Review question

# Can we find the ratio of the areas of these two sectors? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7816

## Solution

The figure shows two circles, centres $O$ and $C$, radii $r_1$ and $r_2$, which touch externally at $P$. Given that $A\hat{O}P=\dfrac{\pi}{4}$ radians, $B\hat{C}P = \dfrac{\pi}{3}$ radians and $AB$ is parallel to $OPC$, show that $\dfrac{r_1}{r_2} = \dfrac{\sqrt{6}}{2}$.

Dropping perpendiculars from $A$ and $B$ gives us the above diagram. Thus we have

$r_1 \sin \dfrac{\pi}{4} = AD= BE = r_2 \sin \frac{\pi}{3}.$

And so we have $\dfrac{r_1}{r_2} = \dfrac{\sin \dfrac{\pi}{3}}{\sin \dfrac{\pi}{4}} = \dfrac{\left(\dfrac{\sqrt{3}}{2}\right)}{\left(\dfrac{1}{\sqrt{2}}\right)} = \dfrac{\sqrt{6}}{2}.$

Hence find the ratio of the areas of the sectors $OAP$, $CBP$.

The area of a sector containing an angle $\theta$ and with radius $r$ is $\dfrac{1}{2}r^2 \theta$. So \begin{align*} A_1 &=\dfrac{1}{2}r_1^2 \times \dfrac{\pi}{4}=\dfrac{\pi r_1^2}{8},\\ A_2 &=\dfrac{1}{2}r_2^2 \times \dfrac{\pi}{3}=\dfrac{\pi r_2^2}{6}.\\ \end{align*}

Hence $\frac{A_1}{A_2} = \dfrac{6}{8} \left( \dfrac{r_1}{r_2} \right)^2 = \dfrac{9}{8}.$