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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

Best Answer 

 #1
avatar+9664 
+1

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

 

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

 

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 \(\dfrac{2 \sqrt {10}}5\)

 Apr 8, 2024
 #1
avatar+9664 
+1
Best Answer

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 \(\dfrac{2 \sqrt {10}}5\)

MaxWong Apr 8, 2024

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