Corresponding sidelengths of similar polygons are in proportion, so

\(\displaystyle \frac{DF}{AD}= \frac{DB}{AB}\)

\(\displaystyle AB = 8, AD = 4\), so

\(\displaystyle \frac{DF}{4}= \frac{DB}{8}\)

\(\displaystyle DF= \frac{DB}{8} \cdot 4\)

\(\displaystyle DF= \frac{DB}{2}\)

We can use the Pythagorean Theorem to find \(\displaystyle DB\):

\(\displaystyle DB = \sqrt{(AD)^{2}+(AB)^{2}}\)

\(\displaystyle DB = \sqrt{4^{2}+8^{2}} = \sqrt{16+64 } = \sqrt{80} = \sqrt{16} \cdot \sqrt{5} = 4 \sqrt{5}\)

\(\displaystyle DF= \frac{DB}{2}= \frac{ 4 \sqrt{5}}{2} = 2\sqrt{5}\)

The area of \(\displaystyle \textup{Rect } DBEF\) is 

\(\displaystyle DB \cdot DF = 4 \sqrt{5} \cdot 2\sqrt{5} = 8 \cdot 5 = 40\)

Polygon \(\displaystyle ABEFD\) can be seen as a composite of right \(\displaystyle \Delta ABD\) and \(\displaystyle \textup{Rect } DBEF\), so we calculate the individual areas and add them.

The area of \(\displaystyle \Delta ABD\) is half the product of legs \(\displaystyle AB\) and \(\displaystyle A D\):

\(\displaystyle \frac{1}{2} \cdot AB \cdot AD = \frac{1}{2} \cdot 12 \cdot 6 = 36\)

Now we find the area of \(\displaystyle \textup{Rect } DBEF\). We can do this by first finding \(\displaystyle DB\) using the Pythagorean Theorem:

\(\displaystyle DB = \sqrt{(AD)^{2}+(AB)^{2}}\)

\(\displaystyle DB = \sqrt{6^{2}+12^{2}} = \sqrt{36+144} = \sqrt{180} = \sqrt{36} \cdot \sqrt{5} = 6 \sqrt{5}\)

The similarity of \(\displaystyle \textup{Rect } DBEF\) to \(\displaystyle \textup{Rect } ABCD\) implies

\(\displaystyle \frac{DF}{AD}= \frac{DB}{AB}\)

so

\(\displaystyle \frac{DF}{6}= \frac{6\sqrt{5}}{12}\)

\(\displaystyle DF= \frac{6 \cdot 6\sqrt{5}}{12} = \frac{3 6\sqrt{5}}{12} = 3\sqrt{5}\)

The area of \(\displaystyle \textup{Rect } DBEF\) is the product of \(\displaystyle DB\) and \(\displaystyle DF\):

\(\displaystyle DB \cdot DF = 6 \sqrt{5} \cdot 3 \sqrt{5}= 6\cdot 3 \cdot \sqrt{5} \cdot \sqrt{5}= 18 \cdot 5 = 90\)

Now add: \(\displaystyle 90 + 36 = 126\), the correct response.

\(\displaystyle AB = 7\) and \(\displaystyle AC = AB + BC = 7 + 3 = 10\), so the similarity ratio of \(\displaystyle \textup{Rect }ACDE\) to \(\displaystyle \textup{Rect }ABGF\) is 10 to 7. The ratio of the areas is the square of this, or 

\(\displaystyle 10^{2}:7^{2}\)

or 

\(\displaystyle 100:49\)

Therefore, \(\displaystyle \textup{Rect }ABGF\) comprises \(\displaystyle \frac{49}{100} = 49 \%\) of \(\displaystyle \textup{Rect }ABGF\), and the remainder of the rectangle - the brown region - is 51% of \(\displaystyle \textup{Rect }ABGF\).

\(\displaystyle \textup{Rect }ABGF \sim \textup{Rect }ACDE\), so the sides are in proportion - that is,

\(\displaystyle \frac{AF}{AB}= \frac{AE}{AC}\)

\(\displaystyle AC = AB + BC = 12 + 3 = 15\)

Set 

\(\displaystyle AB = 12\), \(\displaystyle AC = 15\), \(\displaystyle AE = 10\) and solve for \(\displaystyle AF\):

\(\displaystyle \frac{AF}{12}= \frac{10}{15}\)

\(\displaystyle AF = \frac{10}{15} \cdot 12 = 8\)

\(\displaystyle \textup{Rect }ABGF\) has area 

\(\displaystyle AB \cdot AF = 12 \cdot 8 = 96\)

Given that w = 2x and l = 1.5w + 5, a substitution will show that l = 1.5(2x) + 5 = 3x + 5.  

P = 2w + 2l = 2(2x) + 2(3x + 5) = 4x + 6x + 10 = 10x + 10 = 10(x + 1)

The length of the garden, in feet, is \(\displaystyle 2 \times 7 = 14\) feet less than that of the entire lot, or 

\(\displaystyle L = 2x-14\);

The width of the garden, in feet, is \(\displaystyle 2 \times 7 = 14\) less than that of the entire lot, or 

\(\displaystyle W = x-14\).

The perimeter, in feet, is twice the sum of the two:

\(\displaystyle P = 2 (L + W)\)

\(\displaystyle = 2 (2x-14 + x - 14 )= 2 (3x-28) = 6x - 56\)

The inner square, which represents the garden, has sidelength \(\displaystyle (75 - 2\times 8) = 59\) feet, so its perimeter is four times this:

\(\displaystyle P = 4 \times 59 = 236\) feet.

To solve this problem, find the perimeter of the rectangle. There are two sides that each measure 50 yards and two sides that each measure 40 yards. Together these four sides measure 180 yards.

We define the variables as \(\displaystyle w = \mbox{width}\) and \(\displaystyle l = \mbox{length} = w + 2\).

We substitute these values into the equation for the area of a rectangle and get \(\displaystyle A_{\mbox{rectangle} }= wl = w(w + 2) = 80\). 

\(\displaystyle w^2 +2w-80=0\)

\(\displaystyle (w - 8)(w + 10) = 0\)

\(\displaystyle w = 8\) or \(\displaystyle w = -10\)

Lengths cannot be negative, so the only correct answer is \(\displaystyle w = 8\). If \(\displaystyle w = 8\), then \(\displaystyle l = 10\). 

Therefore, \(\displaystyle \mbox{perimeter }= 2w + 2l = 16 + 20 = 36\:\mbox{m}\).

We have to be careful of our units. The floor is given in feet and the tile in inches. Since the floor is 6ft x 3ft. we can say it is 72in x 36in, because 12 inches equals 1 foot. If the tiles are 9in x 9in we can fit 8 tiles along the length and 4 tiles along the width. To find the total number of tiles we multiply 8 x 4 = 32. Alternately we could find the area of the floor (72 x 36, and divide by the area of the tile 9 x 9)