In the figure we see two shaded regions formed by a quarter circle, a semicircle, and a triangle. Class 9

In the figure we see two shaded regions formed by a quarter circle, a semicircle, and a triangle. Class 9

Question 1.

In the figure we see two shaded regions formed by a quarter circle, a semicircle, and a triangle. Class 9

Show that the areas of the two shaded regions are equal.

Solution:

Given:

A quarter circle, a semicircle on AB, and triangle AOB

To prove:

Area of left shaded region = Area of right shaded region

Proof:

Let OA = OB = r

In right angle ∆AOB:

By Baudhayana-Pythagoras theorem,

AB² = OA² + OB²

AB² = r² + r²

AB = $=\sqrt{r^2+r^2}=r \sqrt{2}$ ...(i)

∵ A(∆AOB) = $\frac{1}{2}$ × base × height

$=\frac{1}{2} \times r \times r=\frac{1}{2} r^2$ ...(ii)

∵ Let R be radius of left semi-circle

$\mathrm{R}=\mathrm{AD}=\frac{\mathrm{AB}}{2}=\frac{r \sqrt{2}}{2}$ ...[From (i)]

∴ Area of left semicircle

$\begin{aligned} & =\frac{1}{2} \pi \mathrm{R}^2 \\ & =\frac{1}{2} \pi\left(\frac{r \sqrt{2}}{2}\right)^2=\frac{1}{4} \pi r^2\end{aligned}$

....(iii)

Area of Quarter circle = $\frac{1}{4}$πr²

∴ Area of left semicircle = Area of quarter circle ... [From (iii) and (iv)]

Area of Left shaded region

S₁ = Area of semicircle - segment AFB

= Area of left semicircle - [Area of quarter circle - Area of ∆AOB]

$=\frac{1}{4} \pi r^2-\left(\frac{1}{4} \pi r^2-\frac{1}{2} r^2\right)$

... [From(ii), (iii) and (iv)]

$=\frac{1}{2} r^2$ ...(v)

Area of Right shaded region

S₂ = Area of ∆AOB = $\frac{1}{2} r^2$ ...(vi)[From (ii)]

∴ Area(S₁) = Area(S₂) = $\frac{1}{2} r^2$ ...[From (v) and (vi)]