In order for two rectangles to be similar, the ratio of their dimensions must be equal. We can check which dimensions are those of a rectangle similar to the given one by first calculating the ratio of the length to the width for the given rectangle, and then doing the same for each of the answer choices until we find which has an equal ratio between its dimensions:

\(\displaystyle Given:\frac{L}{W}=\frac{16}{10}=\frac{8}{5}\)

So in order for a rectangle to be similar to the given rectangle, this must be the ratio of its length to its width. Now we check the answer choices, in no particular order, for one with this ratio:

\(\displaystyle Option1:\frac{L}{W}=\frac{4}{3}\)

\(\displaystyle Option2:\frac{L}{W}=\frac{18}{6}=3\)

\(\displaystyle Option3:\frac{L}{W}=\frac{20}{12}=\frac{5}{3}\)

\(\displaystyle Option4:\frac{L}{W}=\frac{6}{4}=\frac{3}{2}\)

\(\displaystyle Option5:\frac{L}{W}=\frac{32}{20}=\frac{8}{5}\)

We can see that only the rectangle with a length of  \(\displaystyle 32\)  and a width of  \(\displaystyle 20\)  has the same ratio as the given rectangle, so this is the similar one.

In order for two rectangles to be similar, the ratio of their dimensions must be equal. We can calculate the ratio of length to width for the given rectangle, and then check the answer choices for the rectangle whose dimensions have the same ratio:

\(\displaystyle Given:\frac{L}{W}=\frac{21}{9}=\frac{7}{3}\)

Now we check the answer choices, in no particular order, and the dimensions with the same ratio are those of the rectangle that is similar:

\(\displaystyle Option1:\frac{L}{W}=\frac{10}{4}=\frac{5}{2}\)

\(\displaystyle Option2:\frac{L}{W}=\frac{15}{8}\)

\(\displaystyle Option3:\frac{L}{W}=\frac{16}{9}\)

\(\displaystyle Option4:\frac{L}{W}=\frac{18}{10}=\frac{9}{5}\)

\(\displaystyle Option5:\frac{L}{W}=\frac{14}{6}=\frac{7}{3}\)

We can see that a rectangle with a length of  \(\displaystyle 14\)  and a width of  \(\displaystyle 6\)  has the same ratio of dimensions as the given rectangle, so this is the one that is similar.

The length of the midsegment of a trapezoid - the segment that has as its endpoints the midpoints of its legs - is half the sum of the lengths of its legs. Therefore,  Trapezoid \(\displaystyle ABCD\) has as the length of its midsegment

\(\displaystyle \frac{1}{2} (AB + CD) = \frac{1}{2} (12+16) = \frac{1}{2} \cdot 28 = 14\).

Sidelengths of similar figures are in proportion. If the similarity ratio is \(\displaystyle R\), then the bases of Trapezoid \(\displaystyle EFGH\) have length \(\displaystyle 12R\) and \(\displaystyle 14R\), so their midsegment will have length

\(\displaystyle \frac{1}{2} (EF+ GH) = \frac{1}{2} (12R+16R) = \frac{1}{2} \cdot 28R = 14R\),

meaning that the ratio of the lengths of the midsegments will be the same as the similarity ratio. Since the length of the midsegment of Trapezoid \(\displaystyle EFGH\) is 91, this similarity ratio is

\(\displaystyle \frac{91}{14}= \frac{13}{2}\) .

The ratio of the length of \(\displaystyle \overline{FG}\) to that of corresponding side \(\displaystyle \overline{BC}\) is therefore \(\displaystyle \frac{13}{2}\), so

\(\displaystyle \frac{FG}{BC} = \frac{13}{2}\)

\(\displaystyle \frac{FG}{3} = \frac{13}{2}\)

\(\displaystyle \frac{FG}{3} \cdot 3= \frac{13}{2} \cdot 3\)

\(\displaystyle FG= \frac{39}{2} = 19 \frac{1}{2}\)

The similarity ratio of the trapezoids is the ratio of the length of one side to that of the corresponding side of the other. For these trapezoids, we take the ratio of the lengths of corresponding sides \(\displaystyle \overline{EF}\) and \(\displaystyle \overline{AB}\):

\(\displaystyle \frac{EF}{AB} = \frac{66}{12} = \frac{11}{2}\).

The ratio of the areas is the square of this, or 

\(\displaystyle \left ( \frac{11}{2} \right )^{2} = \frac{121}{4}\).

The area of Trapezoid \(\displaystyle ABCD\) is one half the product of the height \(\displaystyle BC\) and the sum of bases \(\displaystyle AB\) and \(\displaystyle CD\):

\(\displaystyle A _{1} = \frac{1}{2} \cdot BC \cdot (AB + CD)\)

\(\displaystyle = \frac{1}{2} \cdot 3 \cdot (12+ 16)\)

\(\displaystyle = \frac{1}{2} \cdot 3 \cdot 28\)

\(\displaystyle = 42\)

Multiply this by the area ratio:

\(\displaystyle A _{2} = A _{1} \cdot \frac{121}{4}\)

\(\displaystyle A _{2} = 42 \cdot \frac{121}{4} = 1,270\frac{1}{2}\).

The correct response is 1,271.

The area of a rhombus is half the product of the lengths of its diagonals, so we can take advantage of this to find \(\displaystyle \overline{HM}\) from the known measurements of Rhombus \(\displaystyle RHOM\):

\(\displaystyle A_{1} = \frac{1}{2} (RO)(HM)\)

\(\displaystyle 90= \frac{1}{2} \cdot 15 \cdot HM\)

\(\displaystyle 90= \frac{15}{2} \cdot HM\)

\(\displaystyle \frac{2} {15}\cdot 90= \frac{2} {15}\cdot \frac{15}{2} \cdot HM\)

\(\displaystyle HM = 12\)

The ratio of the area of Rhombus \(\displaystyle SIPN\) to that of Rhombus \(\displaystyle RHOM\)is 4, so their similarity ratio is the square root of this, or 2. We can calculate \(\displaystyle IN\) now:

\(\displaystyle \frac{IN}{HM} = 2\)

\(\displaystyle \frac{IN}{12} = 2\)

\(\displaystyle \frac{IN}{12}\cdot 12 = 2 \cdot 12\)

\(\displaystyle IN = 24\)

The Quadrilateral \(\displaystyle KITE\) with its diagonals is shown below. We call the point of intersection \(\displaystyle X\):

The diagonals of a quadrilateral with two pairs of adjacent congruent sides - a kite - are perpendicular. \(\displaystyle \overline{KT}\), the diagonal that connects the common vertices of the pairs of adjacent sides, bisects the other diagonal, making \(\displaystyle X\) the midpoint of \(\displaystyle \overline{IE}\). Therefore, \(\displaystyle IX= \frac{1}{2} \cdot IE = \frac{1}{2} \cdot 24 = 12\).

\(\displaystyle \overline{KT}\) also bisects the \(\displaystyle 60 ^{\circ }\) and \(\displaystyle 90^{\circ }\)angles of the kite, so the result is that two 30-60-90 triangles, \(\displaystyle \bigtriangleup KXE\) and \(\displaystyle \bigtriangleup KXI\), and two 45-45-90 triangles , \(\displaystyle \bigtriangleup TXE\) and \(\displaystyle \bigtriangleup TXI\), are formed; also, \(\displaystyle \bigtriangleup TIE\), being isosceles, is a 45-45-90 triangle. 

Examine \(\displaystyle \bigtriangleup KXI\). Since its short leg \(\displaystyle \overline{IX}\) has length 12, by the 30-60-90 Theorem, its hypotenuse, \(\displaystyle \overline{KI}\), has twice this length, or 24.

Examine \(\displaystyle \bigtriangleup TXI\). Since a leg \(\displaystyle \overline{IX}\) has length 12, by the 45-45-90 Theorem, its hypotenuse, \(\displaystyle \overline{IT}\), has length \(\displaystyle \sqrt{2}\) times this, or \(\displaystyle 12\sqrt{2}\).

Since by similarity, corresponding sides are in proportion,

\(\displaystyle \frac{LJ}{KI}= \frac{JU}{IT}\) 

\(\displaystyle \frac{LJ}{24}= \frac{36}{12\sqrt{2}}\)

\(\displaystyle \frac{LJ}{24} \cdot 24 = \frac{36}{12\sqrt{2}} \cdot 24\)

\(\displaystyle LJ = \frac{72}{ \sqrt{2}} = \frac{72\cdot \sqrt{2}}{ \sqrt{2}\cdot \sqrt{2}} = \frac{72 \sqrt{2}}{2}= 36 \sqrt{2}\)

Construct the perpendicular segment from \(\displaystyle A\) to \(\displaystyle \overline{CD}\) and let \(\displaystyle Z\) be its point of intersection with \(\displaystyle \overline{CD}\). By construction, the trapezoid is divided into Rectangle \(\displaystyle ABCZ\) and right triangle \(\displaystyle \bigtriangleup AZD\). Since opposite sides of a rectangle are congruent, \(\displaystyle AZ= BC = 3\) and \(\displaystyle CZ = AB = 12\); as a consequence of the latter, \(\displaystyle ZD = CD - CZ = 16-12 = 4\). By the Pythagrean Theorem, the length of the hypotenuse \(\displaystyle \overline{AD}\) of right triangle  \(\displaystyle \bigtriangleup AZD\) can be calculated from the length of legs \(\displaystyle \overline{AZ}\) and \(\displaystyle \overline{ZD}\):

\(\displaystyle AD = \sqrt{(AZ)^{2}+(DZ)^{2}} = \sqrt{3^{2}+4^{2}} =\sqrt{9+16} = \sqrt{25} = 5\)

The figure, with the segment and the calculated measurements, is below.

Since Trapezoid \(\displaystyle ABCD \sim\) Trapezoid \(\displaystyle EFGH\), by proportionality of corresponding sides of similar figures:

\(\displaystyle \frac{EF}{AB}=\frac{EH}{AD}\)

\(\displaystyle \frac{EF}{12}=\frac{60}{5} =12\)

\(\displaystyle \frac{EF}{12} \cdot 12 =12 \cdot 12\)

\(\displaystyle EF = 144\)

A rhombus has four sides the same length, so each side of Rhombus \(\displaystyle RHOM\) has length one fourth of 80, or 20; each side of Rhombus \(\displaystyle SIPN\) has length one fourth of 180, or 45. 

The diagonals of a rhombus are each other's perpendicular bisectors, so, if we let \(\displaystyle X\) be the point of intersection of the diagonals of Rhombus \(\displaystyle RHOM\), \(\displaystyle \overline{RO}\) and \(\displaystyle \overline{HM}\), we form four congruent right triangles. 

We will examine \(\displaystyle \bigtriangleup RXM\). \(\displaystyle RM = 20\); and,\(\displaystyle XM = \frac{1}{2} \cdot HM = \frac{1}{2} \cdot 32 = 16\).

By the Pythagorean Theorem, 

\(\displaystyle RX = \sqrt{(RM)^{2}-(RX)^{2}}\)

\(\displaystyle = \sqrt{20^{2}-16^{2}}\)

\(\displaystyle = \sqrt{400-256}\)

\(\displaystyle = \sqrt{144}\)

\(\displaystyle = 12\)

and \(\displaystyle RO = 2 \cdot 12 = 24\)

Since corresponding sides of similar figures are in proportion, so are corresponding diagonals. Therefore,

\(\displaystyle \frac{SP}{RO} = \frac{SN}{RM}\)

\(\displaystyle \frac{SP}{24} = \frac{45}{20}\)

\(\displaystyle \frac{SP}{24} \cdot 24= \frac{45}{20} \cdot 24\)

\(\displaystyle SP=54\)

The angles of the rhombus measure:

\(\displaystyle m \angle R = 60 ^{\circ }\);

\(\displaystyle m \angle O = 60 ^{\circ }\), since opposite angles of a rhombus, as in any other parallelogram, are congruent;

\(\displaystyle m \angle H = 120^{\circ }\) and \(\displaystyle m \angle M = 120^{\circ }\), since consecutive angles of a rhombus are supplementary (the sum of their degree measures is 180).

The diagonals of a rhombus are each other's perpendicular bisectors as well as the bisectors of the angles, so \(\displaystyle \overline{HM}\) and \(\displaystyle \overline{RO}\), whose point of intersection we will call \(\displaystyle X\), divide Rhombus \(\displaystyle RHOM\) into four 30-60-90 triangles. If we examine \(\displaystyle \bigtriangleup RXH\), we see that its short leg \(\displaystyle \overline{HX}\) has length half that of \(\displaystyle \overline{HM}\); \(\displaystyle HM = 12\), so \(\displaystyle HX = 6\). By the 30-60-90 Triangle Theorem, long leg \(\displaystyle \overline{RX}\) has length \(\displaystyle \sqrt{3}\) this, or \(\displaystyle 6 \sqrt{3}\), and the diagonal \(\displaystyle \overline{RO}\) has measure twice this, or \(\displaystyle 12\sqrt{3}\).

The ratio of the lengths of corresponding diagonals of the rhombuses is the same as the similarity ratios of the sides, so the similarity ratio of Rhombus \(\displaystyle SIPN\) to Rhombus \(\displaystyle RHOM\) is 

\(\displaystyle \frac{SP }{RO} = \frac{72}{12\sqrt{3}}= \frac{6}{\sqrt{3}}\)

The ratio of the areas is the square of the similarity ratio, which is

\(\displaystyle \left ( \frac{6}{\sqrt{3}} \right )^{2}= \frac{36}{3} = \frac{12}{1}\)

That is, 12 to 1.

The Quadrilateral \(\displaystyle KITE\) with its diagonals is shown below. We call the point of intersection \(\displaystyle X\):

The diagonals of a quadrilateral with two pairs of adjacent congruent sides - a kite - are perpendicular; also, \(\displaystyle \overline{KT}\) bisects the \(\displaystyle 60 ^{\circ }\) and \(\displaystyle 90^{\circ }\)angles of the kite. Consequently, two congruent 30-60-90 triangles, \(\displaystyle \bigtriangleup KXE\) and \(\displaystyle \bigtriangleup KXI\), and two congruent 45-45-90 triangles , \(\displaystyle \bigtriangleup TXE\) and \(\displaystyle \bigtriangleup TXI\), are formed; also, \(\displaystyle \bigtriangleup TIE\), being an isosceles right triangle, is a 45-45-90 triangle. \(\displaystyle \overline{KE}\), the hypotenuse of \(\displaystyle \bigtriangleup KXE\), has length 16, so by the 30-60-90 Triangle Theorem, its shorter leg \(\displaystyle \overline{XE}\) has length half this, or 8. Also, \(\displaystyle \overline{XE}\) is a leg of \(\displaystyle \bigtriangleup TXE\), so by the 45-45-90 Theorem, the hypotenuse \(\displaystyle \overline{TE}\) has length \(\displaystyle \sqrt{2}\) times this, or \(\displaystyle 8\sqrt{2}\).

Corresponding sides of similar quadrilaterals are in proportion, so

\(\displaystyle \frac{LF}{KE} = \frac{UF}{TE}\)

\(\displaystyle \frac{LF}{16} = \frac{24}{8 \sqrt{2}}\)

\(\displaystyle \frac{LF}{16} \cdot 16 = \frac{24}{8 \sqrt{2}} \cdot 16\)

\(\displaystyle LF= \frac{48}{ \sqrt{2}} = \frac{48 \cdot \sqrt{2}}{ \sqrt{2} \cdot \sqrt{2}}= \frac{48 \sqrt{2}}{2} = 24 \sqrt{2}\)