In Fig. 6.52, semicircles have been drawn on all the sides of a right-angled triangle as shown. Class 9

In Fig. 6.52, semicircles have been drawn on all the sides of a right-angled triangle as shown. Class 9

Question 24.

In Fig. 6.52, semicircles have been drawn on all the sides of a right-angled triangle as shown. Class 9

Show that Area (A) + Area (B) = Area (C).

Solution:

Given:

A right angled triangle with semicircles drawn on all three sides.

To prove:

Area (A) + Area(B) = Area(C)

Proof:

Let the sides of the right angled triangle be a, b and hypotenuse c units.

∵ Area of semicircle = $\frac{1}{2} \pi r^2$

∴ Area of semicircle with diameter

$a=\frac{1}{2} \pi\left(\frac{a}{2}\right)^2=\frac{\pi a^2}{8}$

Area of semicircle with diameter $b=\frac{\pi b^2}{8}$

Area of semicircle with diameter $b=\frac{\pi c^2}{8}$

By Baudhayana-Pythagoras theorem,

c² = a² + b²

$\therefore \quad \frac{\pi c^2}{8}=\frac{\pi a^2}{8}+\frac{\pi b^2}{8}$ ...(i)

Let the unshaded area of semicircle with diameter a be 'x' sq units unshaded area of semicircle with diameter 'b' be 'y' sq. units. Area of Region A

= Area of semicircle with diameter 'a' - x

$=\frac{\pi a^2}{8}-x$

Also Area of Region B

= Area of semicircle with diameter 'b' -y

$=\frac{\pi b^2}{8}-y$

Area of Region A + Area of Region B

$\begin{aligned} & =\frac{\pi a^2}{8}-x+\frac{\pi b^2}{8}-y \\ & =\frac{\pi a^2}{8}+\frac{\pi b^2}{8}-(x+y) \\ & =\frac{\pi c^2}{8}-(x+y)\end{aligned}$

....(ii) [From (i)]

Also, Area of semicircle withdiameter ‘c’ = $\frac{\pi c^2}{8}$

Area of Region C + x + y = $\frac{\pi c^2}{8}$

∴ Area of Region C = $\frac{\pi c^2}{8}$ - (x + y) ... (iii)

From (ii) and (iii)

Area of Region (A) + Area of Region (B) = Area of Region (C)