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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...")
 
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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>
 +
 
 +
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>  
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 +
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>
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== See Also ==
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{{AHSME 50p box|year=1951|num-b=48|num-a=50}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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