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There is a right triangle $\triangle ABC$. Medians $t_A$, $t_B$ and $t_C$ join the vertices $A$, $B$ and $C$ to the midpoints of their opposite sides, respectively (for example, vertex $A$ is connected to the midpoint of the side $a$, which is its opposite side since we name sides after the vertex a side opposes, by line $t_A$). If medians have the values of $t_A=7$ and $t_B=4$, what is the length of the side $c$ (hypotenuse)?

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closed as off-topic by Namaste, Saad, Shailesh, JonMark Perry, Taroccoesbrocco Jun 21 '18 at 10:26

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Since $\triangle ABC$ is right-angled, you should have noticed that $t_A$ and $t_B$ themselves are hypotenuses as well. Also, these two lines intersect with each other at centroid.

From these two observations, by Pythagorean theorem, you have $$(t_A)^2=7^2=b^2+\left(\dfrac{a}{2}\right)^2\tag1$$ and $$(t_B)^2=4^2=a^2+\left(\dfrac{b}{2}\right)^2\tag2.$$ From $(1)$, $b^2=7^2-(\dfrac{a}{2})^2$. And putting into $(2)$ yields $$4^2=a^2+\dfrac{1}{4}\left(7^2-\left(\dfrac{a}{2}\right)^2\right)$$ Solving you'll get $a=2$, the only possible solution. By $(1)$ or $(2)$, $b=4\sqrt3$. Using Pythagorean theorem, you obtain $c=\sqrt{2^2+({4\sqrt{3}})^2}=2\sqrt{13}.$

Sometimes it's much easier to reason it out by drawing a triangle:

like this

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  • $\begingroup$ Please, show how to get $c$ using $a$ and $b$. I know how to do it, but your answer is almost complete and there is no reason to be left without the final part. Also, parentheses look a bit weird. Fix that too. $\endgroup$ – Hanlon Jun 20 '18 at 20:05
  • $\begingroup$ @Hanlon I've made my answer complete, though there's a reason: On MSE, people tend not to provide full answer to a question with only the statement of the question. I believe one can easily figure the answer out with my steps; therefore I made my answer kind of "incomplete" in the first place. $\endgroup$ – poyea Jun 20 '18 at 20:59
  • $\begingroup$ When solving problems like this, I like to replace the numbers (in this case 7 and 4) with variables and then only substitute the values at the very end. That way, an infinite number of problems are solved instead of just one. $\endgroup$ – marty cohen Jun 20 '18 at 21:04
  • $\begingroup$ @poyea Thank you! $\endgroup$ – Hanlon Jun 20 '18 at 22:13

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