The perimeter of a rectangle is 30 inches and its area is 56 square inches. What are its dimensions?

Answer 1

2x + 2y = 30

xy = 56

x = 56/y

2(56/y) + 2y = 30

112/y + 2y = 30

multiply both sides by y

112 + 2y^2 = 30y

2y^2 - 30y + 112 = 0

y^2 - 15y + 56 = 0

(y - 8)(y - 7)

16 x 14 inches

Answer 2

The first thing you'll want to do is write equations for perimeter and area. I'll be using x as the short side and y as the long side. For a rectangle, perimeter will be P=2x+2y and area will be A=x*y. We can now solve for the dimensions (our variables) using these two equations. Because both equations have two unknowns, we need to use one to solve the other. We want to make it so that we have an equation with only one unknown, or one variable. I'm going to start by solving the perimeter equation for x. It doesn't matter whether you start with area or perimeter or whether you solve for x or y first, so there are a lot of options. Starting with the perimeter equation and solving for x we get:

P=2x+2y

x=(P-2y)/2

So now I'm going to replace the x in the area equation with the (P-2y)/2:

A=[(p-2y)/2]*y

Notice that now there is only one unknown to deal with, because P and A are given in the problem. I recommend rearranging this and then plugging in the numbers.

2A=Py-2y^2

2y^2-Py+2A=0 **We have a quadratic equation so we'll need to foil**

2y^2-30y+112=0 **Plugging in 30 for P and 56 for A**

y^2-15y+56=0 **Simply by dividing all terms by 2**

(y-7)(y-8)=0

y=7 and y=8

There are two possible solutions for y so that means there are two possible solutions for x as well:

A=x*y

x=A/y

x=56/7 and x=56/8

x=8 and x=7

These are our solutions sets (x,y): (7,8) and (8,7)

Let's test to see which ones work out.

30=2(7)+2(8) 30=2(8)+2(7)

30=30 Good! 30=30 Good!

56=(7*8) 56=(8*7)

56=56 Good! 56=56 Good!

So, both solution sets work, meaning that the dimensions are 7 and 8.