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Given the circle O with perpendicular diameters and a chord, find the area of the circle if EF = 8" and DE = 20" (DF = 12"). (integer)

DE is a chord that intersects one of the diameters and shares a point on the circle with another (click on hyperlink below to see a picture).

It is not a 30-60-90 right triangle; it has been tried and does not produce the correct answer.

The correct answer: 377 sq. inches

If you understand how to do this problem or know the proof behind it, please help me with this.

Link to the image which is the circle to use for reference for this problem

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Triangle $DOF$ is similar to triangle $DEB$ since they share an angle, and angle $\angle DEB$ is right. Therefore $${DF\over r} = {2r\over DE},$$ where $r$ is the radius of the circle. From this you can compute $\pi r^2$.

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  • $\begingroup$ Thank you so much. This opened my eyes. $\endgroup$ – dguerra21 Jun 3 '16 at 3:06
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Hint: Triangles BED and FOD are similar.

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