Question

In given figure, ABPC is a quadrant of a cirle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.
please answer fast.​

Answer 1

$\huge{\boxed{\bf{\pink{Solution:-}}}}$

$\sf \: Radius \: of \: the \: quadrant \:$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: = \: 14 \: cm \: \: \: \: \: \: \: \: \: \: \: \: (given)$

$\sf \: In \: right \: triangle \: BAC,$

$\: \: \: \: \: \: \: \: \: \sf \: BC {}^{2} = AB {}^{2} + AC {}^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf | \: By \: Pythagoras \: Theorem$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = \: (14) {}^{2} + (14) {}^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \sf = \: 196 + 196 = 196 \times 2$

$\implies \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: BC = \: \sqrt{196 \times 2} \: = \: 14 \sqrt{2} \: cm$

$\implies \: \: \: \: \: \: \: \: \: \sf \: Radius \: of \: the \: semicircle = \frac{Diameter}{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = \: \frac{BC}{2} = \frac{14 \sqrt{2} }{2} \: cm = 7 \sqrt{2} .$

$\sf \: \: \: \: Area \: of \: the \: shaded \: region$

$\: \: \: \: \: \sf = \: Area \: of \: semi - circle \: on \: BC \: as \: diameter - Area \: of \: \bf{segment} \sf \: BPC$

$= \sf\frac {1}{2}\pi(7 \sqrt{2} ) {}^{2} - [Area \: of \: quadrant \: ABPC - Area \: of \: \triangle \: ABC]$

$= \sf \frac{1}{2}\pi \times 49 \times 2 - \bigg [ \frac{1}{4}\pi(14) {}^{2} - \frac{1}{2} \times AB \times AC \bigg ]$

$= \sf \frac{1}{2} \times \frac{22}{7} \times 49 \times 2 - \bigg [ \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 - \frac{1}{2} \times 14 \times 14 \bigg]$

$\sf = 22 \times 7 - \bigg[ \frac{22 \times 2 \times 14}{4} - 7 \times 14 \bigg ]$

$\sf = 154 - (154 - 98)$

$= \sf154 - 154 + 98 = 98 \: cm{}^{2}$