SOLUTION: A semicircle with center O and diameter CD is drawn. AB is a chord in it and also another semicircle drawn with AB as diameter intersects CD at E and F respectively. CE=3, EF=7 and

Question 1120521: A semicircle with center O and diameter CD is drawn. AB is a chord in it and also another semicircle drawn with AB as diameter intersects CD at E and F respectively. CE=3, EF=7 and FD=2, find AB^2.
Found 2 solutions by Alex.33, greenestamps:
Answer by Alex.33(110) (Show Source): You can put this solution on YOUR website!

since CE=3, EF=7, FD=2
the radius of circle O is (3+7+2)/2=6
that is to say OC=OD=OA=6.
Therefore, EH=3, OF=4.
since O' is the center of the smaller circle(circle O')
O'A=O'B=O'E=O'F.
hence EH=HF=1/2EF=3.5
So OH=5.
Assume AB=x. O'A=x/2.
In right triangle AOO': OO'^2+(x/2)^2=6^2.
In right triangle O'OH: OO'^2=O'H^2+(0.5)^2
In right triangle O'HF: O'H^2+(3.5)^2=(x/2)^2.
Solve the equation set for x(as you can see OO' and O'H can be eliminated through substitution)
And then you get x. square it and you obtain AB^2.
Enjoy!

Answer by greenestamps(13203) (Show Source): You can put this solution on YOUR website!

Let P be the midpoint of chord AB -- i.e., it is the center of the semicircle with AB as a diameter.

Let r be the radius of that semicircle. That means AB = 2r; so what we are looking for is the value of (2r)^2 = 4r^2.

Segment OP is perpendicular to chord AB, so triangle OPB is a right triangle. PB=r and OB=6, so

(1)

We can use Stewart's Theorem to find (OP)^2 in terms of r:

(2)

Substituting (2) in (1)...

And, remembering that the number we are looking for is 4r^2, the answer is

AB^2 = 96
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