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(D) Since $ABC$ is a quadrant of the circle,$\angle BAC = 90^{\circ}$.In $	riangle ABC$,by Pythagoras theorem:$BC^2 = AC^2 + AB^2 = (14)^2 + (14)^2 =

Vedclass · Accepted Answer

$98$

Vedclass · Answer

(D) Since $ABC$ is a quadrant of the circle,$\angle BAC = 90^{\circ}$.In $	riangle ABC$,by Pythagoras theorem:$BC^2 = AC^2 + AB^2 = (14)^2 + (14)^2 = 196 + 196 = 392$.$BC = \sqrt{392} = 14\sqrt{2} \, cm$.Radius $(r_1)$ of the semicircle drawn on $BC = \frac{14\sqrt{2}}{2} = 7\sqrt{2} \, cm$.Area of $	riangle ABC = \frac{1}{2} 	imes AB 	imes AC = \frac{1}{2} 	imes 14 	imes 14 = 98 \, cm^2$.Area of quadrant $ABC = \frac{1}{4} 	imes \pi 	imes r^2 = \frac{1}{4} 	imes \frac{22}{7} 	imes 14 	imes 14 = 154 \, cm^2$.Area of segment $BC$ (the region between chord $BC$ and arc $BC$) = Area of quadrant $ABC - 	ext{Area of } 	riangle ABC = 154 - 98 = 56 \, cm^2$.Area of semicircle on $BC = \frac{1}{2} 	imes \pi 	imes r_1^2 = \frac{1}{2} 	imes \frac{22}{7} 	imes (7\sqrt{2})^2 = \frac{1}{2} 	imes \frac{22}{7} 	imes 98 = 154 \, cm^2$.Area of shaded region = Area of semicircle on $BC - 	ext{Area of segment } BC = 154 - 56 = 98 \, cm^2$.

In the figure,$ABC$ is a quadrant of a circle of radius $14 \, cm$ and a semicircle is drawn with $BC$ as diameter. Find the area of the shaded region. (in $cm^2$) [Use $\pi = \frac{22}{7}$]

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