6. The diagram shows a square $A B C D$ and a right-angled triangle $A B E$. The length of $B C$ is 3 . The length of $B E$ is 4 .
What is the area of the shaded region?
A $5 \frac{1}{4}$
B $5 \frac{3}{8}$
C $5 \frac{1}{2}$
D $5 \frac{5}{8}$
E $5 \frac{3}{4}$
Solution
D
We let $F$ be the point where the lines $A E$ and $C D$ meet.
Because $A B C D$ is a square, $\angle B C F=90^{\circ}$. It follows, because angles on a line have sum $180^{\circ}$, that $\angle F C E=90^{\circ}$.

Hence, in the triangles $A B E$ and $F C E$ we have $\angle A B E=\angle F C E=90^{\circ}$. Also, we see that $\angle B E A=\angle C E F$. It follows that these triangles are similar.

Therefore
$$
\frac{F C}{A B}=\frac{C E}{B E} .
$$

Because $A B C D$ is a square $A B=B C=3$. We also have $B E=4$. Therefore $C E=B E-B C=$ $4-3=1$.

It follows from equation (1) that
$$
\frac{F C}{3}=\frac{1}{4} .
$$

Hence $F C=\frac{3}{4}$.
The shaded region $A B C F$ is a trapezium. The area of a trapezium is given by the formula $\frac{1}{2} h(a+b)$, where $a$ and $b$ are the lengths of the parallel sides, and $h$ is the distance between these sides. It follows that the area of the trapezium $A B C F$ is given by
$$
\frac{1}{2} B C(A B+F C)=\frac{1}{2}\left(3 \times\left(3+\frac{3}{4}\right)\right)=\frac{1}{2}\left(3 \times \frac{15}{4}\right)=\frac{45}{8}=5 \frac{5}{8} .
$$