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# Geometry

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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?

Apr 8, 2024

#1
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Suppose $$\overline{AC} = x$$ and $$\overline{BC} = y$$. From $$\overline{AM} = \overline{BN} = 1$$, we have

$$\begin{cases} \left(\dfrac x2\right)^2 + y^2 = 1\\ x^2 + \left(\dfrac y2\right)^2 = 1 \end{cases}$$

$$\dfrac 54 \left(x^2 + y^2\right) = 2\\ x^2 + y^2 = \dfrac 85\\ \sqrt{x^2 + y^2} = \dfrac{2 \sqrt {10}}5$$

Hence, length of the hypotenuse is $$\dfrac{2 \sqrt {10}}5$$

Apr 8, 2024

#1
+9664
+1

Suppose $$\overline{AC} = x$$ and $$\overline{BC} = y$$. From $$\overline{AM} = \overline{BN} = 1$$, we have

$$\begin{cases} \left(\dfrac x2\right)^2 + y^2 = 1\\ x^2 + \left(\dfrac y2\right)^2 = 1 \end{cases}$$

$$\dfrac 54 \left(x^2 + y^2\right) = 2\\ x^2 + y^2 = \dfrac 85\\ \sqrt{x^2 + y^2} = \dfrac{2 \sqrt {10}}5$$
Hence, length of the hypotenuse is $$\dfrac{2 \sqrt {10}}5$$