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Another question from my test:

Question

The area of the circle is $157$

$ AE=EB $

$ ABCD$ is a square

The diagonal of the square ($AC$) is equal to the diameter of the circle.

Find the area of BCFE?

I managed to find the area of $ACB$, but I am having a trouble finding $AFE$. Any suggestions?

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closed as off-topic by user21820, JMP, MickG, zhoraster, tomi Jan 15 '17 at 22:08

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An idea:

Draw the diagonal $\;BD\;$ and take a close look at triangle $\;\Delta ABD\;$. Deduce that $\;F\;$ is the intersection point of the medians of this triangle and thus

$$AF=\frac23r\;\left(\implies FC=\frac43r\;\right)\;,\;\;r=\text{ the circle's radius}$$

Likewise, using Pythagoras and with $\;x=\sqrt2\,r=$ the square's side:

$$DE=\frac{\sqrt5}2x\implies FE=\frac13DE=\frac13\frac{\sqrt5}2\sqrt2\,r=\frac{\sqrt{10}}6\,r\;,\;\;AE=\frac12x=\frac r{\sqrt2}$$

So now you know all the sides of $\;\Delta AEF\;$ and you can calculate its area, for example with Heron's formula for the semiperimeter.

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Hint:

Clearly,

$$\Delta\ ADE=\frac{1}{4}\ \square\ ABCD$$

$$\Delta\ ABC=\frac{1}{2}\ \square\ ABCD$$

Since base $AE=\frac{1}{2}AD$, we get

$$\Delta\ AFE=\frac{1}{2}\ \Delta\ AFD=\frac{1}{3}\ \Delta\ ADE=\frac{1}{12}\ \square\ ABCD$$

Therefore

$$\square\ BCEF=\Delta\ ABC-\Delta\ AFE=\frac{5}{12}\ \square\ ABCD\\ $$

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