒ `A_1 = (piy^2)/8` ......(ii)

And area of semi-circle drawn on BC, `A_2 = pi/2((BC)/2)^2 = pi/2(x/2)^2`

⇒ `A_2 = (pix^2)/8`

On adding equations (ii) and (ii), we get

`A_1 + A_2 = (piy^2)/8 + (pix^2)/8`

= `(pi(y^2 + x^2))/8`

= `A_3` .....[From equation (i)]

⇒ A_{1} + A_{2} = A_{3}

Hence proved.