• # question_answer In the given figure, O is the centre of the circle with $AC=24\text{ }cm,\,\,AB=7\text{ }cm$ and $\angle BOD=90{}^\circ$. Find the area of the shaded region.

 Given, C (O, OB) with $AC=24\text{ }cm\text{ }AB=7\text{ }cm$ and $\angle BOD=90{}^\circ$ $\angle CAB=90{}^\circ$ (Angle in semi-circle) Using pythagoras theorem in $\Delta \text{ }CAB$ $B{{C}^{2}}=A{{C}^{2}}+A{{B}^{2}}$ $={{(24)}^{2}}+{{(7)}^{2}}$ $=576+49$ $=625$ $BC=25\,\,cm$ Radius of circle $=OB=OD=OC=\frac{25}{2}cm$ Area of shaded region = Area of semi-circle with diameter BCArea of $\Delta \text{ }CAB$+Area of sector $BOD$ $=\frac{1}{2}\pi {{\left( \frac{25}{2} \right)}^{2}}-\frac{1}{2}\times 24\times 7+\frac{90}{360}\pi {{\left( \frac{25}{2} \right)}^{2}}$ $=\frac{3}{4}\times \frac{22}{7}\times \frac{25}{2}\times \frac{25}{2}-84$ $=\frac{20625}{56}-84$ $=\frac{20625-4704}{56}$ $=\frac{15921}{56}=284.3\,\,c{{m}^{2}}$     (approx..)