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In a square $ABCD$, say there is a point $P$ which lies inside it, the point $P$ is located at distances $x$, $y$ and $z$ meters from $A,B$ and $C$ respectively.

Using this information how could we compute a form for the area of the square?Please explain your approach.

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    How about *hacking* it out? Assume that the vertices are $(0,a), (0,0), (a,0), (a,a)$ and P is $(u,v)$. Then $u^2+(v-a)^2=x^2, u^2+v^2=y^2, (u-a)^2+v^2=z^2$. You can subtract (2) from (1) and (3) to solve for $(u,v)$ in terms of $a$. Plugging that in (2) should give you $a$. :-)2011-09-16

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Using coordinate geometry. Assume that the vertices (in order) are $(0,a), (0,0), (a,0), (a,a)$ and that $P$ is $(u,v)$. Then $ \begin{align} u^2+(v-a)^2 &= x^2,\tag{1} \\ u^2+v^2 &= y^2, \tag{2} \\ (u-a)^2+v^2 &= z^2 \tag{3}. \end{align} $ We can "solve" for $v$ in terms of $a$ by subtracting $(2)$ from $(1)$; similarly for $u$. Since the steps are straightforward, I am just posting the answer here: $ (u, v) = \left( \frac{a^2 + y^2 - z^2}{2a}, \frac{a^2 + y^2 - x^2}{2a} \right). $ Plugging this in $(2)$, we get: $ (a^2+y^2 - z^2)^2 + (a^2 + y^2 - x^2)^2 = 4a^2y^2. $ $ \implies a^4 + y^4 - (a^2+y^2)(x^2+z^2)+\frac{x^4+z^4}{2}=0. $ This is a quadratic equation in $a^2$(=area of the square) and can be solved.

Note. I do not yet understand whether both the roots of the quadratic are true solutions or not.

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    @Srivatsan : I couldn't come up with something other than more hack using law of cosines, angles etc. But I posted$a$picture maybe that helps too.2011-09-17
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This is just a supplement to Srivatsan's answer. I knew that this was a consequence of something but didn't quite get it but this is just Varignon's theorem.

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(I wish one day we can insert tikzpicture environment inside the answers.) I also agree with Srivatsan that there must a shorter/simpler observation and maybe someone can provide it.

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    I didn't work it out completely but [this quote from here](http://www.vias.org/comp_geometry/geom_quad_varignon.html) *"The center of the parallelogram is the geometric centroid of four point masses placed on the vertices of the quadrilateral."* seems to shorten the manipulations.2011-09-18