This problem can be solved in a variety of ways, however you do it, it may be worth eliminating any options shorter than either side of the quad. The diagonal distance is the hypotenuse of a triangle, so it must be longer than 25 or 60 meters. 

Then, we can either find our hypotenuse via Pythagorean Theorem, or our knowledge of Pythagorean Triples. 

Using Pythagorean Theorem:

\(\displaystyle c^2=a^2+b^2\)

\(\displaystyle c=\sqrt{25^2+60^2}=\sqrt{4225}=65\)

So 65 meters.

Alternatively, recognize the triangle as a 5x/12x/13x triangle

\(\displaystyle 5x=25\)

\(\displaystyle 12x=60\)

\(\displaystyle 13x=?\)

\(\displaystyle x=5\) so \(\displaystyle 13x=65\)

Write the formula for finding the area of a square given the diagonal.

\(\displaystyle A=\frac{d^2}{2}\)

Rearrange the equation so that the diagonal term is isolated.

\(\displaystyle d=\sqrt{2A}\)

Substitute the known area and simplify.

\(\displaystyle d=\sqrt{2(9x^2)} = \sqrt{18x^2} = 3x\sqrt2\)

The diagonal of a square is simply the hypotenuse of a right triangle whose other two sides are the length and width of the square. Because all sides of a square are equal in length, this means the length and width are both  \(\displaystyle 4\),  which gives us a right triangle with a base of  \(\displaystyle 4\)  and a height of  \(\displaystyle 4\),  for which the hypotenuse is the diagonal of the square. Applying the Pythagorean Theorem to find the length of the diagonal, we have:

\(\displaystyle a^2+b^2=c^2\)

\(\displaystyle 4^2+4^2=c^2\)

\(\displaystyle c^2=32\rightarrow c=\sqrt{32}=\sqrt{2*16}=4\sqrt{2}\)

So the length of the diagonal for a square with a side length of  \(\displaystyle 4\)  is  \(\displaystyle 4\sqrt{2}\).  In general, we could check the length of the diagonal for any square with side length  \(\displaystyle a\),  and we would see that the diagonal length is always  \(\displaystyle a\sqrt{2}\).

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 

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