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Suppose that there is a right triangle \(ABC\) such that the legs \(AB\) and \(AC\) have lengths \(4\sqrt2\) and \(2\) respectively. What is the length of the median \(BM\)?

 

This question is posted here by @CPhill: https://web2.0calc.com/questions/suppose-that-there-is-a-right-triangle-abc-such-that, but that answer is wrong. 

 

Here is the correct answer:

 

 

Median \(BM \) splits side \(AC \) into two segments with equal length. Since the midpoint of the hypotenuse of a right triangle is also the circumcenter of that triangle, \(AM\) and \(BM \) are circumradii of triangle \(ABC\). Moreover, since the circumcircle of a triangle intersects each corner point of a triangle, \(BM\) is also a circumradius. Thus, \(AM = MC = BM\).

 

We can easily find the length of \(BM \) now. By the Pythagorean theorem, 

 

\((4\sqrt2)^2 + 2^2 = AC^2, \)
\(32 + 4 = AC^2, \)
\(\sqrt36 = AC, \)

\(AC = 6.\)

 

Since \(BM = AM = MC\), \(AC = 2BM\). And since \(AC = 6\), we have

 

\(\boxed{BM=3}\)

 

\(Q.E.D.\)

 Mar 29, 2019
 #1
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Well, the median is half the hypotenuse in a right triangle. And, just divide that length by 2.

 Mar 29, 2019
 #2
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Applesoup,

Did you become a member just so that you could take points off CPhill?

He works hard for those points !

 Mar 29, 2019
edited by Melody  Mar 29, 2019

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