In the figure (iii) given below, the diameter of the semicircle is equal to 14 cm. Calculate the area of the shaded region. Take π = $\dfrac{22}{7}$.

From figure,

BD = 14 cm (Diameter of semi-circle)

AF = FE = x (let)

⇒ BD = AF + FE

⇒ 14 = x + x

⇒ 2x = 14

⇒ x = 7 cm.

Area of semi-circle BCD = $\dfrac{πr^2}{2}$

$= \dfrac{\dfrac{22}{7} \times (7)^2}{2} \\[1em] = \dfrac{22 \times 7}{2} \\[1em] = 77 \text{ cm}^2.$

Area of quadrant ABF = Area of quadrant EDF = $\dfrac{πr^2}{4}$

$= \dfrac{\dfrac{22}{7} \times (7)^2}{4} \\[1em] = \dfrac{154}{4} \\[1em] = 38.5 \text{ cm}^2.$

From figure, AB = ED = AF = FE = 7 cm.

Area of rectangle ABDE= AB × BD = 7 × 14 = 98 cm².

Area of shaded region = Area of rectangle ABDE + Area of semi-circle BCD - Area of quadrant ABF - Area of quadrant EDF

= 98 + 77 - 38.5 - 38.5

= 98 cm².

Hence, area of shaded region = 98 cm².