# Find the hypotenuse of a right triangle given medians [closed]

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)?

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Saad, Shailesh, JonMark Perry, Taroccoesbrocco
If this question can be reworded to fit the rules in the help center, please edit the question.

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 $c=\sqrt{2^2+({4\sqrt{3}})^2}=2\sqrt{13}.$
• 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. – Hanlon Jun 20 '18 at 20:05