# Difference between revisions of "1987 AHSME Problems/Problem 2"

## Problem

A triangular corner with side lengths $DB=EB=1$ is cut from equilateral triangle ABC of side length $3$. The perimeter of the remaining quadrilateral is $[asy] draw((0,0)--(2,0)--(2.5,.87)--(1.5,2.6)--cycle, linewidth(1)); draw((2,0)--(3,0)--(2.5,.87)); label("3", (0.75,1.3), NW); label("1", (2.5, 0), S); label("1", (2.75,.44), NE); label("A", (1.5,2.6), N); label("B", (3,0), S); label("C", (0,0), W); label("D", (2.5,.87), NE); label("E", (2,0), S); [/asy]$ $\text{(A)} \ 6 \qquad \text{(B)} \ 6\frac{1}{2} \qquad \text{(C)} \ 7 \qquad \text{(D)} \ 7\frac{1}{2} \qquad \text{(E)} \ 8$

## Solution $\triangle DBE$ is similar to $\triangle ABC$ by AA, so $\overline{DE}$ = 1 by similarity, and $\overline{CE} = \overline{AD} = 2$, by subtraction. Thus the perimeter is $3+2+2+1 = 8$, or $\boxed{E}$. -slackroadia

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