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