In right triangle $ABC,$ $\angle C = 90^\circ.$ Median $\overline{AM}$ has a length of $19,$ and median $\overline{BN}$ has a length of $13.$ What is the length of the hypotenuse of the triangle?
In right triangle ABC angle C is 90 degrees. Median AM has a length of 19 and median BN has a length of 13. What is the hypotenuse of the traingle.
+9519 Let BC = a, AC = b.
Now, \(\overline{AM} = \sqrt{b^2 + \left(\dfrac a2\right)^2} = \dfrac12 \sqrt{4b^2 + a^2}\)
and \(\overline{BN} = \sqrt{a^2+ \left(\dfrac{b}{2}\right)^2} = \dfrac12 \sqrt{4a^2 + b^2}\)
According to the question, AM = 19 and BN = 13.
\(4b^2 + a^2 = 38^2 \\4a^2 + b^2 = 26^2\)
Adding gives
\(a^2 + b^2 = \dfrac{38^2 + 26^2}{5}\)
The length of the hypotenuse is \(\sqrt{a^2+ b^2} = 2\sqrt{106}\)
