A diagonal is a line segment joining two non-adjacent vertices of a polygon.  A regular hexagon has six sides and six vertices.  One vertex has three diagonals, so a hexagon would have three diagonals times six vertices, or 18 diagonals.  Divide this number by 2 to account for duplicate diagonals between two vertices. The formula for the number of vertices in a polygon is:

$\displaystyle \frac{n\cdot (n-3)}{2}$

where $\displaystyle n=\#\ of\ sides$.

A diagonal connects two non-consecutive vertices of a polygon.  A hexagon has six sides.  There are 3 diagonals from a single vertex, and there are 6 vertices on a hexagon, which suggests there would be 18 diagonals in a hexagon.  However, we must divide by two as half of the diagonals are common to the same vertices. Thus there are 9 unique diagonals in a hexagon. The formula for the number of diagonals of a polygon is:

$\displaystyle \frac{n\cdot (n-3)}{2}$

where n = the number of sides in the polygon.

Thus a pentagon thas 5 diagonals.  An octagon has 20 diagonals.

The key is to examine $\displaystyle \Delta AED$ in thie figure below:

Each interior angle of a regular hexagon, including $\displaystyle \angle F$, measures $\displaystyle 120^{\circ }$, so it can be easily deduced by way of the Isosceles Triangle Theorem that $\displaystyle m \angle FEA = 30^{\circ }$. $\displaystyle m \angle FED = 120^{\circ }$, so by angle addition, 

$\displaystyle m \angle AED = 90^{\circ }$.

Also, by symmetry,

$\displaystyle m \angle EDA = \frac{1}{2} \cdot 120^{\circ } = 60^{\circ }$,

so $\displaystyle m \angle EAD= 30^{\circ }$,

and $\displaystyle \Delta AED$ is a $\displaystyle 30^{\circ}-60^{\circ}-90^{\circ}$ triangle whose long leg $\displaystyle \overline{AE}$ has length $\displaystyle AE= 1$.

By the $\displaystyle 30^{\circ}-60^{\circ}-90^{\circ}$ Theorem, $\displaystyle \overline{AD}$, which is the hypotenuse of $\displaystyle \Delta AED$, has length $\displaystyle \frac{2\sqrt{3}}{3}$ times that of the long leg, so $\displaystyle AD =\frac{2\sqrt{3}}{3}$.

The key is to examine $\displaystyle \Delta AED$ in thie figure below:

Each interior angle of a regular hexagon, including $\displaystyle \angle F$, measures $\displaystyle 120^{\circ }$, so it can be easily deduced by way of the Isosceles Triangle Theorem that $\displaystyle m \angle FEA = 30^{\circ }$. $\displaystyle m \angle FED = 120^{\circ }$, so by angle addition, 

$\displaystyle m \angle AED = 90^{\circ }$.

Also, by symmetry,

$\displaystyle m \angle EDA = \frac{1}{2} \cdot 120^{\circ } = 60^{\circ }$,

so $\displaystyle m \angle EAD= 30^{\circ }$,

and $\displaystyle \Delta AED$ is a $\displaystyle 30^{\circ}-60^{\circ}-90^{\circ}$ triangle whose hypotenuse $\displaystyle \overline{AD}$ has length $\displaystyle AD = 1$.

By the $\displaystyle 30^{\circ}-60^{\circ}-90^{\circ}$ Theorem, the long leg $\displaystyle \overline{AE}$ of $\displaystyle \Delta AED$ has length $\displaystyle \frac{\sqrt{3}}{2}$ times that of hypotenuse $\displaystyle \overline{AD}$, so $\displaystyle AE = \frac{\sqrt{3}}{2}$.

The key is to examine $\displaystyle \Delta AED$ in thie figure below:

Each interior angle of a regular hexagon, including $\displaystyle \angle F$, measures $\displaystyle 120^{\circ }$, so it can be easily deduced by way of the Isosceles Triangle Theorem that $\displaystyle m \angle FEA = 30^{\circ }$. To find  $\displaystyle m \angle AED$ we can subtract $\displaystyle m \angle FEA = 30^{\circ }$ from $\displaystyle m \angle FED = 120^{\circ }$. Thus resulting in:

$\displaystyle m \angle AED = 90^{\circ }$

Also, by symmetry,

$\displaystyle m \angle EDA = \frac{1}{2} \cdot 120^{\circ } = 60^{\circ }$,

so $\displaystyle m \angle EAD= 30^{\circ }$.

Therefore, $\displaystyle \Delta AED$ is a $\displaystyle 30^{\circ}-60^{\circ}-90^{\circ}$ triangle whose short leg $\displaystyle \overline{ED}$ has length  $\displaystyle ED = 1$.

The hypotenuse $\displaystyle \overline{AD}$ of this  $\displaystyle 30^{\circ}-60^{\circ}-90^{\circ}$ triangle measures twice the length of short leg $\displaystyle \overline{ED}$, so $\displaystyle AD = 2$.

The key is to examine $\displaystyle \Delta AED$ in thie figure below:

Each interior angle of a regular hexagon, including $\displaystyle \angle F$, measures $\displaystyle 120^{\circ }$, so it can be easily deduced by way of the Isosceles Triangle Theorem that $\displaystyle m \angle FEA = 30^{\circ }$. To find $\displaystyle m \angle AED$ we subtract $\displaystyle m \angle FEA = 30^{\circ }$ from  $\displaystyle m \angle FED = 120^{\circ }$. Thus resullting in $\displaystyle m \angle AED=90^{\circ}$

Also, by symmetry,

$\displaystyle m \angle EDA = \frac{1}{2} \cdot 120^{\circ } = 60^{\circ }$,

so $\displaystyle m \angle EAD= 30^{\circ }$,

and $\displaystyle \Delta AED$ is a $\displaystyle 30^{\circ}-60^{\circ}-90^{\circ}$ triangle whose short leg $\displaystyle \overline{ED}$ has length $\displaystyle ED = 1$.

The long leg $\displaystyle \overline{AE}$ of this $\displaystyle 30^{\circ}-60^{\circ}-90^{\circ}$ triangle measures $\displaystyle \sqrt{3}$ times the length of short leg $\displaystyle \overline{ED}$, so $\displaystyle AE= \sqrt{3}$.

The perimeter of the figure can be expressed in terms of the variables by adding:

$\displaystyle A + B + A + B + \left (2A + 2B \right ) + \left (2A + 2B \right ) = P$

Simplify and set $\displaystyle P = 888$:

$\displaystyle A + A + 2A + 2A + B+ B +2B + 2B= 888$

$\displaystyle 6A + 6B= 888$

$\displaystyle 6\left (A + B \right )= 888$

$\displaystyle A + B = 148$

Along with $\displaystyle A-B = 10$, we now have a system of equations to solve for $\displaystyle A$ by adding:

$\displaystyle A + B = 148$

$\displaystyle \underline{A-B =\; 10}$

$\displaystyle 2A = \; \; \; \; \; \; 158$

$\displaystyle A = 79$

The perimeter of the figure can be expressed in terms of the variables by adding:

$\displaystyle A + B + A + B + \left (2A + 2B \right ) + \left (2A + 2B \right ) = P$

Simplify and set $\displaystyle P = 132$:

$\displaystyle A + A + 2A + 2A + B+ B +2B + 2B= 132$

$\displaystyle 6A + 6B= 132$

$\displaystyle 6\left (A + B \right )= 132$

$\displaystyle A + B = 22$

The perimeter of the figure can be expressed in terms of the variables by adding:

$\displaystyle A + B + A + B + \left (2A + 2B \right ) + \left (2A + 2B \right ) = P$

Simplify and set $\displaystyle P = 600$:

$\displaystyle A + A + 2A + 2A + B+ B +2B + 2B= 600$

$\displaystyle 6A + 6B= 600$

Since the ratio of $\displaystyle A$ to $\displaystyle B$ is equivalent to $\displaystyle 3:2$ - or

$\displaystyle \frac{A}{B} = \frac{3}{2}$,

then 

$\displaystyle A = \frac{3}{2} B$

and we can substitute as follows:

$\displaystyle 6 \cdot \frac{3}{2}B + 6B= 600$

$\displaystyle 9B + 6B= 600$

$\displaystyle 15B = 600$

$\displaystyle B = 40$