$\displaystyle \overline{AD}$ is a diameter of the regular hexagon. Examine the diagram below, which shows the hexagon with all three diameters:

Each interior angle of a hexagon measures $\displaystyle 120^{\circ }$, so, by symmetry, each base angle of the triangle formed is $\displaystyle 60^{\circ }$; also, each central angle measures one sixth of $\displaystyle 360^{\circ}$, or $\displaystyle 60^{\circ }$. Each triangle is equilateral, so if $\displaystyle AB = 12$, it follows that $\displaystyle AO = OD = 12$, and $\displaystyle AD = AO+OD = 12 + 12 = 24$.

Construct two other diagonals as shown.

Each of the interior angles of a regular octagon have measure $\displaystyle 135^{\circ }$, so it can be shown that $\displaystyle \bigtriangleup ABX$ is a 45-45-90 triangle. Its hypotenuse is $\displaystyle \overline{AB}$, whose length is 8, so, by the 45-45-90 Triangle Theorem, the length of $\displaystyle \overline{AX}$ is 8 divided by $\displaystyle \sqrt{2}$:

$\displaystyle AX= \frac{8}{\sqrt{2}}= \frac{8\cdot \sqrt{2}}{\sqrt{2}\cdot \sqrt{2}} = \frac{8 \sqrt{2}}{2} = 4 \sqrt{2}$

Likewise, $\displaystyle YD = 4 \sqrt{2}$.

Since Quadrilateral $\displaystyle BCYX$ is a rectangle, $\displaystyle XY = BC = 8$.

$\displaystyle AD = AX+XY+YD = 4\sqrt{2}+ 8 + 4 \sqrt{2}= 8+8\sqrt{2}$

By dividing the figure into rectangles and taking advantage of the fact that opposite sides of rectangles are congruent, we have the following sidelengths:

$\displaystyle \overline{AC}$ is the hypotenuse of a triangle with legs of lengths 8 and 16, so its length can be calculated using the Pythagorean Theorem:

$\displaystyle AC = \sqrt{(AB)^{2}+(BC)^{2}} = \sqrt{16^{2}+8^{2}} = \sqrt{256+64} = \sqrt{320}$

The question can now be answered by noting that $\displaystyle 17^{2} = 289$ and $\displaystyle 18^{2}= 324$. 

$\displaystyle 17^{2}< 320 < 18^{2}$,

so $\displaystyle \sqrt{320}$ falls between 17 and 18.

A regular pentagon has five diagonals of equal length, each formed by a line going from one vertex of the pentagon to another. We can see that for one of these diagonals, an isosceles triangle is formed where the two equal side lengths between the vertices joined by the diagonal are the other two sides. If we draw a line bisecting the angle between those two sides perpendicular with the diagonal that forms the other side of the triangle, we will have two congruent right triangles whose hypotenuse is the side length, $\displaystyle 6$, and whose adjacent angle is half the measure of one interior angle of a pentagon. Using these two values, we can solve for the length of the opposite side, which is half of the diagonal, so we can them multiply the result by $\displaystyle 2$ to calculate the full length of the diagonal. We start by determining the sum of the interior angles of a pentagon using the following formula, where $\displaystyle n$ is the number of sides of the polygon:

$\displaystyle 180^{\circ}(n-2)=180^{\circ}(5-2)=540^{\circ}$

So to get the measure of each of the five angles in a pentagon, we divide the result by $\displaystyle 5$:

$\displaystyle \frac{540^{\circ}}{5}=108^{\circ}$

So each interior angle of a regular pentagon has a measure of $\displaystyle 108^{\circ}$. As explained earlier, we can find the length of half the diagonal by bisecting this angle to form two right triangles. If the hypotenuse is $\displaystyle 6$ and the adjacent angle is half of an interior angle, or $\displaystyle 54^{\circ}$, then the length of the opposite side will be the hypotenuse times the sine of that angle. This only gives half of the diagonal, however, as there are two of these congruent right triangles, so we multiply the result by $\displaystyle 2$ and we get the full length of the diagonal of a pentagon as follows:

$\displaystyle D=2(6\sin 54^{\circ})=9.7$

Extend sides $\displaystyle \overline{CD}$ and $\displaystyle \overline{ED}$ as shown to divide the polygon into three rectangles:

Taking advantage of the fact that opposite sides of a rectangle are congruent, we can find $\displaystyle AX$ and $\displaystyle DX$:

$\displaystyle AX= AB - BX = AB - CD = (2n+6)-(n+2)= n+4$

$\displaystyle DX= EX-ED = FA-ED = 2n - n = n$

$\displaystyle \bigtriangleup DXA$ is right, so by the Pythagorean Theorem,

$\displaystyle AD = \sqrt{(AX)^{2}+(DX)^{2}}$

$\displaystyle =\sqrt{\left ( n+4\right )^{2}+n^{2}}$

$\displaystyle =\sqrt{\left ( n^{2}+8n+16\right )+n^{2}}$

$\displaystyle =\sqrt{2 n^{2}+8n+16 }$

The measures of the interior angles of a convex pentagon total

$\displaystyle 180 ^{\circ } \times (5-2) = 540 ^{\circ }$,

so

$\displaystyle m \angle C = 540 ^{\circ } - (m \angle A+m \angle B + m\angle D + m \angle E)$

$\displaystyle = 540 ^{\circ } - (105 ^{\circ }+105 ^{\circ }+105 ^{\circ }+105 ^{\circ })$

$\displaystyle = 540 ^{\circ } - 420 ^{\circ }$

$\displaystyle = 120^{\circ }$

The pentagon referenced is the one below. Note that the diagonal $\displaystyle \overline{BD}$, along with congruent sides $\displaystyle \overline{BC}$ and $\displaystyle \overline{CD}$, form an isosceles triangle $\displaystyle \bigtriangleup BCD$.

Now construct the altitude from $\displaystyle C$ to $\displaystyle \overline{BD}$:

$\displaystyle \overline{CM}$ bisects $\displaystyle \angle BCD$ and $\displaystyle \overline{BD}$ to form two 30-60-90 triangles. Therefore, $\displaystyle CM = \frac{1}{2} BC = \frac{1}{2} \cdot 12 = 6$,

and $\displaystyle BM = CM \cdot \sqrt{3}= 6\sqrt{3}$.

$\displaystyle BD = 2 \cdot BM = 2 \cdot 6 \sqrt{3}= 12 \sqrt{3}$