Angles

*BFE* and

*BCE* intercepted the same arc

*BE*, therefore these angles are equal. It follows that triangle

*BOF* is similar to triangle

*COE*.

For more information about the equality of the two angles, see the relationship between inscribed angle and central angle.

**From triangle ***COE* and triangle *FOB*

$\tan \theta = \dfrac{R - 5}{R} = \dfrac{R - 9}{R - 5}$

$(R - 5)^2 = R \, (R - 9)$

$R^2 - 10R + 25 = R^2 - 9R$

$R = 25 \, \text{ cm }$

$9 + 2r = 2R$

$9 + 2r = 2(25)$

$r = 20.5 \, \text{ cm }$

**Area of the shaded region:**

$A = \pi \, R^2 - \pi \, r^2$

$A = (R^2 - r^2) \, \pi$

$A = (25^2 - 20.5^2) \, \pi$

$A = 204.75 \, \pi \, \text{ cm}^2$ *answer*