, so ; this is the similarity ratio of  to .

 and  are not corresponding sides, nor are they congruent to corresponding sides, so it may or may not be true that .

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 .

The ratio of the areas of similar rectangles is the square of the similarity ratio, so the quotient in the given choice is .

However, by similarity, and the fact that opposite sides  and  are congruent,

.

The correct choice is .

, so

The area of the rectangle is 

The two rectangles are similar with similarity ratio 2.

Corresponding sides of similar rectangles are in proportion, so

.

Since opposite sides of a rectangle are congruent, , so

.

 and  are diagonals of the rectangle. If they are constructed, then, since  and  (both are right angles), by the Side-Angle-Side Similarity Theorem, . By similarity, .

The ratio of the perimeters of the rectangles is

,

It follows from  and a property of proportions that this ratio is equal to .

However, 

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:

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:

A rectangle with a length of  and a width of  has the same ratio of dimensions as a rectangle with a length of  and a width of , 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  and a length of . If the engineer's scale model has a width of , 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:

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:

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:

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:

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 , where  is a constant. Since AD is 15 and BD is 9, the triangle must be a Pythagorean Triple with , 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 , which gives us 42, our final answer.

Given the information we are provided, we could set up an quadratic equation, however, this equation would be really complicated to solve, instead we should try to find whether the diagonal of this rectangle can be the hypotenuse of Pythagorean triangle. The hypotenuse AD of triangle ADC, has length 10, therefore if it is a Pythagorean Triple, its other lengths must be 6 and 8, since the sides are in ratio  where  is a constant. By testing, we see that if  and  we get an area of 48 for this rectangle, by multiplying DC by AC. This is the original area of our rectangle, therefore, these must be the right lenghts and the final answer is given by , which is 28.

Alternatively, we could have found the missing lengths with trial and error, by doing so, we would have picked length for which, the area is 48 and tried to see whether the diagonal would remain 10, until we get the working set 6 and 8.