How to find dimensions of a rectangle with area and perimeter

Question

Rectangles have two dimensions, a length which we will call L, and a width that we will call W. For problems like this, it might be useful to draw a rectangle and label all its sides, like below:

W

_________________________________

| |

L | | L

| |

|________________________________|

W

Because these values are unknown right now, we have two variables, L and W. To solve for any number of variables, you'll need at least the same number of independent equations, which means we need to write two equations to be able to find L and W. Your question gives us two important pieces of info: the perimeter is equal to 48, and the area is equal to 135. Let's use this info to write our two equations.

The perimeter of a shape is the length of its outline. To get the perimeter of a rectangle, you'll add up the length of all its sides. To write our perimeter, P in math terms, it could look like this:

P = W + L + W + L

= 2W + 2L

Because our perimeter is equal to 48, we can say:

P = 2W + 2L

48 = 2W + 2L

Now let's use our area equation. The area of a shape is the amount of space contained in its outline. To get our area, A you need to multiply a rectangle's length by its width, like below:

A = LW

Because we know our area is 135, we can also say:

A = LW

135 = LW

Now we have two equations with two variables, L and W. To be able to solve for a variable, you need to use one equation to solve what a variable is equal to in terms of the other, then use the second equation to rewrite everything in terms of one variable. To write this in math terms, let's start with the second equation and solve for W:

135 = LW

W = 135/L

Now we know what W is equal to in terms of L. We can use this in the first equation to get:

48 = 2W + 2L

= 2(135/L) + 2L

= 270/L + 2L. Because we're dividing 270/L, let's multiply everything by L so there are no fractions

48L = 270 + 2L2. Now because we have L2,L, and a number without L, we have what's called a quadratic. You can solve these by moving everything to one side. I prefer to keep the L2 part positive, so let's subtract both sides by 48L.

48L = 270 + 2L2

0 = 270 + 2L2 - 48L

2L2 - 48L + 270 = 0. This is the most common way to write a quadratic, with the variable squared part first, the variable part, then the number without the variable.

Now you can solve a quadratic two ways, either by factoring or using the quadratic formula. The most direct way would be to use the quadratic formula. The quadratic formula is given below:

L = [-b ± √(b2 - 4ac)] / 2a

Here, our variable is L instead of x, and our equation is 2L2 - 48L + 270, so

a = 2

b = -48

c = 270

Plug these numbers into the quadratic formula to get:

L = [-(-48) ± √((-48)2 - 4(2)(270))] / 2(2)

= [48 ± √(2304 - 2160)] / 4

= [48 ± √(144)] / 4

= (48 ± 12) / 4

= 48/4 ± 12/4

= 12 ± 3

For our last step, we have a ±, which tells us we have two answers: 12 + 3, and 12 - 3. Knowing this, we have L = 15, and L = 9.

Now we have take both of these values, and plug them into our area equation to get a corresponding W. For L = 15,

A = LW

135 = (15)W

W = 135/15

= 9

For L = 9,

A = LW

135 = (9)W

W = 135/9

= 15

So if L = 15, W = 9, and if L = 9, W = 15. Notice these make the same rectangle, so you could probably just choose one of these for your answer.

by a visitor

Question

The area of a rectangle is 45 square cm. If the length is 4 cm greater than the width, what is the dimensions of the rectangle?

Answer

The picture below shows the rectangle with the area of 45cm2. Now, let the width be w. Since, the length is 4cm greater than the width, the length will be w+4 cm.

Before we can find the dimensions of the rectangle, we need find w first. Here's how:

1) Write an equation that relates 45cm2, w+4 and w.
To do so, we know that the area of the rectangle, 45cm2 can be found by multiplying w with w+4. Hence, we have:

To continue, we need to remove the bracket and simplify the equation. This is shown below:

2) Solve the Quadratic Equation
Notice that, now we have a quadratic equation:

To find w, we need to solve the quadratic equation. One way to do so is to factorize the quadratic equation. Hence, we have:

Now, since w is the width of a rectangle. There is no way it can be a negative number. Therefore, w must be 5.

With w = 5. The...

...and the...

Below is the rectangle with its dimensions: