A non-square rectangle has integer dimensions. The number of square units in its area is four times the number of units in its perimeter. What is the smallest possible length for the perimeter?

Guest Jan 1, 2021

#1**0 **

Hello Guest!

Let's start by setting the sides of our rectangle as x and y.

2x + 2y = perimeter

x*y = area

Thus, our equation is x*y = 4(2x + 2y)

x*y = 4(2x + 2y)

x*y = 8x + 8y # Distribute

x*y - 8x - 8y # Subtract 8x + 8y to both sides

The next part is kinda complicated.

(x-8)(y-8) = xy - 8x - 8y + 64

So (x-8)(y-8) = -64

The smallest possible perimeter will be when x and y are very close, but since it's not a square, x and y can't be equal, x-8 and y-8 both must be positive integers too.

Out of our factors of -64, -8 and 8 are the closest to each other. However, this doesn't work because for y - 8 = -8, y is equal to 0.

The next closest factor is -4 and 16.

x - 8 = 16

x = 24

y - 8 = -4

y = 4

Thus, our side lengths are 4 and 16 making the perimeter 40. :)))

I hope this helped.

=^._.^=

catmg Jan 1, 2021

#2**0 **

That did not help and here's why:

If the sides are 4 and 16, then the perimeter is 40, and the area of a rectangle is **only 64 **square units.

Important: **"***The number of square units in its area is four times the number of units in its perimeter." *

*a = 10 b = 40*

*Perimeter P = 100*

*Area A = 400 *

Guest Jan 1, 2021