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

$\displaystyle 2x\cdot2y=4xy$