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2007 AMC 12B Problems/Problem 7

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Problem

All sides of the convex pentagon $ABCDE$ are of equal length, and $\angle A = \angle B = 90^{\circ}$ . What is the degree measure of $\angle E$ ?

$\mathrm {(A)}\ 90 \qquad \mathrm {(B)}\ 108 \qquad \mathrm {(C)}\ 120 \qquad \mathrm {(D)}\ 144 \qquad \mathrm {(E)}\ 150$

Solution

Since $A$ and $B$ are right angles, and $AE$ equals $BC$ , $AECB$ is a square, and $EC$ is 5. Since $ED$ and $CD$ are also 5, triangle $CDE$ is equilateral. Angle $E$ is therefore $90+60=150 \Rightarrow \mathrm {(E)}$

See Also

2007 AMC 12B (Problems • Resources)
Preceded by Problem 6	Followed by Problem 8
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

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2007_AMC_12B_Problems/Problem_7"

Category: Introductory Geometry Problems

Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.

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