Difference between revisions of "1958 AHSME Problems/Problem 19"

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== Problem ==
 
== Problem ==
  
The sides of a right triangle are <math> a</math> and <math> b</math> and the hypotenuse is <math> c</math>. A perpendicular from the vertex divides <math> c</math> into segments <math> r</math> and <math> s</math>, adjacent respectively to <math> a</math> and <math> b</math>. If <math> a : b \equal{} 1 : 3</math>, then the ratio of <math> r</math> to <math> s</math> is:
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The sides of a right triangle are <math> a</math> and <math> b</math> and the hypotenuse is <math> c</math>. A perpendicular from the vertex divides <math> c</math> into segments <math> r</math> and <math> s</math>, adjacent respectively to <math> a</math> and <math> b</math>. If <math> a : b = 1 : 3</math>, then the ratio of <math> r</math> to <math> s</math> is:
  
 
<math> \textbf{(A)}\ 1 : 3\qquad  
 
<math> \textbf{(A)}\ 1 : 3\qquad  

Latest revision as of 03:30, 29 June 2017

Problem

The sides of a right triangle are $a$ and $b$ and the hypotenuse is $c$. A perpendicular from the vertex divides $c$ into segments $r$ and $s$, adjacent respectively to $a$ and $b$. If $a : b = 1 : 3$, then the ratio of $r$ to $s$ is:

$\textbf{(A)}\ 1 : 3\qquad  \textbf{(B)}\ 1 : 9\qquad  \textbf{(C)}\ 1 : 10\qquad  \textbf{(D)}\ 3 : 10\qquad  \textbf{(E)}\ 1 : \sqrt{10}$

Solution

$\fbox{}$

See Also

1958 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AHSME Problems and Solutions

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