The circular window shown in the diagram has nine panes of equal area. The inner circular pane has radius $20 \mathrm{~cm}$ and the same centre, $O$, as the outer circle. The eight lines separating the outer panes are of equal length, $x \mathrm{~cm}$, and all, if extended, would pass through $O$. What is the value of $x$, to the nearest tenth?
(A) 40.0
(B) 36.6
(C) 30.0
(D) 20.0
(E) 43.2

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-094.jpg?height=363&width=374&top_left_y=2144&top_left_x=1342)
Since the radius of the inner circular pane is $20 \mathrm{~cm}$, then its area is $\pi 20^{2}=400 \pi \mathrm{cm}^{2}$.

Therefore, the area of each of the outer panes is also $400 \pi \mathrm{cm}^{2}$, so the total area of the circular window is $9(400)=3600 \pi \mathrm{cm}^{2}$.

If the radius of the larger circle is $R$, then $\pi R^{2}=3600 \pi$, so $R^{2}=3600$ or $R=60$, since $R>0$. Since each of the outer lines can be extended to form a radius by joining its inner end to the centre using a radius of the inner circle, then the radius of the larger circle is $x+20=60$, so $x=40=40.0$, to the nearest tenth.

(It was not actually necessary to calculate the area of the circles. Since the large circle is formed from 9 pieces of equal area, its area is 9 times that of the inner circle. Thus, its radius is $\sqrt{9}=3$ times that of the inner circle.)

ANSWER: (A)