The diagonal of a rectangle can be thought of as the hypotenuse of a right triangle whose base and height are the length and width of the rectangle, respectively.  This means we can use the Pythagorean Theorem to calculate the length of the diagonal for a rectangle:

\(\displaystyle l^2+w^2=d^2\)

\(\displaystyle 9^2+3^2=d^2\)

\(\displaystyle d^2=90\rightarrow d=\sqrt{90}=\sqrt{9\cdot10}=3\sqrt{10}\)

The area of a rhombus is half the product of the lengths of its diagonals, which here are \(\displaystyle \overline{RO}\) and \(\displaystyle \overline{HM}\). This means

\(\displaystyle \frac{1}{2} \cdot RO \cdot HM = 72\)

\(\displaystyle 2 \cdot \frac{1}{2} \cdot RO \cdot HM =2 \cdot 72\)

\(\displaystyle RO \cdot HM =144\)

Since both diagonals have whole numbers as their lengths, and \(\displaystyle RO > HM\), we are looking for \(\displaystyle HM\) to be a whole number that can be divided into 144 to yield a quotient less than \(\displaystyle HM\). Equivalently, 

The quotient is \(\displaystyle HM< 144 \div HM\)

Since we can multiply both sides by \(\displaystyle HM\) to yield the inequality

\(\displaystyle (HM)^{2}< 144\),

we know that

\(\displaystyle HM < 12\), 

so we can eliminate 12 and 16.

Also, since \(\displaystyle 144 \div 10 = 14 .4\), 10 is not correct, as \(\displaystyle RO\) would not be a whole number.

If \(\displaystyle HM = 9\), then \(\displaystyle RO = 144 \div 9 = 16\). Both diagonals have lengths that are whole numbers, and \(\displaystyle \overline{RO}\) is the longer diagonal. 9 is the correct choice.

A rhombus has four sides of equal length. Since Rhombus \(\displaystyle RHOM\) has a right angle \(\displaystyle \angle R\), it follows that, the rhombus being a parallelogram, all four angles are right angles, and, by definition, Rhombus \(\displaystyle RHOM\) is a square. The length of a diagonal of a square is \(\displaystyle \sqrt{2}\) times the length of a side; since the rhombus has perimeter 80, each side measures one fourth of this, or 20, and diagonal \(\displaystyle \overline{RO}\) has length \(\displaystyle 20 \sqrt{2}\).

The area of a rhombus is half the product of the lengths of its diagonals, which here are \(\displaystyle \overline{RO}\) and \(\displaystyle \overline{HM}\). This means

\(\displaystyle \frac{1}{2} \cdot RO \cdot HM = 56\)

Therefore, we need to test each of the choices to find the pair of diagonal lengths for which this holds.

\(\displaystyle RO = 28, HM = 28\):

Area: \(\displaystyle \frac{1}{2} \cdot RO \cdot HM = \frac{1}{2} \cdot 28 \times 28 = 392\)

\(\displaystyle RO = 14, HM = 14\)

Area: \(\displaystyle \frac{1}{2} \cdot RO \cdot HM = \frac{1}{2} \cdot 14 \times 14 =98\)

\(\displaystyle RO = 4, HM = 14\)

Area: \(\displaystyle \frac{1}{2} \cdot RO \cdot HM = \frac{1}{2} \cdot 4 \times 14 = 28\)

\(\displaystyle RO = 16, HM = 7\)

Area: \(\displaystyle \frac{1}{2} \cdot RO \cdot HM = \frac{1}{2} \cdot16 \times 7 =56\)

\(\displaystyle RO = 16, HM = 7\) is the correct choice.

The sides of a rhombus are all congruent; since the perimeter of Rhombus \(\displaystyle RHOM\) is 64, each side measures one fourth of this, or 16. 

The referenced rhombus, along with diagonal \(\displaystyle \overline{HM}\), is below:

Since consecutive angles of a rhombus, as with any other parallelogram, are supplementary, \(\displaystyle \angle RHO\) and \(\displaystyle \angle RMO\) have measure \(\displaystyle 120^{\circ }\); \(\displaystyle \overline{HM}\) bisects both into \(\displaystyle 60 ^{\circ }\) angles, making \(\displaystyle \bigtriangleup RHM\)equilangular and, as a consequence, equilateral. Therefore, \(\displaystyle HM = RH = 16\).

The referenced rhombus, along with diagonals \(\displaystyle \overline{RO}\) and \(\displaystyle \overline{HM}\), is below.

The four sides of a rhombus have equal measure, so each side has measure one fourth of the perimeter of 48, which is 12.

Since consecutive angles of a rhombus, as with any other parallelogram, are suplementary, \(\displaystyle \angle RHO\) and \(\displaystyle \angle RMO\) have measure \(\displaystyle 120^{\circ }\); the diagonals bisect \(\displaystyle \angle MRH\) and \(\displaystyle \angle RHO\) into \(\displaystyle 30 ^{\circ }\) and \(\displaystyle 60 ^{\circ }\) angles, respectively, to form four 30-60-90 triangles. \(\displaystyle \bigtriangleup HXR\) is one of them; by the 30-60-90 Triangle Theorem, \(\displaystyle HX = \frac{1}{2} \cdot RH = \frac{1}{2} \cdot 12 =6\),

and

\(\displaystyle RX = HX \cdot \sqrt{3} = 6 \sqrt{3}\).

Since the diagonals of a rhombus bisect each other, \(\displaystyle RO = 2 \cdot RX = 2 \cdot 6 \sqrt{3} = 12\sqrt{3}\).

The Quadrilateral \(\displaystyle KITE\) is shown below with its diagonals \(\displaystyle \overline{I E}\) and \(\displaystyle \overline{KT}\).

. 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, \(\displaystyle \bigtriangleup KXI\) is a 30-60-90 triangle and \(\displaystyle \bigtriangleup TXI\) is a 45-45-90 triangle. Also, 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, 

\(\displaystyle IX = \frac{1}{2} \cdot IE = \frac{1}{2} \cdot 24 = 12\).

By the 30-60-90 Theorem, since \(\displaystyle \overline{IX}\) and \(\displaystyle \overline{KX}\) are the short and long legs of \(\displaystyle \bigtriangleup KXI\), 

\(\displaystyle KX = IX \cdot \sqrt{3}= 12 \sqrt{3}\)

By the 45-45-90 Theorem, since \(\displaystyle \overline{IX}\) and \(\displaystyle \overline{XT}\) are the legs of a 45-45-90 Theorem, 

\(\displaystyle XT = IX = 12\).

The diagonal \(\displaystyle \overline{KT}\) has length 

\(\displaystyle KT =XT + KX= 12+ 12\sqrt{3}\).

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

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

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

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

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