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A rectangle that meets those conditions can be worked out like so:

Let x and y be it's width and height respectively. Given the equations for the perimeter and area of a rectangle, we can say:

2x + 2y = 16

∴x + y = 8

xy = 15

∴ x = 15/y

∴ 15/y + y = 8

∴ 15 + y2 = 8y

∴ y2 - 8y + 15 = 0

∴ y2 - 8y + 16 = 1

∴ (y - 4)2 = 1

∴ y - 4 = 1

∴ y = 5

Now that we have y, we can plug it into one of our first equations to calculate x:

xy = 15

∴ x = 15/y

∴ x = 15/5 = 3

So the rectangle has a width of 3 and a height of 5

Another shape with that relationship would be a rhombus. We know that the sides of the rhombus will all be equal, so they will each be 16/4, or 4 units long.

l ≡ length of one side = 4

We also know that the area of a rhombus is equal to it's base times it's height. The base in this case will be the length of a side, 4, and we are given the area, 15. So we can say:

15 = 4h

∴ h = 15/4

where h is the height of the rhombus.

Now that we have it's base, height, and side length, we can work out it's angles. If drawn out in a graph, these numbers will give us a convenient right triangle to find the smaller angle in the rhombus (note that all angles here are in radians):

sin(α) = h/l

∴ sin(α) = 15/16

∴ α= sin-1(15/16)

∴ α ≈ 1.215375

To get the other angle, all you need to realize is that it will be equal to π - α :

β = π - α

∴ β ≈ 3.141593 - 1.215375

∴ β ≈ 1.926218

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Q: Can you draw 2 different shapes with an area of 15 and a perimeter of 16?
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