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September 30, 2014

September 30, 2014

Posted by **Me too** on Tuesday, March 8, 2011 at 8:43pm.

- geometry -
**tchrwill**, Wednesday, March 9, 2011 at 11:13amHere is the same problem with different numbers.

The medians of a right triangle which are drawn from the vertices of the acute angles are 5 and root 40. What is the length of the hypotenuse? Can someone show me how to do it? thanks

Draw triangle ABC where A is the upper left vertex, B is the lower right vertex and C is the right angle at the lower left. Call AC x and BC y. Draw the medians BD and AE, D being the midpoint of x and E the midpoint of y.

From triangle ACE, x^2 + (y/2)^2 = 25 or 4x^2 + y^2 -100 = 0.---(1)

From triangle DCB, (x/2)^2 + y^2 = 40 or x^2 + 4y^2 - 160 = 0.---(2). (2)

From (1), y^2 = 100 - 4x^2

Substituting into (2) yields x = 4 and y = 6.

Therefore, the hypotenuse is sqrt(4^2 + 6^2 = 7.211.

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