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:

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

\(\displaystyle 16\)\(\displaystyle ft\) \(\displaystyle =\)\(\displaystyle 2l\)\(\displaystyle +\) \(\displaystyle 10\)\(\displaystyle ft\)

\(\displaystyle 2l\)\(\displaystyle =\)\(\displaystyle 6\)\(\displaystyle ft\)

\(\displaystyle l\)\(\displaystyle =\)\(\displaystyle 3\)\(\displaystyle ft\)

To find the width, multiply the length that you have been given by 2, and subtract the result from the perimeter. You now have the total length for the remaining 2 sides.  This number divided by 2 is the width. 

\(\displaystyle 36-(12\cdot 2)=12\)

\(\displaystyle 12/2=6\)

If we call the width of this rectangle \(\displaystyle w\), then its length \(\displaystyle L\) can be restated as\(\displaystyle L=4\frac{1}{2}w\), or, equivalently, \(\displaystyle L=4.5w\).

The perimeter can then be written as:

\(\displaystyle P=2L+2w=2(4.5w)+2w\)

Since the perimeter of the rectangle is 272 cm, we can set up the following equation:

\(\displaystyle 2(4.5w)+2w=272\)

\(\displaystyle 9w+2w=272\)

\(\displaystyle 11w=272\)

\(\displaystyle w=\frac{272}{11}=24.7\)

To find the width, we must take half of the length, which means we must divide the length by 2. Then we must take 2 more than that number, which means we must add 2 to the number. Combining these, we get:

w = ½ l + 2

The width equals 3 times the length, so 3l, plus an additional two inches, so + 2, = 3l + 2

We can solve this by setting up a proportion and solving for x,the length of the shorter side of the house. If the drawing is scale and is 6 : 8, then the actual house is x : 64. Then we can cross multiply so that 384 = 8x. We then divide by 8 to get x = 48. 

Recall how to find the perimeter of a rectangle.

\(\displaystyle \text{Perimeter}=2(\text{length}+\text{width})\)

We can then manipulate this equation to find the width.

\(\displaystyle \text{length}+\text{width}=\frac{\text{Perimeter}}{2}\)

\(\displaystyle \text{width}=\frac{\text{Perimeter}}{2}-\text{length}\)

Now, plug in the information given by the question to find the width.

\(\displaystyle \text{width}=\frac{14}{2}-5=7-5=2\)

Recall how to find the perimeter of a rectangle.

\(\displaystyle \text{Perimeter}=2(\text{length}+\text{width})\)

We can then manipulate this equation to find the width.

\(\displaystyle \text{length}+\text{width}=\frac{\text{Perimeter}}{2}\)

\(\displaystyle \text{width}=\frac{\text{Perimeter}}{2}-\text{length}\)

Now, plug in the information given by the question to find the width.

\(\displaystyle \text{width}=\frac{18}{2}-3=9-3=6\)