# 2007 AMC 10B Problems/Problem 7

## 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 $\angle E?$ $\textbf{(A) } 90 \qquad\textbf{(B) } 108 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 144 \qquad\textbf{(E) } 150$

## Solution $[asy] unitsize(1.5cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair A=(0,2), B=(0,0), C=(2,0), D=(2+sqrt(3),1), E=(2,2); draw(A--B--C--D--E--cycle); draw(E--C,gray); draw(rightanglemark(B,A,E)); draw(rightanglemark(A,B,C)); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,E); label("E",E,NE); [/asy]$ $AB = EC$ because they are opposite sides of a square. Also, $ED = DC = AB$ because all sides of the convex pentagon are of equal length. Since $ABCE$ is a square and $\triangle CED$ is an equilateral triangle, $\angle AEC = 90$ and $\angle CED = 60.$ Use angle addition: $$\angle E = \angle AEC + \angle CED = 90 + 60 = \boxed{\textbf{(E)} 150}$$

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