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Is there a shape that I can draw with area and perimeter equal to sum of area and perimeter of 2 shapes.

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If the two shapes to be summed can be different:

  • Draw a $1\times1$ square and a $2\times2$ square.
  • Draw a $1\times5$ rectangle.

If the two shapes to be summed must be the same:

  • Draw two $3\times4$ rectangles.
  • Draw a $2\times12$ rectangle.
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  • $\begingroup$ How did you get to this values. I could get some too but those were irrationals. With same formula of the others guys. $\endgroup$ – Ursescu Ionut Apr 5 '17 at 19:03
  • $\begingroup$ I wondered if there might be some small integer-sided rectangular solutions, so I just did trial and error on the first few possibilities. $\endgroup$ – nickgard Apr 5 '17 at 20:17
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As your two shapes take two $1\times 1$ squares. The total area enclosed by these shapes is $2$ and the total perimeter is $8$.

As your single shape having area $2$ and total perimeter $8$, use a rectangle of length $a$ and width $2/a$, where $a$ is a positive solution to the equation $8 = 2a + 2/a$.

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  • $\begingroup$ Yes but the solution is not rational, I need something that I can actually draw $\endgroup$ – Ursescu Ionut Apr 5 '17 at 15:42
  • $\begingroup$ I misunderstood what you meant by `draw'. $\endgroup$ – BindersFull Apr 5 '17 at 15:46
  • $\begingroup$ @BindersFull, shouldn't the very last term be 4/a (2/a+2/a)? $\endgroup$ – electronpusher Apr 5 '17 at 17:44
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Taking a page from @Bindersfull, take two squares of side length l. The total area is $2l^2$, and the total perimeter is $8l$.

A rectangle with this area can be constructed by taking sides of length $a$ and $2l^2/a$. The side lengths will be restricted by the perimeter condition (see @Binderfull), which after applying the quadratic formula gives:

$$a=(2\pm\sqrt2)l$$

If you desire both $a$ and $l$ to be integers, or even rational numbers, I think you're S.O.L. because the coefficient $(2\pm\sqrt2)$ is irrational.

Therefore, for a rational solution, you will need to consider a pair of shapes that are not squares.

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  • $\begingroup$ Yes, that's what I am looking for. An example that I can draw. $\endgroup$ – Ursescu Ionut Apr 5 '17 at 17:57

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