The fence is going around the backyard, so this is a perimeter problem. 

We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown. 

Subtract  from both sides

Divide  by both sides

In order to find the area of a rectangle we use the formula . 

In this problem, we know the width is .  We also know that the length is twice as long as the width, which can be written as .  This means that in order to find the length, we must multiply the width by .

Now that we know that our length is , we simply multiply it by our width of .

The area of the rectangle is .

This type of problem reminds us to be wary of simply plugging in numbers (which works with certain problems). If you were to choose 1 and 7 here, the perimeter would be larger; if you chose 10 and 70, the area would be much larger.

To solve this problem with variables:

From here we can see that smaller values of will lead to a larger perimeter, while larger values of will lead to a larger area.

The answer cannot be determined.

(a) The surface of a rectangular prism comprises six rectangles, so we can take the sum of their areas.

Two rectangles have area: .

Two rectangles have area: .

Two rectangles have area: .

Add the areas: 

(b) The surface of a cube comprises six squares, so we can square the sidelength - which we rewrite as 30 centimeters - and multiply the result by 6:

.

The first figure has the greater surface area.

First, you must calculate Column A. The formula for the area of a rectangle is . Plug in the values given to get , which gives you . Then, calculate the area of the square. Since all of the sides of a square are equal, the formula is , or . Therefore, the area of the square is , which gives you . Therefore, the quantity in Column A is greater.

The area of a rectangle is the product of its length and its width. 

Since the area of Rectangle  is 1,000, 

Substitute 40 for  in the height of Rectangle  and calculate the area as follows:

The area of a rectangle is the product of its length and its width.

The rectangle described in (a) has area 

The rectangle described in (b) has area 

, and  is positive, so , and . The rectangle from (a) has the greater area.

Note that the value of  has no bearing on the answer, except for the fact that it is positive.

We can fill in a few edge lengths below:

All six sides are rectangles, so their areas are equal to the products of their dimensions. We specifically notice that the top, bottom. front, and back each have area . Since the total of these four areas is  . Since the left and right sides have not been included, the total surface area must be more than .

To find the area of the figure above, we need to split the figure into two rectangles. 

Using our area formula, , we can solve for the area of both of our rectangles

To find our final answer, we need to add the areas together.