Question

In the given figure, AB = BC, ZAВС = 90°, AC = 14`sqrt2` cm and BPC (shaded portion) is semi-circle. If π = `22/7` the area of shaded portion is:

Options

• 77 cm²

• 308 cm²

• 231 cm²

• 154 cm²

Answer 1

154 cm²

Explanation:

Since triangle ABC is a right-angled isosceles triangle at B with AB = BC, using Pythagoras theorem:

\[\text{AB}^2 + \text{BC}^2 = \text{AC}^2 \implies 2\text{AB}^2 = (14\sqrt{2})^2 = 392 \implies \text{AB} = 14\text{ cm}\]

The semicircle has diameter BC = 14 cm, so its radius is r = 7 cm. The area of the shaded semicircle is:

\[\frac{1}{2}\pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 7^2 = 154\text{ cm}^2\]