### Find shaded irregular area in a circle by subtracting triangle from sector

Question Sample Titled 'Find shaded irregular area in a circle by subtracting triangle from sector'

In the figure, ${O}$ is the centre of the circle. ${C}$ and ${D}$ are points lying on the circle. ${O}{B}{C}$ and ${B}{A}{D}$ are straight lines. If ${O}{C}={36}$ $\text{cm}$ and ${O}{A}={A}{B}={18}$ $\text{cm}$, find the area of the shaded region ${B}{C}{D}$ correct to the nearest  $\text{cm}^{{2}}$ .

${O}$${A}$${C}$${D}$${B}$
A
${745}$ $\text{cm}^{{2}}$
B
${864}$ $\text{cm}^{{2}}$
C
${236}$ $\text{cm}^{{2}}$
D
${907}$ $\text{cm}^{{2}}$

${O}$${A}$${C}$${D}$${B}$$\alpha$$\beta$${18}$${18}$${36}$

 Let $\angle{D}{O}{A}$ and $\angle{A}{O}{B}$ be $\alpha$ and $\beta$ respectively. ${O}{D}$ $={O}{C}={36}$ $\text{cm}$ radii Consider $\triangle{D}{O}{A}$ , ${\cos{\alpha}}$ $=\dfrac{{18}}{{36}}$ $\alpha$ $={60}^{\circ}$ Consider $\triangle{O}{A}{B}$ , ${O}{A}$ $={O}{B}$ given $\beta$ $=\dfrac{{{180}^{\circ}-{90}^{\circ}}}{{2}}={45}^{\circ}$ ∠ sum of △ ∴   The area of the shaded region $=\text{area of sector}$ ${O}{D}{C}-\text{area of}$ $\triangle{D}{O}{A}-\text{area of}$ $\triangle{A}{O}{B}$ $={\left(\dfrac{{{60}^{\circ}+{45}^{\circ}}}{{360}^{\circ}}\right)}\pi{\left({36}\right)}^{{2}}-\dfrac{{1}}{{2}}{\left({36}\right)}{\left({18}\right)}{\left({\sin{{60}}}^{\circ}\right)}-\dfrac{{1}}{{2}}{\left({18}\right)}{\left({18}\right)}$ $={745}$ $\text{cm}^{{2}}$(cor. to the nearest $\text{cm}^{{2}}$ )

 The area of the shaded region $=\text{area of sector}$ ${O}{D}{C}-\text{area of}$ $\triangle{D}{O}{B}$ $={\left(\dfrac{{{60}^{\circ}+{45}^{\circ}}}{{360}^{\circ}}\right)}\pi{\left({36}\right)}^{{2}}-\dfrac{{1}}{{2}}{\left({36}\right)}{\left(\sqrt{{{18}^{{2}}+{18}^{{2}}}}\right)}{\left({\sin{{\left({60}^{\circ}+{45}^{\circ}\right)}}}\right)}$ ${O}{B}=\sqrt{{{18}^{{2}}+{18}^{{2}}}}$ $={745}$ $\text{cm}^{{2}}$(cor. to the nearest $\text{cm}^{{2}}$ )

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