In the following figure, 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 π =]`22/7` [CBSE 2014]

#### Solution

In right triangle AED

AD^{2} = AE^{2} + DE^{2}

= (9)^{2} + (12)^{2}

= 81 + 144

= 225

∴ AD^{2} = 225

⇒ AD = 15 cm

We know that the opposite sides of a rectangle are equal

AD = BC = 15 cm

Area of the shaded region = Area of rectangle − Area of triangle AED + Area of semicircle

\[= AB \times BC - \frac{1}{2} \times AE \times DE + \frac{1}{2}\pi \left( \frac{BC}{2} \right)^2 \]

\[ = 20 \times 15 - \frac{1}{2} \times 9 \times 12 + \frac{1}{2} \times \frac{22}{7} \times \left( \frac{15}{2} \right)^2 \]

\[ = 300 - 54 + 88 . 3928\]

\[ = 334 . 3928 {cm}^2\]

Hence, the area of shaded region is 334.39 cm^{2}