question_answer

ABCMA is a quadrant of a circle of radius 14 cm. With AC as diameter, a semicircle ANC is drawn. The area of the shaded portion is

A)  154 sq cm        

B)  77 sq cm

C)  98 sq cm         

D)  196 sq cm

Correct Answer: C

Solution :

            Area of shaded portion             \[=\Delta ABC+ANC-ABCMA\] \[=\Delta ABC+\frac{1}{2}\pi {{\left( \frac{14\sqrt{2}}{2} \right)}^{2}}-\frac{1}{4}\pi {{\left( 14 \right)}^{2}}\] \[=\frac{1}{2}\times 14\times 14+\frac{1}{2}\pi \times \frac{14\times 14\times 2}{\left( 2\times 2 \right)}\] \[-\frac{1}{4}\times \pi \times 14\times 14\] \[=\frac{1}{2}\times 14\times 14=98sq\,\,cm\]