In the diagram, $A B C D$ is a trapezoid with $\angle D A B=\angle A D C=90^{\circ}$. Also, $A B=7, D C=17$, and $A D=10$. Point $P$ is on $A D$ and point $Q$ is on $B C$ so that $P Q$ is parallel to $A B$ and to $D C$. Point $F$ is on $D C$ so that $B F$ is perpendicular to $D C$. $B F$ intersects $P Q$ at $T$.

![](https://cdn.mathpix.com/cropped/2024_04_17_1deb74b62155eda4bdbfg-3.jpg?height=437&width=666&top_left_y=535&top_left_x=1052)

(a) Determine the area of trapezoid $A B C D$.

(b) Determine the measure of $\angle B Q P$.

(c) If $P Q=x$, determine the length of $A P$ in terms of $x$.

(d) Determine the length of $P Q$ for which the areas of trapezoids $A B Q P$ and $P Q C D$ are equal.
(a) Since $A B$ is parallel to $D C$ and $A D$ is perpendicular to both $A B$ and $D C$, then the area of trapezoid $A B C D$ is equal to $\frac{1}{2} \cdot A D \cdot(A B+D C)$ or $\frac{1}{2} \cdot 10 \cdot(7+17)=120$.

Alternatively, we could separate trapezoid $A B C D$ into rectangle $A B F D$ and right-angled triangle $\triangle B F C$.

We note that $A B F D$ is a rectangle since it has three right angles.

Rectangle $A B F D$ is 7 by 10 and so has area 70 .

$\triangle B F C$ has $B F$ perpendicular to $F C$ and has $B F=A D=10$.

Also, $F C=D C-D F=D C-A B=17-7=10$.

Thus, the area of $\triangle B F C$ is $\frac{1}{2} \cdot F C \cdot B F=\frac{1}{2} \cdot 10 \cdot 10=50$.

This means that the area of trapezoid $A B C D$ is $70+50=120$.

(b) Since $P Q$ is parallel to $D C$, then $\angle B Q P=\angle B C F$.

We note that $A B F D$ is a rectangle since it has three right angles. This means that $B F=A D=10$ and $D F=A B=7$.

In $\triangle B C F$, we have $B F=10$ and $F C=D C-D F=17-7=10$.

Therefore, $\triangle B C F$ has $B F=F C$, which means that it is right-angled and isosceles.

Therefore, $\angle B C F=45^{\circ}$ and so $\angle B Q P=45^{\circ}$.

(c) Since $P Q$ is parallel to $A B$ and $A P$ and $B T$ are perpendicular to $A B$, then $A B T P$ is a rectangle.

Thus, $A P=B T$ and $P T=A B=7$.

Since $P T=7$, then $T Q=P Q-P T=x-7$.

Since $\angle B Q T=45^{\circ}$ and $\angle B T Q=90^{\circ}$, then $\triangle B T Q$ is right-angled and isosceles.

Therefore, $B T=T Q=x-7$.

Finally, $A P=B T=x-7$.

(d) Suppose that $P Q=x$.

In this case, trapezoid $A B Q P$ has parallel sides $A B=7$ and $P Q=x$, and height $A P=x-7$.

The areas of trapezoid $A B Q P$ and trapezoid $P Q C D$ are equal exactly when the area of trapezoid $A B Q P$ is equal to half of the area of trapezoid $A B C D$.

Thus, the areas of $A B Q P$ and $P Q C D$ are equal exactly when $\frac{1}{2}(x-7)(x+7)=\frac{1}{2} \cdot 120$, which gives $x^{2}-49=120$ or $x^{2}=169$.

Since $x>0$, then $P Q=x=13$.

Alternatively, we could note that trapezoid $P Q C D$ has parallel sides $P Q=x$ and $D C=17$, and height $P D=A D-A P=10-(x-7)=17-x$.

Thus, the area of trapezoid $A B Q P$ and the area of trapezoid $P Q C D$ are equal exactly when $\frac{1}{2}(x-7)(x+7)=\frac{1}{2}(17-x)(x+17)$, which gives $x^{2}-49=17^{2}-x^{2}$ or $x^{2}-49=289-x^{2}$ and so $2 x^{2}=338$ or $x^{2}=169$.

Since $x>0$, then $P Q=x=13$.