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In the given figure, diameters AC and BD of the circle intersect at O. If $\angle AOB = 60^\circ$ and OA = 10 cm, then :
(i) find the length of the chord AB.
(ii) find the area of shaded region.
(Take $\pi = 3.14$ and $\sqrt{3} = 1.73$)
(i) find the length of the chord AB.
(ii) find the area of shaded region.
(Take $\pi = 3.14$ and $\sqrt{3} = 1.73$)
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(i) $\triangle OAB$ is an equilateral triangle.
$\therefore AB = OA = 10$ cm
(ii) Area of segment APB ($A_1$) = $3.14 \times 100 \times \frac{60}{360} - \frac{1.73}{4} \times 100$
$= 9.08 \text{ cm}^2 \text{ approx.}$
Area of sector OBC ($A_2$) = $3.14 \times 100 \times \frac{120}{360}$
$= 104.67 \text{ cm}^2 \text{ approx.}$
Area of shaded region = $A_1 + A_2 = 113.75 \text{ cm}^2 \text{ approx.}$
$\therefore AB = OA = 10$ cm
(ii) Area of segment APB ($A_1$) = $3.14 \times 100 \times \frac{60}{360} - \frac{1.73}{4} \times 100$
$= 9.08 \text{ cm}^2 \text{ approx.}$
Area of sector OBC ($A_2$) = $3.14 \times 100 \times \frac{120}{360}$
$= 104.67 \text{ cm}^2 \text{ approx.}$
Area of shaded region = $A_1 + A_2 = 113.75 \text{ cm}^2 \text{ approx.}$