As shown in the diagram,overline{AC} is a dia | vedclass

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(C) The area of the shaded region is equal to the area of the semicircle minus the area of $\Delta ABC$.Since $\overline{AC}$ is the diameter,$\angle

Vedclass · Accepted Answer

$187.25$

Vedclass · Answer

(C) The area of the shaded region is equal to the area of the semicircle minus the area of $\Delta ABC$.Since $\overline{AC}$ is the diameter,$\angle ABC = 90^{\circ}$.Radius $r = \frac{AC}{2} = \frac{35}{2} = 17.5 \, cm$.Area of the semicircle $= \frac{1}{2} \pi r^2 = \frac{1}{2} 	imes \frac{22}{7} 	imes 17.5 	imes 17.5 = 481.25 \, cm^2$.Area of $\Delta ABC = \frac{1}{2} 	imes AB 	imes BC = \frac{1}{2} 	imes 21 	imes 28 = 294 \, cm^2$.Area of the shaded region $= 481.25 - 294 = 187.25 \, cm^2$.

As shown in the diagram,$\overline{AC}$ is a diameter of the circle with centre $O$. $\Delta ABC$ is inscribed in the circle. If $AC = 35 \, cm$,$AB = 21 \, cm$ and $BC = 28 \, cm$,find the area of the shaded region in $cm^2$.

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