1987 AHSME Problems/Problem 2

Revision as of 17:37, 23 April 2017 by Slackroadia (talk | contribs) (Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

See also

1987 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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. AMC logo.png

Invalid username
Login to AoPS