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.
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 $8 = 2a + 2/a$.
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:
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.