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

AplayerDanny (talk | contribs) (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> | + | == Problem == |

+ | |||

+ | 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: | ||

+ | |||

+ | <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> | ||

+ | |||

+ | ==Solution== | ||

+ | We will proceed by coordinate bashing. | ||

+ | |||

+ | 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) | ||

+ | |||

+ | Notice that we want to find <math>\sqrt{(2a)^2+(2b)^2}</math> | ||

+ | |||

+ | Two equations can be written for the two medians: <math>a^2 + 4b^2 = 40</math> and <math>4a^2+b^2= 25</math>. | ||

+ | |||

+ | Add them together and we get <math>5a^2+5b^2=65</math>, | ||

+ | |||

+ | Dividing by 5 gives <math>x^2+y^2=13</math> | ||

+ | |||

+ | 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 and . The value of the hypotenuse is:

## Solution

We will proceed by coordinate bashing.

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

Notice that we want to find

Two equations can be written for the two medians: and .

Add them together and we get ,

Dividing by 5 gives

Multiply them by 4 gives , just what we need to find the hypotenuse. Recall that he hypotenuse is . The value inside the radical is equal to , so the hypotenuse is equal to