1968 AHSME Problems/Problem 35
Problem
In this diagram the center of the circle is , the radius is inches, chord is parallel to chord . ,,, are collinear, and is the midpoint of . Let (sq. in.) represent the area of trapezoid and let (sq. in.) represent the area of rectangle Then, as and are translated upward so that increases toward the value , while always equals , the ratio becomes arbitrarily close to:
Solution
Let , where . Since the areas of rectangle and trapezoid are both half of rectangle and trapezoid , respectively, the ratios between their areas will remain the same, so let us consider rectangle and trapezoid .
Draw radii and , both of which obviously have length . By the Pythagorean Theorem, the length of is , and the length of is . It follows that the area of rectangle is while the area of trapezoid is
Now, we want to find the limit, as approaches , of . Note that this is equivalent to finding the same limit as approaches . Substituting into yields that trapezoid has area and that rectangle has area Our answer thus becomes
See also
1968 AHSME (Problems • Answer Key • Resources) | ||
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Followed by Problem 35 | |
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