Question

In Figure 4, from a rectangular region ABCD with AB = 20 cm, a right triangle AED with AE = 9 cm and DE = 12 cm, is cut off. On the other end, taking BC as diameter, a semicircle is added on outside the region. Find the area of the shaded region.\[[Use\pi = 3 . 14]\]

Answer 1

In right-angled ∆AED:
AD² = AE² + ED²
⇒ AD² = (9² + 12²) cm²
           = (81 + 144) cm²
           = 225 cm²
⇒ AD = 15 cm
Now, area of the rectangular region ABCD

= AB x AD 
= (20 X15 ) cm²

= 300 cm²

Area of ∆AED

\[= \frac{1}{2} \times AE \times DE\]
\[ = \left( \frac{1}{2} \times 9 \times 12 \right) {cm}^2 \]
\[ = 54 {cm}^2\]

We have:
AD = BC = 15 cm
Since, BC is the diameter of the circle, therefore radius of the circle = 152 cm152 cm
Now, area of the semi-circle

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

\[ = \left( \frac{1}{2} \times 3 . 14 \times \frac{15}{2} \times \frac{15}{2} \right) {cm}^2 \]

\[ = 88 . 3125 {cm}^2\]

Area of the shaded region = Area of the rectangle + Area of the semi-circle −- Area of the triangle

                                           = (300 + 88.3125 −- 54) cm²
                                           = 334.3125 cm²