In the diagram, the circles with centres $P$, $Q$ and $R$ have radii 3,2 and 1 respectively. Each circle touches the other two as shown. The area of $\triangle P Q R$ is
(A) 12
(B) 6
(C) 7.5
(D) 10
(E) 4

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-093.jpg?height=399&width=599&top_left_y=1470&top_left_x=1099)
Since $P$ and $Q$ are the centres of their respective circles, then line segment $P Q$ passes through the point of tangency between these two circles. Therefore, the length $P Q$ is the sum of the radii of these two circles, or $P Q=3+2=5$.

Similarly, $P R=3+1=4$ and $Q R=2+1=3$.

Therefore, $\triangle P Q R$ has side lengths 3,4 and 5 , so is right-angled since $3^{2}+4^{2}=5^{2}$. In fact, the right-angle is between the sides of length 3 and 4 , so the area of $\triangle P Q R$ is $\frac{1}{2}(3)(4)=6$.