+1439 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?
+1897 We can use the great pythaogrean theorem to our advantage.
First off, let's set some variables.
Let's let BC = a and AC = b.
Now, using pythagoras, let's right some equations. We have
\(\overline{AM} = \sqrt{b^2 + \left(\dfrac a2\right)^2} = \dfrac12 \sqrt{4b^2 + a^2}\) since AM is the median and we also have
\(\overline{BN} = \sqrt{a^2+ \left(\dfrac{b}{2}\right)^2} = \dfrac12 \sqrt{4a^2 + b^2}\) since BN is also the median of the triangle.
Now, we're trying to isolate a^2+b^2 because that's how we find the hypotenuse of the triangle.
Thus, we now have
\(4b^2 + a^2 = 2^2 \\4a^2 + b^2 = 2^2\)
Now, adding the two equations up with each other and dividing both sides by 5, we find that
\(a^2 + b^2 = \dfrac{4+4}{5} = 8/5 \)
So the hypotenuse is that square rooted, which is \(\sqrt{\frac{8}{5}}\)
So \(\sqrt{\frac{8}{5}}\) is our final answer.
This might be wrong...not sure
Thanks! :)
