Suppose AB and CD are two perpendicular
chords of the same circle that intersect at the point
E,AE = 12, DE = 4 and CE = 6. The area of the
circle, is
Refer to the above figure.
In triangle OFA , 12+x2=r2
In triangle OGC, (12-x)2+52=r2
Hence, (12-x)2+52=12+x2
Solving we get x=7.
Hence, 12+72=r2=50
Area of the circle is
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