Difference between revisions of "1987 AHSME Problems/Problem 2"
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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 DBE</math> is similar to <math>\triangle ABC</math> 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>. -slackroadia | ||
== See also == | == See also == |
Latest revision as of 18:37, 23 April 2017
Problem
A triangular corner with side lengths is cut from equilateral triangle ABC of side length . The perimeter of the remaining quadrilateral is
Solution
is similar to by AA, so = 1 by similarity, and , by subtraction. Thus the perimeter is , or . -slackroadia
See also
1987 AHSME (Problems • Answer Key • Resources) | ||
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 |
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