1992 AHSME Problems/Problem 27
Problem
A circle of radius has chords of length and of length 7. When and are extended through and , respectively, they intersect at , which is outside of the circle. If and , then
Solution
Applying Power of a Point on , we find that and thus . Observing that and that , we conclude that is a right triangle with right angle at . Thus, and triangle is also right. Using that fact that the circumcircle of a right triangle has its diameter equal to the hypotenuse, we compute using the Pythagorean Theorem . From here we see that . The answer is thus .
See also
1992 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 26 |
Followed by Problem 28 | |
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