A quadrilateral's side's midpoints form a square. The quadrilateral's area is 50 and two sides opposite to each other are 5 and $\sqrt{85}$ long. How long are the other two sides of the quadrilateral?
How can I find out what's the size of the other two sides?
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Let $ABCD$ be the quadrilateral with $AB=5,CD=\sqrt{85}$. And let $E,F,G,H$ be the midpoint of the side $AB,BC,CD,DA$ respectively. Also, let $I,J,K,L$ be the intersection point between $AC$ and $EH$, $BD$ and $EF$, $AC$ and $FG$, $BD$ and $HG$ respectively.
Note here that $EH, FG$ are parallel to $BD$, and that $HG,EF$ are parallel to $AC$. So, if we let $a$ be the side length of the square $EFGH$, then we have $BD=AC=2a$.
Since $AC$ is perpendicular to $BD$, we have $$50=[\text{quadrilateral $ABCD$}]=\frac 12\times 2a\times 2a\quad \Rightarrow \quad a=5.$$
Here, if we let $AI=b,EI=c$ (WLOG, $b\lt c$), then we have $$DL=HI=EH-EI=5-c$$ $$LG=JF=EF-EJ=5-AI=5-b$$
So, considering the right triangles $AEI,LDG$ gives $$b^2+c^2=\left(\frac 52\right)^2\quad \text{and}\quad (5-b)^2+(5-c)^2=\left(\frac{\sqrt{85}}{2}\right)^2,$$ $$\Rightarrow \quad bc=3,\quad b+c=\frac 72\quad\Rightarrow \quad (b,c)=\left(2,\frac 32\right).$$
Hence, $$AD=2AH=2\sqrt{b^2+(5-c)^2}=\sqrt{65},$$ $$BC=2BF=2\sqrt{c^2+(5-b)^2}=3\sqrt{5}.$$