006 Review Problem - Area illuminated by light placed in front of the globe

$\cos \theta = \dfrac{1.5 - h}{1.5} = \dfrac{1.5}{5}$

$5(1.5 - h) = 1.5^2$

$7.5 - 5h = 2.25$

$5h = 5.25$

$h = 1.05 ~ \text{ft}$
 

$A_z = 2\pi rh$

$A_z = 2\pi (1.5)(1.05)$

$A_z = 9.8960 ~ \text{ft.}^2$           answer