Question 1060632: In the figure, ACB is an arc of a circle, and CD is the perpendicular bisector of chord AB. If CD =18 and AB = 12, find the area of the entire circle.
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Answer by ikleyn(52788)   (Show Source):
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In the figure, ACB is an arc of a circle, and CD is the perpendicular bisector of chord AB. If CD =18 and AB = 12, find the area of the entire circle.
Image: http://prntscr.com/dgqi9q
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First, notice that perpendicular bisector of EVERY chord in a circle passes through the center of the circle.
This statement is easy to prove and it belongs to the area of "common knowledge", so I will not prove it.
In particular, our perpendicular bisector CD passes through the center of the circle.
Let O be the center of the circle. Then CD passes through O. So, CD is the part of the diameter of the circle.
Second, let us proceed the segment CD till the intersection with the circle at the point E.
It is not visible in the Figure, but I recommend you to make your own sketch to follow my arguments.
Then the well known property of chords intersecting inside a circle says that
|CD|*|DE| = |BD|*|DA|, or 18*|DE| = 6*6. ( here 6 =  )
Hence, |DE| =  = 2.
Then the get the remarkable fact: the diameter of the circle is |CD| + |DE| = 18 + 2 = 20 units.
Hence, the radius of the circle is 10 units.
Can you complete the solution from this point on your own?
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