The medians to the two legs of a right angled triangle are $10$ and $4\sqrt{10}$. Find the hypotenuse of the right angled triangle.
I'm confused. Can somebody please explain to me how to do this step by step? Not just the answer I want to know what you did to get the answer and why you did it please. Thank you!
$\endgroup$2 Answers
$\begingroup$Let the two legs have lengths $a,b$, and let the hypotenuse have length $c=\sqrt{a^2+b^2}$. Now note that by the Pythagorean theorem, $$a^2+(\dfrac{b}{2})^2=10^2=100$$ and $$(\dfrac{a}{2})^2+b^2=(4\sqrt{10})^2=160$$ Adding these two equations, we get $$ \dfrac{5}{4}(a^2+b^2)=260 $$ So $c=\sqrt{a^2+b^2}=4\sqrt{13}$.
$\endgroup$ 1 $\begingroup$Let the sides be $a,b,c$ and $c$ being the hypotenuse. Let the medians on $a$ and $b$ be $10$ and $4\sqrt{10}$. Then by applying the Appolonius theorem for sides $a$ and $b$ respectively, we have:
$$b^2+c^2=2(10^2+\dfrac{a^2}{4})$$
$$a^2+c^2=2((4\sqrt{10})^2+\dfrac{b^2}{4})$$ Adding the two equations and rearranging, we have, $$\dfrac{a^2}{2}+\dfrac{b^2}{2}+2c^2=520$$
Also noting that $a^2+b^2=c^2$, we get $c=4\sqrt {13}$.
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