+826 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?
+1908 We can use a handy equation and the fact that it's a right triangle to solve this problem.
First, let's note some equations. Since AM and BN is the median, let's set a and b to be the two legs of the triangle. we have the equation
\(\overline{AM} = \sqrt{b^2 + \left(\dfrac a2\right)^2} = \dfrac12 \sqrt{4b^2 + a^2}\)
\(\overline{BN} = \sqrt{a^2+ \left(\dfrac{b}{2}\right)^2} = \dfrac12 \sqrt{4a^2 + b^2}\)
Now, these two equations are quite important, since we want to know what a^2+b^2 is. Plugging in 1 and 1, and isolating all terms except for a and b, we find that
\(2^2 = 4a^2+b^2\\ 2^2 = 4b^2 + a^2\)
Now, adding the two equations, we have the formula
\(5a^2+5b^2 = 8\\ a^2+b^2 = 8/5\)
Thus, the hypotenuse, which is equal to the square root of a^2 + b^2 is just \(\sqrt{8/5}\)
So our answer is just \(\sqrt{8/5}\)
Thanks! :)
+1908 We can use a handy equation and the fact that it's a right triangle to solve this problem.
First, let's note some equations. Since AM and BN is the median, let's set a and b to be the two legs of the triangle. we have the equation
\(\overline{AM} = \sqrt{b^2 + \left(\dfrac a2\right)^2} = \dfrac12 \sqrt{4b^2 + a^2}\)
\(\overline{BN} = \sqrt{a^2+ \left(\dfrac{b}{2}\right)^2} = \dfrac12 \sqrt{4a^2 + b^2}\)
Now, these two equations are quite important, since we want to know what a^2+b^2 is. Plugging in 1 and 1, and isolating all terms except for a and b, we find that
\(2^2 = 4a^2+b^2\\ 2^2 = 4b^2 + a^2\)
Now, adding the two equations, we have the formula
\(5a^2+5b^2 = 8\\ a^2+b^2 = 8/5\)
Thus, the hypotenuse, which is equal to the square root of a^2 + b^2 is just \(\sqrt{8/5}\)
So our answer is just \(\sqrt{8/5}\)
Thanks! :)
