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A cyclic quadrilateral $ABCD$ has sides of magnitude $1,2,3,4$ respectively. One of its diagonals is $2.5$. Find the magnitude of the other diagonal.

This is how I tried to solve it:

For a cyclic quadrilateral,using

$ac+bd=mn$ where $a,b,c,d$ are the magnitude of sides respectively and $m,n$ are diagonals.

$AB\cdot CD+BC\cdot DA=AC\cdot BD$ (Here I don't know what is $2.5$ whether its $AC$ or $BC$). however the other diagonal is found to be $4.4$

Question $1$

Is my solution correct?

Question $2$

What values should I assign to $AC$ and $BD$?

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I suspect that the question itself is incorrect. Let $a,b,c,d$ denote the lengths of the sides ; $p,q$ lengths of diagonals, then

$$p=\sqrt{\frac{(ac+bd)(ad+bc)}{ab+cd}}$$

$$q=\sqrt{\frac{(ac+bd)(ab+cd)}{ad+bc}}$$

$p=\sqrt\frac{55}{7}$ and $q=\sqrt\frac{77}{5}$

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