2007 AMC 12B Problems/Problem 7

Revision as of 17:31, 17 June 2023 by Justin.ziyu.sun (talk | contribs) (Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

2007 12B AMC-7.png

Since $A$ and $B$ are right angles, and $AE$ equals $BC$, and $AECB$ is a square. 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 (ProblemsAnswer KeyResources)
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
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png