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In the interior of a rectangle ABCD we pick a point K for which AK=1, BK=2 and CK=3.

Find the area of the rectangle ABCD. See drawing

It seems to me that something is missing.

Obviously, sum of areas of triangles ABK and CKD = half the area of the rectangle.

Same also for the other two triangles. But we are missing one side (KD). Or not?

If we had this side, we could calculate both triangles using Heron's formula and then calculate the two sides of the rectangle.

Am I wrong?

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  • $\begingroup$ You are correct that you need another piece of information to reduce to a uniquely dimensioned rectangle. But perhaps the area is invariant with just these three pieces of information $\endgroup$ – Henry Mar 23 '18 at 17:18
  • $\begingroup$ though perhaps not: if $K$ lies on $AB$ then I think the area may be $3\sqrt{5}$ which is not David G. Stork's square answer $\endgroup$ – Henry Mar 23 '18 at 17:24
  • $\begingroup$ Can we say that $AK^2+KC^2=BK^2+KD^2$ so $KD=\sqrt{6}$? $\endgroup$ – Pradeep Suny Mar 23 '18 at 18:26
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You need more information, but assuming the figure is a square:

$h^2 + L_1^2 = 1$

$h^2 + L2^2 = 4$

$L2^2 + (L1 + L2 - h)^2 = 9$

Solve to find $(L1 + L2)^2 = 5 + 2 \sqrt{2}$

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