$ABCD$, $P$ is any interior point, $PA=24, PB=32, PC=28, PD=45$

Question

could anyone tell me how to solve it?

I have a convex quadrilateral $ABCD$, $P$ is any interior point, $PA=24, PB=32, PC=28, PD=45$ cm, I need to know the perimeter of $ABCD$. Thanks for helping.

Area of the $ABCD$ was given $2002$ sq unit

Answer 1

$AC \le AP + PC = 24 + 28 = 52$ and $BD \le BP + PD=32+45 = 77$

adding $AB + BC < AC \le 52, BC + CD < BD \le 77,CD + DA < CA \le 52$ and $DA + AB < DB \le 77$

we find that the perimeter is bounded above by $129.$

writing $AB = a, BC = b, Cd = c, DA = d,$ we have $$a + b + c + d \le 129, ac+bd \le 52*77=4004$$

i am stuck and not sure if the ptolemy inequality helps.

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edit after question was edited:

with the area of the quadrilateral being $2002$ which is twice of what the ptolemy bound, you can see that the diagonals are collinear and orthogonal to each other. so the perimeter is $$perimeter = \sqrt{24^2 + 32^2} + \sqrt{32^2 +28^2} + \sqrt{28^2 + 45^2 } + \sqrt{45^2 + 24^2}$$

ptolemy inequality sure helped. but an even easier method is pointed out by user orangekid in the comments below.

Answer 2

There is insufficient information to find the perimeter.

Draw point P and make 4 line segments from it, of given lengths. There is no way of finding the angles between them, and there are many cases that give a convex quadrilateral.