If the area of the rectangle is equal to the area of the square, then it must have an area of \(\displaystyle \dpi{100} \small 144in^{2}\) \(\displaystyle \dpi{100} \small (12\times 12)\). If the rectangle has an area of \(\displaystyle \dpi{100} \small 144 in^{2}\) and a side with a lenth of 3 inches, then the equation to solve the problem would be \(\displaystyle \dpi{100} \small 144=3x\), where \(\displaystyle \dpi{100} \small x\) is the length of the rectangle. The solution:

\(\displaystyle \frac{144}{3} = 48\).

Let \(\displaystyle W\) be the width of the rectangle; then its length is \(\displaystyle L = 2W\), and its perimeter is 

\(\displaystyle 2W + 2 (2W) = 2W + 4W = 6W\)

Set this equal to \(\displaystyle 6N - 12\) and solve for \(\displaystyle W\):

\(\displaystyle 6W = 6N - 12\)

\(\displaystyle \frac{6W}{6} =\frac{ 6N-12}{6}\)

\(\displaystyle W = N - 2\)

The width is \(\displaystyle W = N - 2\) and the length is  \(\displaystyle L = 2W = 2 \left (N - 2 \right ) = 2N - 4\),  so multiply these expressions to get the area:

\(\displaystyle A = LW = (N-2) \cdot \left (2N-4 \right ) = 2N^{2} -8N + 8\)

A rectangle with vertices \(\displaystyle (-4,-3), (-4,7), (1,7), (1,-3)\) has width \(\displaystyle 1 - (-4) = 5\) and height \(\displaystyle 7 - (-3) = 10\) , thereby having area \(\displaystyle 10 \times 5 = 50\).

The portion of the rectangle in Quadrant III is a rectangle with vertices

\(\displaystyle (-4,-3), (-4,0), (0,0), (0,-3)\).

It has width \(\displaystyle 0-(-4) = 4\) and height \(\displaystyle 0-(-3) = 3\), thereby having area \(\displaystyle 4 \times 3 = 12\) .

Therefore, \(\displaystyle \frac{12}{50 }\) of the rectangle is in Quadrant III; this is equal to 

\(\displaystyle \frac{12}{50 } \times 100 = 24 \%\)

To find the area of a rectangle, you must use the following formula:

\(\displaystyle A=lw\)

\(\displaystyle A=(10)(5)\)

\(\displaystyle A=50\)

The area of a rectangle is the product of the length and the width. The expression \(\displaystyle (A+4)(A-4)\) can be multplied by noting that this is the product of the sum and the difference of the same two terms; its product is the difference of the squares of the terms, or 

\(\displaystyle (A+4)(A-4) = A^{2}-4^{2}= A^{2}-16\)

A rectangle with vertices \(\displaystyle (-4,-3), (-4,7), (1,7), (1,-3)\) has width \(\displaystyle 1 - (-4) = 5\) and height \(\displaystyle 7 - (-3) = 10\) ; it follows that its area is \(\displaystyle 10 \times 5 = 50\). 

The portion of the rectangle in Quadrant IV has vertices \(\displaystyle (0,0),(1,0), (1, -3), (0, -3)\). Its width is \(\displaystyle 1-0 = 1\), and its height is \(\displaystyle 0- (-3 ) = 3\), so its area is \(\displaystyle 1 \times 3 = 3\).

Therefore, \(\displaystyle \frac{3}{50 }\), or  \(\displaystyle \frac{3}{50} \times 100 \% = 6\%\), of this rectangle is in Quadrant IV.

The perimeter of a rectangle is the sum of all four sides, that is: \(\displaystyle 104= 2l + 2w\)

since \(\displaystyle l=3w\), we can rewrite the equation as: 

\(\displaystyle 104= 2(3w) + 2w\)

\(\displaystyle 104= 8w\)

\(\displaystyle w=13\textup{ in}\)

We are being asked for the area so we still aren't done. The area of a rectangle is the product of the width and length. We know what the width is so we can find the length and then take their product.

\(\displaystyle l=3w=3(13)=39\textup{ in}\)

\(\displaystyle A= l \cdot w = 39 \cdot 13\)

\(\displaystyle A = 507\textup{ in}^2\)

Mark is building a garden with raised beds. One side of the garden will be 10 feet long and the other will be 5 less than three times the first side. What area of will Mark's garden be?

This problem asks us to find the area of a rectangle. We are given one side and asked to find the other. To find the other, we need to use the provided clues.

"...five less..." \(\displaystyle -5\)

"...three times the first side..." \(\displaystyle 3x\) or \(\displaystyle 3(10)\)

So put it together:

\(\displaystyle 3(10)-5=30-5=25ft\)

Next, find the area via the following formula:

\(\displaystyle A=l*w=25ft*10ft=250ft^2\)

So our answer is:

\(\displaystyle 250ft^2\)

To calculate area, multiply width times height. Thus,

\(\displaystyle 7*9=63\)

To find area, simply multiply length times width. Thus

\(\displaystyle 2x\cdot2y=4xy\)