In the given figure, AB = BC, ∠ABC = 90°, AC = 14 ${\sqrt2}$ cm and BPC (shaded portion) is semi-circle. If $π = \dfrac{22}{7}$, the area of shaded portion is :

1. 77 cm²

2. 308 cm²

3. 231 cm²

4. 154 cm²

Let AB = BC = a cm and triangle ABC is a right triangle,

Using the Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ BC² + AB² = AC²

⇒ a² + a² = ($14\sqrt{2}$)²

⇒ 2a² = 392

⇒ a² = $\dfrac{392}{2}$

⇒ a² = 196

⇒ a = $\sqrt{196}$

⇒ a = 14 cm

Thus, AB = BC = 14 cm.

The diameter of the semicircle is BC = 14 cm, so the radius r is:

r = $\dfrac{d}{2}$ = $\dfrac{14}{2}$ = 7 cm

Area of semi-circle = $\dfrac{1}{2}$ πr²

$= \dfrac{1}{2} \times \dfrac{22}{7} \times 7^2\\[1em] = \dfrac{1 \times 22}{2 \times 7}\times 49\\[1em] = \dfrac{22}{14}\times 49\\[1em] = \dfrac{1,078}{14}\\[1em] = 77 \text{cm}^2$

Thus, the area of the shaded portion (the semicircle) is 77 cm².

Hence, option 1 is the correct option.