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> | ||
− | + | 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 == | ||
+ | {{AHSME 50p box|year=1951|num-b=48|num-a=50}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 12:44, 22 February 2015
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
Multiplying this 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
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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