Question

The area of a rectangular ceiling tile is 756 square inches. The perimeter is 114 inches. What are the dimensions of the tile?

Answer 1

Here we need to use what we know about rectangles to make a system of equations.

By solving that system we found that the tile has a length of 36 inches and a width of 21 inches.

Remember that for a rectangle of length L and width W, the perimeter is:

P = 2*(L + W)

And the area is:

A = W*L

Here we know that the perimeter is 114 inches, then we can write:

114in = 2*(L + W)

We also know that the area is 756 in^2, then we can write:

756 in^2 = L*W

So we found two equations, which means that we have a system of two equations with two variables:

114in = 2*(L + W)

756 in^2 = L*W

To solve this, the first step is to isolate one of the variables in one of the equations, we can isolate L in the first equation:

114in = 2*(L + W)

114in/2 = (L + W)

57in = L + W

57in - W = L

Now that we have an expression equivalent to L, we can replace it in the other equation to get:

756 in^2 = L*W

756 in^2 = (57in - W)*W

Now we can solve this for W.

756 in^2 = W*57in - W^2

W^2 - W*57in + 756 in^2 = 0

The solutions are given by the Bhaskara's formula:

$W = \frac{57in \pm \sqrt{(-57in)^2 - 4*1*(756in^2)} }{2*1} = \frac{57in \pm 15in}{2}$

Then the two possible values of the width will be:

W = (57in + 15in)/2 =  36 in

W = (57in - 15in)/2 = 21 in

Suppose that we choose the second solution, W = 21in

Now using the equation 57in - W = L we can find the value of L

L = 57in - W = 57in - 21in = 36in

L = 36in

Then we found that the tile has a length of 36 inches and a width of 21 inches.

If you want to learn more, you can read:

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