The ratio of the areas of similar rectangles is the square of the similarity ratio, so the ratio of the area of \(\displaystyle \textup{Rect }ABCD\) to that of \(\displaystyle \textup{Rect }WXYZ\) is 

\(\displaystyle \left ( \frac{7}{5} \right )^{2} = \frac{49}{25}\).

So if the area of \(\displaystyle \textup{Rect }WXYZ\) is \(\displaystyle N\), the area of \(\displaystyle \textup{Rect }ABCD\) is \(\displaystyle \frac{49}{25} N\)

The area of the shaded region is the difference between the areas of the rectangles, making this area

\(\displaystyle \frac{49}{25} N - N = \frac{24}{25} N\).

The desired ratio is 24 to 25.

\(\displaystyle \frac{AB}{WX} = 3\), so \(\displaystyle \frac{WX}{AB} = \frac{1}{3}\); this is the similarity ratio of \(\displaystyle \textup{Rect }WXYZ\) to \(\displaystyle \textup{Rect }ABCD\).

\(\displaystyle \overline{WZ}\) and \(\displaystyle \overline{AB}\) are not corresponding sides, nor are they congruent to corresponding sides, so it may or may not be true that \(\displaystyle \frac{WZ}{AB} = \frac{1}{3}\).

The ratio of the perimeters of similar rectangles is the same as their actual similarity ratio, but the choice gives the rectangles in the original order; the quotient of the perimeters as given is 3, not \(\displaystyle \frac{1}{3}\).

The ratio of the areas of similar rectangles is the square of the similarity ratio, so the quotient in the given choice is \(\displaystyle \left (\frac{1}{3} \right )^{2} = \frac{1}{9}\).

However, by similarity, and the fact that opposite sides \(\displaystyle \overline{BD }\) and \(\displaystyle \overline{AC}\) are congruent,

\(\displaystyle \frac{WY}{BD} = \frac{WY}{AC} = \frac{WX}{AB} = \frac{1}{3}\).

The correct choice is \(\displaystyle \frac{WY}{BD}\).

\(\displaystyle \textup{Rect }ABCF \sim \textup{Rect }CDEF\), so

\(\displaystyle \frac{AF}{CF} = \frac{CF}{EF}\)

\(\displaystyle \frac{AF}{24} = \frac{24}{12}\)

\(\displaystyle AF = \frac{24}{12} \cdot 24 = 48\)

The area of the rectangle is 

\(\displaystyle AE \cdot DE\)

\(\displaystyle = \left (AF + EF \right ) \cdot CF\)

\(\displaystyle = \left (48+12 \right ) \cdot 24\)

\(\displaystyle = 60 \cdot 24\)

\(\displaystyle = 1,440\)

The two rectangles are similar with similarity ratio 2.

Corresponding sides of similar rectangles are in proportion, so

\(\displaystyle \frac{BC}{XY} = \frac{AB}{WX} = 2\).

Since opposite sides of a rectangle are congruent, \(\displaystyle XY = WZ\), so

\(\displaystyle \frac{BC}{WZ}= 2\).

\(\displaystyle \overline{AC}\) and \(\displaystyle \overline{WY}\) are diagonals of the rectangle. If they are constructed, then, since \(\displaystyle \frac{AB}{WX} = \frac{BC}{XY}\) and \(\displaystyle \angle B \cong \angle X\) (both are right angles), by the Side-Angle-Side Similarity Theorem, \(\displaystyle \bigtriangleup ABC \cong \bigtriangleup WXY\). By similarity, \(\displaystyle \frac{AC}{WY} = \frac{AB}{WX} =2\).

The ratio of the perimeters of the rectangles is

\(\displaystyle \frac{2 \cdot AB + 2 \cdot BC }{2\cdot WX + 2 \cdot XY}\)

\(\displaystyle = \frac{ AB + BC }{ WX + XY}\),

It follows from \(\displaystyle \frac{AB}{WX} =\frac{BC}{XY} = 2\) and a property of proportions that this ratio is equal to \(\displaystyle 2\).

However, 

\(\displaystyle \frac{CD}{YX}\)

is not a ratio of corresponding sides of the rectangle, so it does not have any restrictions on it. This is the correct choice.

In order for two rectangles to be similar, the ratio of their dimensions must be equal. We can first check the ratio of the length to the width for the given rectangle, and then see which option has the same ratio, which will tell us whether or not the rectangle is similar:

\(\displaystyle Given:\frac{L}{W}=\frac{15}{6}=\frac{5}{2}\)

So in order for a rectangle to be similar, the ratio of its length to its width must be the same. All we must do then is check the answer options, in no particular order, for the rectangle with the same ratio:

\(\displaystyle Option1:\frac{L}{W}=\frac{18}{7}\) 

\(\displaystyle Option2:\frac{L}{W}=\frac{12}{5}\)

\(\displaystyle Option3:\frac{L}{W}=\frac{24}{10}=\frac{12}{5}\)

\(\displaystyle Option4:\frac{L}{W}=\frac{32}{12}=\frac{8}{3}\)

\(\displaystyle Option5:\frac{L}{W}=\frac{20}{8}=\frac{5}{2}\)

A rectangle with a length of \(\displaystyle 20\) and a width of \(\displaystyle 8\) has the same ratio of dimensions as a rectangle with a length of \(\displaystyle 15\) and a width of \(\displaystyle 6\), so these two rectangles are similar.

An engineer is making a scale model of a building. The real building needs to have a width of \(\displaystyle 54m\) and a length of \(\displaystyle 164m\). If the engineer's scale model has a width of \(\displaystyle 0.05m\), what does the length of the model need to be?

To begin, we need to know what a scale model is. A scale model is a smaller version of something that is "to scale." In other words, it is similar but not congruent. 

So, we find a length that will make the model accurate, we need a ratio. Try the following:

\(\displaystyle \frac{0.05}{54}=\frac{l_m}{164}\)

\(\displaystyle l_m=\frac{.05}{54}*164=0.15m\)

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

\(\displaystyle P=2(10)+2(5)\)

\(\displaystyle P=20+10\)

\(\displaystyle P=30\)

The fence is 2 feet around the field. Therefore the area actually enclosed by the fence is 4 feet longer and 4 feet wider than the field.

The length of fencing needed is the perimeter of the area enclosed which is calculated as follows:

\(\displaystyle 2\times((21+4)+(16+4))=2\times(25+20)=2\times45=90\)

The fencing should be 90 feet.

To find the perimeter, we need the length and the width. We are only given the length, so first we must find the width using the given area:

\(\displaystyle A=LW\)

\(\displaystyle 63=9W\rightarrow W=7\)

The perimeter is simply two times the length plus two times the width, so we can now use the known length and width to calculate the perimeter:

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

\(\displaystyle P=2(9)+2(7)=32\)

We can see that the hypotenuse AD of triangle ABD is 15 cm. We should check whether ABD is a Pythagorean Triple, whose sides are in the ratio \(\displaystyle 3n,4n,5n\), where \(\displaystyle n\) is a constant. Since AD is 15 and BD is 9, the triangle must be a Pythagorean Triple with \(\displaystyle n=3\), therefore AB must be 12 cm long. Now we know all the lengths of the sides of our triangle, we can find the perimeter which will be given by \(\displaystyle 2\cdot (AB+BD)\), which gives us 42, our final answer.