1995 AHSME Problems/Problem 28
Problem
Two parallel chords in a circle have lengths and , and the distance between them is . The chord parallel to these chords and midway between them is of length where is
Solution
Solution 1
We let be the center, , represent the chords with length respectively (as shown below). Connecting the endpoints of the chords with the center, we have several right triangles. However, we do not know whether the two chords are on the same side or different sides of the center of the circle.
By the Pythagorean Theorem on , we get , where is the length of the other leg. Now the length of the leg of is either or depending whether or not are on the same side of the center of the circle:
Only the negative works here (thus the two chords are on opposite sides of the center), and solving we get . The leg formed in the right triangle with the third chord is , and by the Pythagorean Theorem again
Solution 2
Let be the chord of length , be the chord of length , be the point on such that , be the point on such that . Then . Extend intersecting the circle at and let . By the Power of a Point Theorem, Since is a right angle and and are on the circle, the diameter by Pythagorean Theorem, so the radius is . Let be the center of the circle, be the midpoint of . Then and is a right angle. Then , again by Pythagorean Theorem. The chord midway between and is a distance of 3 away from each, so 2 away from . Using the Pythagorean Theorem one more time, half the length of the chord is equal to . Then .
See also
1995 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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