- Thread starter
- #1
karush
Well-known member
- Jan 31, 2012
- 2,811
find the area of the shaded part
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 ....

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 ....

