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I have this geometry problem.

enter image description here

Supose any $\triangle{ABC}$, where $\overline{CE} \perp \overline{AB}$; $\overline{CM}$ is median; $n$ is $proy_{\overline{CM}}\overline{AB}$; $\angle{CMA}$ is obtuse.

Find:

$c^2;b^2;c^2+b^2$ in terms of $a$ and $x$

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    $\begingroup$ Um. This is... unreadable. I'll assume a,b,c which you never define, are the lengths of the sides of the triangles opposite A,B,C, but what the heck is x? What the heck is n? And what is proy? From what you have given ABC could be any triangle and the values you ask can be anything. $\endgroup$ – fleablood May 31 '16 at 4:45
  • $\begingroup$ Really what is a proy? $\endgroup$ – Neel May 31 '16 at 6:00
  • $\begingroup$ it is the contraction of projection, in spanish $\endgroup$ – Erincon Jun 1 '16 at 2:24
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By Pythagoras' Theorem, from triangle $CME$, we have that $EM=n=\sqrt{x^2-h^2}$.
Similarly, by Pythagoras' Theorem, from triangle $CBE$, we have that $BE=\sqrt{a^2-h^2}$.
Since $CM$ is a median, $BM=\frac{c}{2}$ and also we have $BM=BE+EM$.
Hence, we get $$\frac{c}{2}=\sqrt{a^2-h^2}+\sqrt{x^2-h^2}$$

See if this helps you even a little bit.

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For $CME$, $EM=\sqrt{x^2-h^2}=n$

for $CBE$, we have that $BE=\sqrt{a^2-h^2}$

We have,$BM=BE+EM$, Therefore,$BM=\sqrt{a^2-h^2}+\sqrt{x^2-h^2}$

As CM is median,we can say, $c=2BM= 2[\sqrt{a^2-h^2}+\sqrt{x^2-h^2}]$

Also, $b^2 = AC^2 = CE^2+AE^2=h^2+(AM+ME)^2 = h^2+(c/2+n)^2$

therefore, $b^2=h^2+[(\sqrt{a^2-h^2}+\sqrt{x^2-h^2})+\sqrt{x^2-h^2}]^2$

$b^2= h^2+(\sqrt{a^2-h^2}+2\sqrt{x^2-h^2})^2$

You get $b,c$ so you can calculate $b^2+c^2$ also.

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