1993 AHSME Problems/Problem 14
Problem
The convex pentagon has and . What is the area of ABCDE?
Solution
First, drop perpendiculars from points and to segment .
Since , .
This implies that is a 30-60-90 Triangle, so and .
Similarly, and .
Since is a rectangle, .
Now, notice that since , triangle is equilateral.
Thus, $[ABCDE] = [EABC]+[DCE] = \frac{2+4}{2}(\sqrt3)+\frac{4^2\cdot\sqrt3}{4} = 3\sqrt3+4\sqrt3=\boxed{7\sqrt3 (B)$ (Error compiling LaTeX. Unknown error_msg)
See also
1993 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.