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# Suppose that there is a right triangle ABC...

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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. 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

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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
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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