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#### karush

##### Well-known member

- Jan 31, 2012

- 3,241

first I tried to find out the area of $OAPB$ which can be found by adding triangles $OAP$ and $OBP$ which are similar, $AP$ and $BP$ are perpendicular to the radius and tangent to the circle

$BP = 12\tan{37.5^0}=9.2 cm^2$

area $\Delta OBP= \frac{1}{2}(12)(9.2)=52.2 cm^2$

area $OAPD = 2(52.2)= 110.4 cm^2$

area of sector $\frac{75}{360}\pi 12^2 = 94.2 cm^2$

so shaded area $= 110.4 - 92.2 = 16.2 cm^2$

just seeing if this is OK i went over it quite a few times...

no ans given ....