Problem:
In the figure, AB and CD are diameters of the circle with center O, ABβ₯CD, and chord DΛFΛ intersects AB at E. If DE=6 and EF=2, then the area of the circle is
Answer Choices:
A. 23Ο
B. 247βΟ
C. 24Ο
D. 249βΟ
E. 25Ο
Solution:
Draw segment FC. Angle CFD is a right angle since arc CFD is a semicircle. Then right triangles DOE and DFC are similar, so
DFDOβ=DCDEβ.
Let DO=r and DC=2r. Substituting, we have
8rβ=2r6β,2r2=48,r2=24
Then the area of the circle is Οr2=24Ο.
OR
Let OA=OB=r and OE=x. Substituting into AEβ
EB=DEβ
EF gives (r+x)(rβx)=6β
2 so In right triangle EOD, r2βx2=12. r2+x2=36.
Add to find 2r2=48. Thus, the area of the circle is Οr2=24Ο.
OR
Construct OGβ₯DF with G on DF. Then DG= 21βDF=4. Since OG is an altitude to the hypotenuse of right triangle EOD, we have DODEβ=DGDOβ. Let DO=r. Then r6β=4rβ, so r2=24, and the area of the circle is Οr2=24Ο.
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