Difference between revisions of "1951 AHSME Problems/Problem 49"

(Created page with "Two equations can be written: <math>x^2 + 4y^2 = 40</math> and <math>4x^2+y^2= 25</math> Add them together and get <math>5x^2+5y^2=65</math> then <math>x^2+y^2=13</math> Multiply...")
 
(fixed a very non-descriptive and hurried solution)
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Two equations can be written: <math>x^2 + 4y^2 = 40</math> and <math>4x^2+y^2= 25</math> Add them together and get <math>5x^2+5y^2=65</math> then <math>x^2+y^2=13</math> Multiply them by 4 to get the square of the value of the hypotenuse which is 52. Take the square root and get the answer <math>2SQRT13</math>
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== Problem ==
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The medians of a right triangle which are drawn from the vertices of the acute angles are <math>5</math> and <math>\sqrt{40}</math>. The value of the hypotenuse is:
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<math> \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} </math>
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==Solution==
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We will proceed by coordinate bashing.
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Call the first leg <math>2a</math>, and the second leg <math>2b</math> (We are using the double of a variable to avoid any fractions)
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Notice that we want to find <math>\sqrt{(2a)^2+(2b)^2}</math>
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Two equations can be written for the two medians: <math>a^2 + 4b^2 = 40</math> and <math>4a^2+b^2= 25</math>.
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Add them together and we get <math>5a^2+5b^2=65</math>,
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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>

Revision as of 21:51, 10 April 2013

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 $x^2+y^2=13$

Multiply them 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 $\sqrt{52}=\boxed{\textbf{(D)}\ 2\sqrt{13}}$

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