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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!

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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 $c=\sqrt{a^2+b^2}=4\sqrt{13}$.

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  • $\begingroup$ pi37, +1 for your solution... $\endgroup$ – Apurv Feb 12 '14 at 4:44
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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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  • $\begingroup$ I think that you made a small mistake-the lengths of the medians should be $10$ and $4\sqrt{10}$, not $4$ and $4\sqrt{10}$. $\endgroup$ – pi37 Feb 12 '14 at 3:28
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    $\begingroup$ @pi37, I have edited my answer.. $\endgroup$ – Apurv Feb 12 '14 at 3:29
  • $\begingroup$ @user128125, please upvote and accept the answer if you liked it. Anyways, your wish..;-) $\endgroup$ – Apurv Feb 12 '14 at 3:31

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