022 Review Problem - Tin required to create a funnel

The funnel is in the form of frustum of right circular cone. Thus, the lateral area of frustum is the required tin.
$R = \frac{1}{2}(28) = 14 ~ \text{in.}$

$r = \frac{1}{2}(14) = 7 ~ \text{in.}$

$h = 24 ~ \text{in.}$
 

$L^2 = (R - r)^2 + h^2$

$L^2 = (14 - 7)^2 + 24^2$

$L = 25 ~ \text{in.}$
 

$A_L = \pi (R + r)L = \pi(14 + 7)(25)$

$A_L = 525\pi ~ \text{in.}^2$

$A_L = 525\pi ~ \text{in.}^2 \left( \dfrac{1 ~ \text{ft.}}{12 ~ \text{in.}} \right)^2$

$A_L = 11.454 ~ \text{ft.}^2$           answer