1998 AHSME Problems/Problem 8
Problem
A square with sides of length is divided into two congruent trapezoids and a pentagon, which have equal areas, by joining the center of the square with points on three of the sides, as shown. Find , the length of the longer parallel side of each trapezoid.
Solution
Solution 1
Then . Let the shorter side of be and the base of be such that ; then implies that , and since it follows that and .
Solution 2
The area of the trapezoid is , and the shorter base and height are both . Therefore,
Solution 3
Divide the pentagon into 2 small congruent trapezoids by extending the common shorter base of the 2 larger trapezoids.
Since each of the smaller trapezoids has its area half each of the larger trapezoids, and each of them has a base , we have
~ Nafer
See also
1998 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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