To solve for the perimeter of the Quidditch field, we must first determine the length of the field. With the provided information, we can solve for this by drafting an algebraic equation:

$\displaystyle \\length=3(width)-10\;yd=3(30\;yd)-10\;yd\\length=90\;yd-10\;yd=80\;yd$

Now that we know our length, we can determine the perimeter of the field:

$\displaystyle Perimeter=80\;yd+80\;yd+30\;yd+30\;yd=220\;yd$

A rectangle has the same length on either side and the same width on either side. Therefore if the football field has 100 yards in length and 50 yards in width you can use this formula.

$\displaystyle 2l + 2 w = p$

$\displaystyle (2 * 100) + (2 * 50) =$

$\displaystyle 200 + 100 =$

$\displaystyle 300 yards$

To solve, simply use the formula for the perimeter of a rectangle. Thus,

$\displaystyle P=e(l+w)=2*(7+2)=2*9=18$

If the formula escapes you, simply draw a picture and add all of the sides together. Rememeber, a rectangle has 4 sides, two are equal and the other 2 are equal.

First, we must solve for $\displaystyle x$. The formula for the area of a rectangle is $\displaystyle A=lw$. Because we know the area is $\displaystyle 27$, and the length and width are $\displaystyle x$ and $\displaystyle 3x$, we can plug these in to solve for $\displaystyle x$, as follows:

$\displaystyle A=lw$

$\displaystyle 27=x(3x)$

$\displaystyle 27=3x^2$

$\displaystyle \frac{27}{3}=\frac{3x^2}{3}$

$\displaystyle 9=x^2$

$\displaystyle \sqrt{9}=\sqrt{x^2}$

$\displaystyle x=3$

Since we know that $\displaystyle x=3$, we can use this to solve for the sides of the rectangle:

$\displaystyle l=x$

$\displaystyle l=3$

and

$\displaystyle w=3x$

$\displaystyle w=3(3)$

$\displaystyle w=9$

We can now use these values to solve for the rectangle's perimeter:

$\displaystyle P=2(l+w)$

$\displaystyle P=2(3+9)$

$\displaystyle P=2(12)$

$\displaystyle P=24cm$

Therefore, the perimeter of the rectangle is $\displaystyle 24 cm$.

Not enough information is given to solve this problem. While $\displaystyle 24 cm^2$ is a possible answer (assuming all of the sides of the rectangle are $\displaystyle 6cm$), because it is a rectangle, you cannot assume that all of the sides are the same. Because $\displaystyle 36$ is not a prime number, there is more than one combination of numbers that could work out to equal an area of $\displaystyle 36 cm^2$. 

Since we know the width and area, we can determine the length by using the equation for area of a rectangle:

$\displaystyle \\area=length\cdot width\\315\;cm^2=length\cdot15\;cm\\length=\frac{315\;cm^2}{15\;cm=21\;cm}$

Now that we know the length of the tile, we can solve for perimeter:

$\displaystyle perimeter=2l+2w=2(21\;cm)+2(15\;cm)=42\;cm+30\;cm=72\;cm$

First we need to know that the square root of 4 is 2, we can know this because because if we work backwards we see that 2 x 2= 4, so we know the width of the rectangle is 2.

$\displaystyle \sqrt{4}=\sqrt{2\times 2}=2$

Since the length is 2 more than the width, we add 2 + 2 and get a length of 4 inches.

$\displaystyle 2+2=4$

Now we can draw the rectangle and label each width (short side) with a measure of 2 inches and each length (long side) with a measure of 4 inches.

Now all we need to do is add up all the sides, so we have 2 + 2 + 4 + 4 which equals 12.

$\displaystyle \\P= 2 + 2 + 4 + 4 \\P=12$

Since we are adding, and the measurements are in inches, our answer is $\displaystyle 12 in$.

For a rectangle, its perimeter is the sum of all for sides. We can write

$\displaystyle 54=20+20+S+S$

Simplify

$\displaystyle 54=40+2S$

Solve for S

$\displaystyle 14=2S$

$\displaystyle S=7$

The given rectangle has a length that is $\displaystyle 10$ units longer than its width. This can be expressed in the following equation, where $\displaystyle l$ is the length and $\displaystyle w$ is the width of the rectangle.

$\displaystyle l = w +10$

Since the area of the rectangle is equal to its length multiplied by its width ($\displaystyle A = l * w$), and the area of the rectangle is given, the following equation must be true.

$\displaystyle 75 = l * w$

Replacing $\displaystyle l$ in this equation with its value stated in the first equation results in the following.

$\displaystyle 75 = (w+10) * w$

Distribute the variable into the parentheses.

$\displaystyle 75 = w^2 +10w$

$\displaystyle w^2 +10w -75 = 0$

Factor the polynomial.

$\displaystyle (w-5)(w+15) = 0$

$\displaystyle 5$ and $\displaystyle -15$ are both solutions for this equation, but $\displaystyle -15$ is not valid as a width for a rectangle. The width of the rectangle is $\displaystyle 5$ units, which is the shorter side since the length is $\displaystyle 10$ units longer $\displaystyle (l = w+10)$.

The formula for the perimeter of a rectangle is:

$\displaystyle P =$$\displaystyle 2($$\displaystyle l$$\displaystyle +$$\displaystyle w$$\displaystyle )$, where $\displaystyle l$ represents the length and $\displaystyle w$ represents the width.

We know the perimeter of the rectange is $\displaystyle 16$$\displaystyle ft$ and the length is $\displaystyle 5$$\displaystyle ft$. Plugging these values into the equation, we get:

$\displaystyle 16$$\displaystyle ft$ $\displaystyle =$$\displaystyle 2($$\displaystyle l$$\displaystyle +$$\displaystyle 5$$\displaystyle ft$$\displaystyle )$

$\displaystyle 16$$\displaystyle ft$ $\displaystyle =$$\displaystyle 2l$$\displaystyle +$ $\displaystyle 10$$\displaystyle ft$

$\displaystyle 2l$$\displaystyle =$$\displaystyle 6$$\displaystyle ft$

$\displaystyle l$$\displaystyle =$$\displaystyle 3$$\displaystyle ft$