1951 AHSME Problems/Problem 49
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 | |
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