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

(Created page with "==Problem== A triangular corner with side lengths <math>DB=EB=1</math> is cut from equilateral triangle ABC of side length <math>3</math>. The perimeter of the remaining quad...")
 
(Problem)
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\text{(C)} \ 7 \qquad  
 
\text{(C)} \ 7 \qquad  
 
\text{(D)} \ 7\frac{1}{2} \qquad  
 
\text{(D)} \ 7\frac{1}{2} \qquad  
\text{(E)} \ 8  </math>
+
\text{(E)} \ 8  </math>
  
 +
==Solution==
  
 +
<math>\triangle</math> DBE is similar to <math>\triangle</math> ABC by AA, so <math>\overline{DE}</math> = 1 by similarity, and <math>\overline{CE} = \overline{AD} = 2</math>, by subtraction. Thus the perimeter is <math>3+2+2+1 = 8</math>, or <math>\boxed{E}</math>
  
 
== See also ==
 
== See also ==

Revision as of 17:36, 23 April 2017

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 $\boxed{E}$

See also

1987 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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