SolUTION
B

Let the lengths of the vertical sides of the shaded trapezium be $p$ and $q$. Using similar triangles, $\frac{p}{3}=\frac{q}{3+5}=\frac{8}{3+5+8}$. Hence $p=\frac{3}{2}$ and $q=4$.
Therefore the area of the trapezium is $\frac{1}{2} \times\left(\frac{3}{2}+4\right) \times 5$ that is $\frac{55}{4} \mathrm{~cm}^{2}$.
7. The diagram shows three squares of side-length $3 \mathrm{~cm}, 5 \mathrm{~cm}$ and $8 \mathrm{~cm}$. What is the area, in $\mathrm{cm}^{2}$, of the shaded trapezium?
A 13
B $\frac{55}{4}$
C $\frac{61}{4}$
D $\frac{65}{4}$
E $\frac{69}{4}$