+937 In right triangle $ABC,$ $\angle C = 90^\circ$. Median $\overline{AM}$ has a length of 1, and median $\overline{BN}$ has a length of 1. What is the length of the hypotenuse of the triangle?
+9589 Suppose \(\overline{AC} = x\) and \(\overline{BC} = y\). From \(\overline{AM} = \overline{BN} = 1\), we have
\(\begin{cases} \left(\dfrac x2\right)^2 + y^2 = 1\\ x^2 + \left(\dfrac y2\right)^2 = 1 \end{cases}\)
Adding, we have
\(\dfrac 54 \left(x^2 + y^2\right) = 2\\ x^2 + y^2 = \dfrac 85\\ \sqrt{x^2 + y^2} = \dfrac{2 \sqrt {10}}5\)
Hence, length of the hypotenuse is \(\dfrac{2 \sqrt {10}}5\)
+9589 Suppose \(\overline{AC} = x\) and \(\overline{BC} = y\). From \(\overline{AM} = \overline{BN} = 1\), we have
\(\begin{cases} \left(\dfrac x2\right)^2 + y^2 = 1\\ x^2 + \left(\dfrac y2\right)^2 = 1 \end{cases}\)
Adding, we have
\(\dfrac 54 \left(x^2 + y^2\right) = 2\\ x^2 + y^2 = \dfrac 85\\ \sqrt{x^2 + y^2} = \dfrac{2 \sqrt {10}}5\)
Hence, length of the hypotenuse is \(\dfrac{2 \sqrt {10}}5\)
