20. In trapezium $P Q R S, S R=P Q=25 \mathrm{~cm}$ and $S P$ is parallel to $R Q$. All four sides of $P Q R S$ are tangent to a circle with centre $C$. The area of the trapezium is $600 \mathrm{~cm}^{2}$. What is the radius of the circle?
A $7.5 \mathrm{~cm}$
B $8 \mathrm{~cm}$
C $9 \mathrm{~cm}$
D $10 \mathrm{~cm}$
E $12 \mathrm{~cm}$
Solution: $\mathbf{E}$
Let the radius of circle be $r \mathrm{~cm}$, and let the points where the trapezium touches the circle be $W, X, Y$ and $Z$ as shown. As the radii $C W, C X, C Y$ and $C Z$ are perpendicular to the edges of the trapezium $P Q, Q R, R S$ and $S P$, respectively, the area of the trapezium, which is the sum of the areas of the triangles $C P Q, C Q R, C R S$ and $C S P$, is
$$
\frac{1}{2} r P Q+\frac{1}{2} r Q R+\frac{1}{2} r R S+\frac{1}{2} S P=\frac{1}{2} r(P Q+Q R+R S+S P) .
$$

Now using the property that the two tangents to a circle from a given point have equal length, it follows that $S P+Q R=S Z+P Z+Q X+R X=S Y+P W+Q W+R Y=(P W+Q W)+(R Y+S Y)$ $=P Q+R S=25+25=50 \mathrm{~cm}$. Hence $P Q+Q R+R S+S P=50+50=100 \mathrm{~cm}$. Therefore, as the area of the trapezium of $600 \mathrm{~cm}^{2}$, we have that $\frac{1}{2} r(100)=600$ and hence $r=12$.