1986 AHSME Problems/Problem 28
Contents
Problem
is a regular pentagon. and are the perpendiculars dropped from onto extended and extended, respectively. Let be the center of the pentagon. If , then equals
Solution 1
To solve the problem, we compute the area of regular pentagon in two different ways. First, we can divide regular pentagon into five congruent triangles.
If is the side length of the regular pentagon, then each of the triangles , , , , and has base and height 1, so the area of regular pentagon is .
Next, we divide regular pentagon into triangles , , and .
Triangle has base and height . Triangle has base and height . Triangle has base and height . Therefore, the area of regular pentagon is also Hence, which means , or . The answer is .
Solution 2
Now, we know that angle has measure . Since Therefore, . Therefore, . Recalling that gives a final answer of .
See also
1986 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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