In the given figure, diameters AC and BD of the circle intersect at O. If ∠ AOB =60^° and OA =10 cm, then: (i) find the length of the chord AB. (ii)…

\(\displaystyle \Delta \mathrm{OAB}\) is an equilateral triangle.

\[\therefore \mathrm{AB}=\mathrm{OA}=10 \mathrm{~cm} \]

Area of segment APB \(\displaystyle \left(\mathrm{A}_{1}\right)=3.14 \times 100 \times \frac{60}{360}-\frac{1.73}{4} \times 100\)

\[=9.08 \mathrm{~cm}^{2} \text { approx. } \]

\[\begin{aligned} \text { Area of sector } \mathrm{OBC}\left(\mathrm{~A}_{2}\right) & =3.14 \times 100 \times \frac{120}{360} \\ & =104.67 \mathrm{~cm}^{2} \text { approx. } \end{aligned} \]

\[\text { Area of shaded region }=\mathrm{A}_{1}+\mathrm{A}_{2}=113.75 \mathrm{~cm}^{2} \text { approx. } \]