Multiply height \(\displaystyle BD\) by base \(\displaystyle CD\) to get the area.

By the 30-60-90 Theorem:

\(\displaystyle BD= \frac{BC}{2} = \frac{30}{2}} = 15\). 

\(\displaystyle CD =BD \sqrt{3} = 15\sqrt{3}\)

The area is therefore

\(\displaystyle BD \cdot CD =15 \cdot 15\sqrt{3} = 225\sqrt{3}\)

Apply the Pythagorean Theorem twice here.

\(\displaystyle BD = \sqrt{6^{2}+ 8^{2}} = \sqrt{36+64 } = \sqrt{100}= 10\)

\(\displaystyle BC = \sqrt{26^{2}-10^{2}} = \sqrt{676-100} = \sqrt{576} = 24\)

The quadrilateral is a composite of two right triangles, each of whose area is half the product of its legs:

Area of \(\displaystyle \Delta ABD\): \(\displaystyle \frac{1}{2} \times 6 \times 8 = 24\)

Area of \(\displaystyle \Delta BCD\): \(\displaystyle \frac{1}{2} \times 10 \times 24= 120\)

Add: 

\(\displaystyle A = 24 + 120 = 144\)

Construct the other diagonal of the rhombus, which, along with the first one, form a pair of mutual perpendicular bisectors.

By the Pythagorean Theorem, 

\(\displaystyle n = \sqrt{12^{2}- 10^{2}} =\sqrt{144- 100} = \sqrt{44}= \sqrt{4}\cdot \sqrt{11} = 2 \sqrt{11}\)

The rhombus can be seen as the composite of four congruent right triangles, each with legs 10 and \(\displaystyle 2 \sqrt{11}\), so the area of the rhombus is 

\(\displaystyle A = 4 \cdot \frac{1}{2}\cdot 10 \cdot 2\sqrt{11} =40 \sqrt{11}\).

Each side of a rhombus is congruent, so if it has perimeter 48, it has sidelength 12. Also, the diagonals of a rhombus are each other's perpendicular bisectors, so if they are both constructed, and their point of intersection is called \(\displaystyle M\), then \(\displaystyle AM = 6\). The following figure is formed by the rhombus and its diagonals.

\(\displaystyle \bigtriangleup AMB\) is a right triangle with its short leg half the length of its hypotenuse, so it is a 30-60-90 triangle, and its long leg measures \(\displaystyle 6\sqrt{3}\) by the 30-60-90 Theorem. Therefore, \(\displaystyle BD = 12\sqrt{3}\). The area of a rhombus is half the product of the lengths of its diagonals:

\(\displaystyle \frac{1}{2} \times 12 \times 12 \sqrt{3} = 72 \sqrt{3}\)

The area of a trapezoid with height \(\displaystyle h\) and bases of length \(\displaystyle B\) and \(\displaystyle b\) is

\(\displaystyle A = \frac{1}{2} (B + b)h\).

Setting \(\displaystyle h= b = 2^{x}, B = 2^{x+1}\):

\(\displaystyle A = \frac{1}{2} ( 2^{x+1} + 2^{x}) \cdot 2^{x}\)

\(\displaystyle = \frac{1}{2} ( 2 \cdot 2^{x } + 2^{x}) \cdot 2^{x}\)

\(\displaystyle = 2^{-1}\cdot 2^{x} \cdot ( 2 \cdot 2^{x } + 2^{x})\)

\(\displaystyle = 2^{x-1} \cdot ( 2 \cdot 2^{x } + 2^{x})\)

\(\displaystyle = 2^{x-1} \cdot 3 \cdot 2^{x }\)

\(\displaystyle = 3 \cdot 2^{x+x-1 }\)

\(\displaystyle = 3 \cdot 2^{2x-1 }\)

The area of a trapezoid with height \(\displaystyle h\) and bases of length \(\displaystyle B\) and \(\displaystyle b\) is

\(\displaystyle A = \frac{1}{2} (B + b)h\).

Setting \(\displaystyle h= b = 2^{x}, B = 3 \cdot 2^{x}\):

\(\displaystyle A = \frac{1}{2} (3 \cdot 2^{x} + 2^{x}) \cdot 2^{x}\)

\(\displaystyle = \frac{1}{2} \cdot 4 \cdot 2^{x} \cdot 2^{x}\)

\(\displaystyle = 2 \cdot 2^{x} \cdot 2^{x}\)

\(\displaystyle = 2^{1} \cdot 2^{x} \cdot 2^{x}\)

\(\displaystyle = 2^{1+x+x}\)

\(\displaystyle = 2^{2x+1}\)

The area of a rhombus is half the product of the lengths of its diagonals, so

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

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

\(\displaystyle = HM \cdot HM\)

\(\displaystyle = 2 ^{x} \cdot 2 ^{x}\)

\(\displaystyle = 2^{x+x}\)

\(\displaystyle = 2^{2x}\)

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  \(\displaystyle a\sqrt{2}\).