5. Three circles with radii 2,3 and 3 touch each other, as shown in the diagram.
What is the area of the triangle formed by joining the centres of these circles?
A 10
B 12
C 14
D 16
E 18
Solution
B

We let $O$ be the centre of the circle with radius 2 , and let $P$ and $Q$ be the centres of the circles with radius 3 .

The line joining the centres of touching circles goes through the point where the circles touch. [You are asked to prove this in Problem 5.2.] It follows that both $O P$ and $O Q$ have length $2+3=5$, and $P Q$ has length $3+3=6$.
Let $K$ be the midpoint of $P Q$.
The triangles $O P K$ and $O Q K$ are congruent (SSS), and therefore the angles $\angle O K P$ and $\angle O K Q$ are equal, and therefore they are both $90^{\circ}$.
By Pythagoras' Theorem, applied to the triangle $O K P$, we have $O K^{2}=O P^{2}-P K^{2}=5^{2}-3^{2}=$ $25-9=16$. Therefore $O K=4$.
The triangle $O P Q$ has base $P Q$ of length 6 , and height $O K$ of length 4 . Therefore the area of this triangle is $\frac{1}{2}(6 \times 4)=12$.