In the given figure, AB = $\dfrac{1}{2}BC$, where BC = 14 cm. Find :

(i) Area of quad. AEFD

(ii) Area of △ABC

(iii) Area of semicircle.

Hence, find the area of shaded region.

Given,

BC = 14 cm

AB = $\dfrac{1}{2}BC = \dfrac{1}{2}$ × 14 = 7 cm.

BC is a diameter of the semicircle

So, radius = $\dfrac{1}{2}$ × 14 = 7 cm.

(i) Area of quadrilateral AEFD:

Height (AE) = AB + BE = 7 + 7 = 14 cm.

Width : AD = BC = 14

Area of quad. AEFD = 14 × 14 = 196 cm².

Hence, area of quad. AEFD = 196 cm².

(ii) Area of △ABC:

Area = $\dfrac{1}{2}$ × base × height

= $\dfrac{1}{2}$ × BC × AB

= $\dfrac{1}{2}$ × 14 × 7

= 7 × 7 = 49 cm².

Hence area of △ABC = 49 cm².

(iii) Calculating,

$\text{Area of semicircle} = \dfrac{1}{2}πr^2 \\[1em] = \dfrac{1}{2} \times \dfrac{22}{7} \times 7^2. \\[1em] = \dfrac{11}{7} \times 49 \\[1em] = 11 \times 7 \\[1em] = 77 \text{ cm}^2.$

Hence, area of semicircle = 77 cm².

Area of shaded region = Area of quad. AEFD - (Area of triangle ABC + Area of semicircle)

= 196 - (49 + 77)

= 196 - 126

= 70 cm².

Hence, area of shaded region = 70 cm².