Finding the Area of Shaded Part

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SUMMARY

The area of the shaded part in the given problem is calculated to be approximately 16.2 cm². The area of triangle OAPB is derived from the sum of the areas of triangles OAP and OBP, both of which are similar. The area of sector OAB is computed as 94.2 cm², leading to the final shaded area calculation of 110.4 cm² minus 94.2 cm². An alternative method confirms the shaded area as approximately 15.33 cm² using the formula A_S = 15·12·sin(37.5°) - (1/2)·12²·(5π/12).

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karush
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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 ...(Cool)(Coffee)
 
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Area of $\displaystyle OAPB$ is:

$\displaystyle 15\cdot12\sin(37.5^{\circ})$

Subtract the area of sector $\displaystyle OAB$:

$\displaystyle A_S=15\cdot12\sin(37.5^{\circ})-\frac{1}{2}12^2\cdot\frac{5\pi}{12}=180\sin(37.5^{\circ})-30\pi\approx15.329277613875924$

I have verified this using another method as well.
 

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