2. (a) In Figure 1, trapezoid $A B E D$ is formed by joining $\triangle B C E$ (which is rightangled at $C$ ) to rectangle $A B C D$. If $A B=3, C E=6$ and the area of trapezoid $A B E D$ is 48 , determine the length of $B E$.
(b) In Figure 2, circles with centres $P$ and $Q$ touch a line $\ell$ (that is, are tangent to $\ell$ ) at points $S$ and $T$, respectively, and touch each other at point $X$. Because the circles touch at $X$, line segment $P Q$ passes through $X$ and so $P Q=P X+X Q$. Because the circles touch $\ell$ at $S$ and $T, P S$ and $Q T$ are perpendicular to $\ell$. If the circle with centre $P$ has a radius of 25 and the circle with centre $Q$ has a radius of 16 , determine the area of trapezoid $P Q T S$.
(c) Figure 3 is formed by adding a circle of radius $r$ to Figure 2 that touches $\ell$ and each of the two other circles, as shown. Determine the value of $r$.
2. (a) Since $A B C D$ is a rectangle and $A B=3$, then $D C=3$.
Thus, $D E=D C+C E=3+6=9$.
Now, the area of trapezoid $A B E D$ is 48 .
Since $A B C D$ is a rectangle, then $A D$ is perpendicular to $D E$ which means that $A D$ is a height of trapezoid $A B E D$.
The area of trapezoid $A B E D$ equals $\frac{1}{2}(A B+D E) \times A D$.
(We could instead note that trapezoid $A B E D$ is made up of a rectangle with base 3 and an unknown height, and a right-angled triangle with base 6 and the same unknown height.) Therefore, $\frac{1}{2}(3+9) \times A D=48$ and so $6 \times A D=48$ which gives $A D=8$.
Since $A B C D$ is a rectangle, then $B C=A D=8$.
Finally, by the Pythagorean Theorem, $B E=\sqrt{B C^{2}+C E^{2}}=\sqrt{8^{2}+6^{2}}=\sqrt{100}=10$.
(b) Draw a perpendicular from $Q$ to $F$ on $P S$.
Quadrilateral $F Q T S$ has three right angles and so it must be a rectangle.
Therefore, $F S=Q T$. Since $Q T$ is a radius, then $F S=Q T=16$.
Since $P S$ is a radius, then $P S=25$.
Therefore, $P F=P S-F S=25-16=9$.
We note that $P Q=P X+X Q=25+16=41$ (they are both radii).
By the Pythagorean Theorem in $\triangle P F Q$, which is right-angled at $F$,
$$
F Q=\sqrt{P Q^{2}-P F^{2}}=\sqrt{41^{2}-9^{2}}=\sqrt{1681-81}=40
$$

Since $F Q T S$ is a rectangle, $S T=F Q=40$.
Finally, the area of trapezoid $P Q T S$ equals
$$
\frac{1}{2}(P S+Q T) \times S T=\frac{1}{2}(25+16) \times 40=41 \times 20=820
$$
(c) Suppose that the centre of the circle of radius $r$ is $O$ and that this circle touches the line $\ell$ at point $V$.
Join $P$ to $O$ and $Q$ to $O$.
Join $V$ to $O$. Since the circle is tangent to $\ell$ at $V, O V$ is perpendicular to $\ell$. Also, draw perpendiculars from $O$ to $G$ on $P S$ and from $O$ to $H$ on $Q T$.

As in (b), GOVS is a rectangle with $G S=O V=r$ and $S V=G O$.
Since $P S=25$, then $P G=P S-G S=25-r$.
Since $P O$ joins the centres of two tangent circles, then the length of $P O$ equals the sum of the radii of the two circles. In other words, $P O=25+r$.
By the Pythagorean Theorem in $\triangle P O G$,
$$
G O^{2}=P O^{2}-P G^{2}=(25+r)^{2}-(25-r)^{2}=\left(625+50 r+r^{2}\right)-\left(625-50 r+r^{2}\right)=100 r
$$

Since $G O>0$, then $G O=\sqrt{100 r}=\sqrt{100} \times \sqrt{r}=10 \sqrt{r}$.
Thus, $S V=G O=10 \sqrt{r}$.
Using a similar analysis,
$$
H O^{2}=Q O^{2}-Q H^{2}=(16+r)^{2}-(16-r)^{2}=\left(256+32 r+r^{2}\right)-\left(256-32 r+r^{2}\right)=64 r
$$

Since $H O>0$, then $H O=\sqrt{64 r}=\sqrt{64} \times \sqrt{r}=8 \sqrt{r}$.
Thus, $T V=H O=8 \sqrt{r}$.
From (b), $S T=40$ and so $S V+T V=40$.
Therefore, $10 \sqrt{r}+8 \sqrt{r}=40$ which gives $18 \sqrt{r}=40$ and so $\sqrt{r}=\frac{40}{18}=\frac{20}{9}$.
Finally, $r=\left(\frac{20}{9}\right)^{2}=\frac{400}{81}$.