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Difference between revisions of "1951 AHSME Problems/Problem 49"
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Dividing by 5 gives <math>x^2+y^2=13</math>  
 
Dividing by 5 gives <math>x^2+y^2=13</math>  
  
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Multiply them by 4 gives <math>4x^2+4y^2=52\implies (2x)^2+(2y)^2=52</math>, just what we need to find the hypotenuse. Recall that he hypotenuse is <math>\sqrt{(2a)^2+(2b)^2}</math>. The value inside the radical is equal to <math>52</math>, so the hypotenuse is equal to <math>\sqrt{52}=\boxed{\textbf{(D)}\ 2\sqrt{13}}</math>
+
Multiplying this by 4 gives <math>4x^2+4y^2=52\implies (2x)^2+(2y)^2=52</math>, just what we need to find the hypotenuse. Recall that he hypotenuse is <math>\sqrt{(2a)^2+(2b)^2}</math>. The value inside the radical is equal to <math>52</math>, so the hypotenuse is equal to <math>\sqrt{52}=\boxed{\textbf{(D)}\ 2\sqrt{13}}</math>
  
 
== See Also ==
 
== See Also ==

Latest revision as of 11:44, 22 February 2015

Problem

The medians of a right triangle which are drawn from the vertices of the acute angles are $5$ and $\sqrt{40}$. The value of the hypotenuse is:

$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 2\sqrt{40}\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 2\sqrt{13}\qquad\textbf{(E)}\ \text{none of these}$

Solution

We will proceed by coordinate bashing.

Call the first leg $2a$, and the second leg $2b$ (We are using the double of a variable to avoid any fractions)

Notice that we want to find $\sqrt{(2a)^2+(2b)^2}$

Two equations can be written for the two medians: $a^2 + 4b^2 = 40$ and $4a^2+b^2= 25$.

Add them together and we get $5a^2+5b^2=65$,

Dividing by 5 gives $x^2+y^2=13$

Multiplying this by 4 gives $4x^2+4y^2=52\implies (2x)^2+(2y)^2=52$, just what we need to find the hypotenuse. Recall that he hypotenuse is $\sqrt{(2a)^2+(2b)^2}$. The value inside the radical is equal to $52$, so the hypotenuse is equal to $\sqrt{52}=\boxed{\textbf{(D)}\ 2\sqrt{13}}$

See Also

1951 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 48 		Followed by
Problem 50
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All AHSME Problems and Solutions

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