In order to find the area of a rectangle we use the formula \(\displaystyle \small A=l\cdot w\). 

In this problem, we know the width is \(\displaystyle \small 3\: in\).  We also know that the length is twice as long as the width, which can be written as \(\displaystyle \small 2w\).  This means that in order to find the length, we must multiply the width by \(\displaystyle \small 2\).

\(\displaystyle \small 2(3\: in)=6\: in\)

Now that we know that our length is \(\displaystyle \small 6\: in\), we simply multiply it by our width of \(\displaystyle \small 3\: in\).

\(\displaystyle \small \small 6\: in\cdot 3\: in=18\: in^{2}\)

The area of the rectangle is \(\displaystyle \small 18\: in^{2}\).

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:

\(\displaystyle perimeter=w+w+7w+7w= 16w\)

\(\displaystyle area=w\cdot (7w)= 7w^{2}\)

From here we can see that smaller values of \(\displaystyle w\) will lead to a larger perimeter, while larger values of \(\displaystyle w\) 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: \(\displaystyle 60 \times 30 = 1,800 \textrm{ cm}^{2}\).

Two rectangles have area: \(\displaystyle 60 \times 15 = 900 \textrm{ cm}^{2}\).

Two rectangles have area: \(\displaystyle 15 \times 30 = 450 \textrm{ cm}^{2}\).

Add the areas: \(\displaystyle A = 1,800 + 1,800 + 900 + 900 + 450 + 450 = 6,300 \textrm{ cm}^{2}\)

(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:

\(\displaystyle A = 6 \times 30 ^{2} = 6 \times 900 = 5,400 \textrm{ cm}^{2}\).

The first figure has the greater surface area.

First, you must calculate Column A. The formula for the area of a rectangle is \(\displaystyle A=lw\). Plug in the values given to get \(\displaystyle 11\cdot 5\), which gives you \(\displaystyle 55cm^2\). Then, calculate the area of the square. Since all of the sides of a square are equal, the formula is \(\displaystyle A=s^2\), or \(\displaystyle A=lw\). Therefore, the area of the square is \(\displaystyle 7\cdot 7\), which gives you \(\displaystyle 49cm^2\). 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 \(\displaystyle ABCD\) is 1,000, 

\(\displaystyle 20 \cdot (x+10) = 1,000\)

\(\displaystyle x+10 = 1,000 \div 20 = 50\)

\(\displaystyle x = 50 - 10 = 40\)

Substitute 40 for \(\displaystyle x\) in the height of Rectangle \(\displaystyle EFGH\) and calculate the area as follows:

\(\displaystyle 10 (x+20) = 10 (40+20) = 10 \cdot 60 = 600\)

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

The rectangle described in (a) has area 

\(\displaystyle 20 \cdot (X+10) = 20 X + 20 \cdot 10 = 20 X +100\)

The rectangle described in (b) has area 

\(\displaystyle 10 \cdot (X+20) = 10 X + 10 \cdot 20 = 10 X +100\)

\(\displaystyle 20 > 10\), and \(\displaystyle X\) is positive, so \(\displaystyle 20 X> 10 X\), and \(\displaystyle 20 X+ 100 > 10 X + 100\). The rectangle from (a) has the greater area.

Note that the value of \(\displaystyle X\) 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 \(\displaystyle 20x\). Since the total of these four areas is  \(\displaystyle 4 \cdot 20x = 80 x\). Since the left and right sides have not been included, the total surface area must be more than \(\displaystyle 80x\).

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

Using our area formula, \(\displaystyle A=l\times w\), we can solve for the area of both of our rectangles

\(\displaystyle A=7\times 5\)            \(\displaystyle A=6\times 4\)

\(\displaystyle A=35in^2\)           \(\displaystyle A=24in^2\)

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

\(\displaystyle 35in^2+24in^2=59in^2\)

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

Using our area formula, \(\displaystyle A=l\times w\), we can solve for the area of both of our rectangles

\(\displaystyle A=6\times 2\)            \(\displaystyle A=5\times 2\)

\(\displaystyle A=12in^2\)           \(\displaystyle A=10in^2\)

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

\(\displaystyle 12in^2+10in^2=22in^2\)

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

Using our area formula, \(\displaystyle A=l\times w\), we can solve for the area of both of our rectangles

\(\displaystyle A=7\times 2\)            \(\displaystyle A=6\times 3\)

\(\displaystyle A=14in^2\)           \(\displaystyle A=18in^2\)

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

\(\displaystyle 14in^2+18in^2=32in^2\)