Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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)
Line 1: Line 1:
−
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>
+
== 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 $5$ 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 $\sqrt{52}=\boxed{\textbf{(D)}\ 2\sqrt{13}}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1951_AHSME_Problems/Problem_49&oldid=52452"