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A. \[101.98c{{m}^{2}}\]

B. \[161.54c{{m}^{2}}\]

C. \[101.54c{{m}^{2}}\]

D. None of these

Answer

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We know that any angle made by the diameter QR in the semicircle is \[{{90}^{\circ }}\]

Hence \[\angle RPQ={{90}^{\circ }}\]

In the right angled \[\Delta RPQ\]

By Pythagoras we know that

\[R{{Q}^{2}}=P{{Q}^{2}}+P{{R}^{2}}\]

\[\begin{align}

& R{{Q}^{2}}={{24}^{2}}+{{7}^{2}} \\

& R{{Q}^{2}}=576+49 \\

& R{{Q}^{2}}=625 \\

& RQ=\sqrt{625} \\

& RQ=25cm \\

\end{align}\]

We know that RQ is the diameter of circle so the radius of circle is \[OQ=\dfrac{RQ}{2}=\dfrac{25}{2}cm\]

The area of right angled \[\Delta RPQ=\dfrac{1}{2}\times base\times height\]

The area of right angled \[\Delta RPQ=\dfrac{1}{2}\times RP\times PQ\]

The area of right angled \[\Delta RPQ=\dfrac{1}{2}\times 7\times 24\] \[=7\times 12=84c{{m}^{2}}\]. . . . . . . . (1)

We know that the area of semicircle is given by the formula \[=\dfrac{\pi {{r}^{2}}}{2}\]

\[=\dfrac{22}{7}\times \dfrac{25}{2}\times \dfrac{25}{2}\times \dfrac{1}{2}\]

\[=\dfrac{11\times 25\times 25}{28}\]

\[=\dfrac{6875}{28}c{{m}^{2}}\]. . . . . . . . . . . (2)

The area of shaded region is given by area of semicircle -area of right angled \[\Delta RPQ\]

\[=\dfrac{6875}{28}-84\]

\[=\dfrac{6875-2352}{28}\]

\[=\dfrac{4523}{28}=161.54c{{m}^{2}}\]

Hence the area of the shaded region is \[161.54c{{m}^{2}}\]