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 diameter AB, BC and AC, respectively.
Again, let the areas of circles with diameters AB, BC and AC are respectively
A1,A2 and
A3.
To prove that
A3=A1+A2
Proof
In
ΔABC, by using Pythagoras theorem,
AC2=AB2+BC2
⇒ AC2=y2+x2
⇒ AC=√y2+x2
We know that, area of a semi – circle with radius,
r=πr22
∴ Area of semi – circle drawn on AC,
A3=π2(AC2)2=π2(√y2+x22)2
⇒ A3=π(y2+x2)8 …..(i)
Now, area of semi – circle drawn on AB,
A1=π2(AB2)2
⇒ A1=π2(y2)2⇒A1=πy28 ….(ii)
And area of semi – circle drawn on BC,
A2=π2(BC2)2=π2(x2)2
⇒ A2=πx28 .......(iii)
On adding Eq.s (ii) and (iii), we get
A1+A2=πy28+πx28
=π(y2+x2)8=A3 [from Eq.(i)]
⇒ A1+A2=A3