Can we find the ratio of the areas of these two sectors?

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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 \[\dfrac{r_1}{r_2} = \dfrac{\sin \dfrac{\pi}{3}}{\sin \dfrac{\pi}{4}} = \dfrac{\left(\dfrac{\sqrt{3}}{2}\right)}{\left(\dfrac{1}{\sqrt{2}}\right)} = \dfrac{\sqrt{6}}{2}.\]

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