Circle Tangent Internally to Another Circle

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