25. In the pentagon $U V W X Y, \angle U=\angle V=90^{\circ}, U Y=V W$ and $Y X=X W$. Four equally spaced points are marked between $U$ and $V$, and perpendiculars are drawn through each point, as shown in the diagram. The dark shaded region has an area of $13 \mathrm{~cm}^{2}$ and the light shaded region has an area of $10 \mathrm{~cm}^{2}$. What is the area, in $\mathrm{cm}^{2}$, of the entire pentagon?
A 45
B 47
C 49
D 58
E 60
Solution
A

Consider the left-hand part of the pentagon divided into congruent rectangles and congruent triangles, as shown. Let the area of each of the congruent rectangles be $r \mathrm{~cm}^{2}$ and let the area of each of the congruent triangles be $t \mathrm{~cm}^{2}$. Since the light shaded region is made up of two rectangles and twelve triangles and since the dark shaded region is made up of two rectangles and eighteen triangles, the information in the question tells us that $2 r+12 t=10$ and that $2 r+18 t=13$. Therefore $6 t=3$ and hence $t=\frac{1}{2}$. Then substitute back into the first equation to obtain $2 r+6=10$ and hence $2 r=4$ and then $r=2$. Therefore the area of the whole pentagon, in $\mathrm{cm}^{2}$, is $2 \times(r+t+r+3 t+r+5 t+r+7 t+r+9 t)=2 \times(5 r+25 t)$. Hence the total area, in $\mathrm{cm}^{2}$, is $10 r+50 t$ which, when you substitute in the values for $r$ and $t$, gives $(20+25) \mathrm{cm}^{2}$, or $45 \mathrm{~cm}^{2}$.