14. The diagram shows a cuboid in which the area of the shaded face is one-quarter of the area of each of the two visible unshaded faces. The total surface area of the cuboid is $72 \mathrm{~cm}^{2}$. What, in $\mathrm{cm}^{2}$, is the area of one of the visible unshaded faces of the cuboid?

A 16
B 28.8
C 32
D 36
E 48
Solution: A
Let the area of the shaded face be $x \mathrm{~cm}^{2}$. Then each of the visible unshaded faces has area $4 x \mathrm{~cm}^{2}$. Hence the total surface area of the cuboid in $\mathrm{cm}^{2}$ is $2(4 x+4 x+x)=18 x$. So $18 x=72$. Therefore $x=4$, and the area of one of the visible shaded faces, in $\mathrm{cm}^{2}$ is $4 x=16$.