Difference between revisions of "1988 AHSME Problems/Problem 2"

(Created page with "==Problem== Triangles <math>ABC</math> and <math>XYZ</math> are similar, with <math>A</math> corresponding to <math>X</math> and <math>B</math> to <math>Y</math>. If <math>AB=3,...")
 
(Solution)
 
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==Solution==
 
==Solution==
 
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Since the triangles are similar, we know that the corresponding sides of the triangle are in ratio to each other.
 
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We have <math>\frac{\overline{AB}}{\overline{XY}}=\frac{\overline{BC}}{\overline{YZ}}</math>. Plugging in values we have:
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<math>\frac{3}{4}=\frac{5}{\overline{YZ}}</math>. Solving for <math>\overline{YZ}</math>, we have <math>\overline{YZ}=6\frac{2}{3}</math>. So, the answer is <math>\boxed{\text{D}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 05:21, 31 August 2015

Problem

Triangles $ABC$ and $XYZ$ are similar, with $A$ corresponding to $X$ and $B$ to $Y$. If $AB=3, BC=4$, and $XY=5$, then $YZ$ is:

$\text{(A)}\ 3\frac{3}{4} \qquad  \text{(B)}\ 6 \qquad  \text{(C)}\ 6\frac{1}{4} \qquad  \text{(D)}\ 6\frac{2}{3} \qquad  \text{(E)}\ 8$


Solution

Since the triangles are similar, we know that the corresponding sides of the triangle are in ratio to each other. We have $\frac{\overline{AB}}{\overline{XY}}=\frac{\overline{BC}}{\overline{YZ}}$. Plugging in values we have: $\frac{3}{4}=\frac{5}{\overline{YZ}}$. Solving for $\overline{YZ}$, we have $\overline{YZ}=6\frac{2}{3}$. So, the answer is $\boxed{\text{D}}$.

See also

1988 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AHSME Problems and Solutions

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