Question

In Figure 5, ABCD is a quadrant of a circle of radius 28 cm and a semi circle BEC is drawn with BC as diameter. Find the area of the shaded region ?\[[Use\pi = \frac{22}{7}]\]

Answer 1

Given:
Radius (r) of the circle = AB = AC = 28 cm
Area of quadrant ABDC:

\[= \frac{1}{4} \times \pi \times r^2 \]
\[ = \left( \frac{1}{4} \times \frac{22}{7} \times 28 \times 28 \right) {cm}^2 \]
\[ = 616 {cm}^2\]

Area of ∆ABC:
\[= \frac{1}{2} \times AC \times AB\]
\[ = \left( \frac{1}{2} \times 28 \times 28 \right) {cm}^2 \]
\[ = 392 {cm}^2\]

Area of segment BDC = Area of quadrant ABDC

                                     =(616 −- 392) cm²
                                    = 224 cm²        ....(i)

Now, in right-angled triangle BAC:
BC² = BA² + AC²     (By Pythagoras theorem)
⇒ BC² = (28² + 28²) cm²
⇒ BC² =\[28 \times 28 \times 2\]  cm²

⇒ BC = \[28\sqrt{2}\] cm

Since BC is the diameter of the semi-circle BEC, its radius will be as follows:

\[\frac{28\sqrt{2}}{2} cm = 14\sqrt{2} cm\]

Area of semi-circle BEC:

\[= \frac{1}{2} \times \pi \times r^2 \]

\[ = \left( \frac{1}{2} \times \frac{22}{7} \times 14\sqrt{2} \times 14\sqrt{2} \right) {cm}^2 \]

\[ = \left( \frac{1}{2} \times \frac{22}{7} \times 14 \times 14 \times 2 \right) {cm}^2 \]

\[ = 616 {cm}^2\]

Area of the shaded portion = Area of semi-circle BEC −-Area of segment BDC
                                      = 616 cm² − 224 cm²
                                      = 392 cm²