Question

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

Answer 1

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 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 $\Delta ABC$, by Pythagoras theorem,
$A{C}^2=A{B}^2+B{C}^2$
$\Rightarrow A{C}^2=y^2+x^2$
$\Rightarrow AC=\sqrt{y^2+x^2}$
We know that, area of a semi – circle with radius, $r=\frac{\pi r^2}{2}$

$\therefore$ Area of semi – circle drawn on AC, $A_3=\frac{\pi }{2}={\left(\frac{AC}{2}\right)}^2=\frac{\pi }{2}{\left(\frac{\sqrt{y^2+x^2}}{2}\right)}^2$
$\Rightarrow A_3=\frac{\pi (y^2+x^2)}{8}$ …..(i)
Now, area of semi – circle drawn on AB, $A_1=\frac{\pi }{2}{\left(\frac{AB}{2}\right)}^2$
$\Rightarrow A_1=\frac{\pi }{2}{\left(\frac{y}{2}\right)}^2\Rightarrow A_1=\frac{\pi y^2}{8}A$ ….(ii)
And area of semi – circle drawn on BC, $A_2=\frac{\pi }{2}{\left(\frac{BC}{2}\right)}^2=\frac{\pi }{2}{\left(\frac{x}{2}\right)}^2$
$\Rightarrow A_2=\frac{\pi r^2}{8}$
On adding Eq.s (ii) and (iii), we get $A_1+A_2=\frac{\pi y^2}{8}+\frac{\pi x^2}{8}$
$=\frac{\pi (y^2+x^2)}{8}=A_3$ [from Eq.(i)]
$\Rightarrow A_1+A_2=A_3$