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
