Difference between revisions of "2017 AMC 12A Problems/Problem 19"

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\qquad\textbf{(D)}\ \frac{37}{35}
 
\qquad\textbf{(D)}\ \frac{37}{35}
 
\qquad\textbf{(E)}\ \frac{13}{12}</math>
 
\qquad\textbf{(E)}\ \frac{13}{12}</math>
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==See Also==
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{{AMC12 box|year=2017|ab=A|num-b=18|num-a=20}}
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{{MAA Notice}}

Revision as of 17:21, 8 February 2017

Problem

A square with side length $x$ is inscribed in a right triangle with sides of length $3$, $4$, and $5$ so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length $y$ is inscribed in another right triangle with sides of length $3$, $4$, and $5$ so that one side of the square lies on the hypotenuse of the triangle. What is $\frac{x}{y}$?

$\textbf{(A)}\ \frac{12}{13} \qquad \textbf{(B)}\ \frac{35}{37} \qquad\textbf{(C)}\ 1 \qquad\textbf{(D)}\ \frac{37}{35} \qquad\textbf{(E)}\ \frac{13}{12}$

See Also

2017 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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
All AMC 12 Problems and Solutions

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