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A farmer has put up a decorative windmill in his farm in which there are eight blades of equal width and equally placed in a circular arrangement. A circular wire goes through them. The diagram shows two blades OAB and OPQ in a quarter circle with centre O. $\angle AOB = \angle POQ = 30^\circ, OA = 28$ cm, $OC = 21$ cm. O is the centre of both the circles.
(i) Determine the measure of $\angle BOP$.
(ii) Find length of arc CD.
(iii) (a) Find the area of region CABD.
OR
(iii) (b) Find perimeter of region CABD.
(i) Determine the measure of $\angle BOP$.
(ii) Find length of arc CD.
(iii) (a) Find the area of region CABD.
OR
(iii) (b) Find perimeter of region CABD.
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Solution: (i) $\angle AOC = 90^\circ$ and blades are equally placed
$\therefore \angle BOP = \frac{1}{2} (90^\circ - 60^\circ) = 15^\circ$
(ii) Length of arc $CD = \frac{30}{360} \times 2 \times \frac{22}{7} \times 21 = 11$ cm
(iii) (a) Area $CABD = \frac{30}{360} \times \frac{22}{7} \times (28^2 - 21^2) = 89.8$ sq. cm
OR
(iii) (b) Length of arc $AB = \frac{30}{360} \times 2 \times \frac{22}{7} \times 28 = \frac{44}{3} = 14.67$ cm
Perimeter of $CABD = 14.67 + 11 + 2 \times (28 - 21) = 39.67$ cm
$\therefore \angle BOP = \frac{1}{2} (90^\circ - 60^\circ) = 15^\circ$
(ii) Length of arc $CD = \frac{30}{360} \times 2 \times \frac{22}{7} \times 21 = 11$ cm
(iii) (a) Area $CABD = \frac{30}{360} \times \frac{22}{7} \times (28^2 - 21^2) = 89.8$ sq. cm
OR
(iii) (b) Length of arc $AB = \frac{30}{360} \times 2 \times \frac{22}{7} \times 28 = \frac{44}{3} = 14.67$ cm
Perimeter of $CABD = 14.67 + 11 + 2 \times (28 - 21) = 39.67$ cm