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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?

Aug 4, 2024

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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.

Thanks! :)

Aug 4, 2024
edited by NotThatSmart  Aug 4, 2024