Geometry shape perimeter and area equal to perimeter sum and area sum of another 2 shapes

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

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.
• How did you get to this values. I could get some too but those were irrationals. With same formula of the others guys. – Ursescu Ionut Apr 5 '17 at 19:03
• I wondered if there might be some small integer-sided rectangular solutions, so I just did trial and error on the first few possibilities. – nickgard Apr 5 '17 at 20:17

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 $8 = 2a + 2/a$.

• Yes but the solution is not rational, I need something that I can actually draw – Ursescu Ionut Apr 5 '17 at 15:42
• I misunderstood what you meant by `draw'. – BindersFull Apr 5 '17 at 15:46
• @BindersFull, shouldn't the very last term be 4/a (2/a+2/a)? – electronpusher Apr 5 '17 at 17:44

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.

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