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

Explanation

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 diametersAb,BC and Acare respectively `A_(1),A_(2) and A_(3)`
To prove `A_(3)=A_(1)+A_(2)`
Proof In `DeltaABC` by pythogoras theorem,
`AC^(2)=AB^(2)+BC^(2)`
`rArrAC^(2)+y^(2)+x^(2)`
`rArr AC=sqrt(y^(2)+x^(2))`
we know that, area of semi circle with radius, r`=(pir^(2))/2`
`therefore `Area of semicircle drawn on AC,`A_(3)=pi/2((AC)/2)^(2)=pi/2((sqrt(y^(2)+x^(2)))/2)^(2)`
`rArrA_(3)=(pi(y^(2)+x^(2)))/8`...(i)
Now, areea of semi circle drawn on AB, `A_(1)=pi/2((AB)/2)^(2)`
`rArrA_(1)=pi/2(y/2)^(2)rArrA_(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)`
`rArr A_(2)=(pix^(2))/8`
On adding Eqs. (ii) and (iii), we get `A_(1)+A_(2)=(piy^(2))/8+(pix^(2))/8`
`=(pi(y^(2)+x^(2)))/8=A_(3)`[from Eq.(i)]
`rArr A_(1)+A_(2)=A_(3)` Hence proved