AB's length is 7 so the area of ABCD is:

$\displaystyle ABCD = 7\cdot 7=49$.

The garden area (EFGH) is equal to the wall area $\displaystyle (ABCD-EFGH)$.

So

$\displaystyle EFGH=ABCD-EFGH$,

therefore 

$\displaystyle EFGH = 0.5 \cdot ABCD$

$\displaystyle EFGH = 0.5 \cdot 49 = 24.5$.

Let $\displaystyle x$ be the common measure of $\displaystyle \angle B$, $\displaystyle \angle C$, $\displaystyle \angle D$, and $\displaystyle \angle E$

Then 

$\displaystyle m \angle A = \frac{7}{5} x$

The sum of the measures of the angles of a pentagon is $\displaystyle 180 (5 - 2) = 540^{\circ }$ degrees; this translates to the equation

$\displaystyle \frac{7}{5} x + x + x + x + x = 540$

or 

$\displaystyle \frac{27}{5} x = 540$

$\displaystyle x = 540 \cdot \frac{5}{27} = 100$

$\displaystyle m \angle A = \frac{7}{5} x = \frac{7}{5} \cdot 100$

$\displaystyle m \angle A = 140$

This can more easily be explained if the shared side is extended in one direction, and the new angles labeled.

$\displaystyle \angle1$ and $\displaystyle \angle 2$ are exterior angles of the regular polygons. Also, the measures of the exterior angles of any polygon, one at each vertex, total $\displaystyle 360^{\circ }$. Therefore, 

 $\displaystyle m \angle1 = \frac{360}{5} = 72^{\circ }$

$\displaystyle m \angle2 = \frac{360}{6} = 60^{\circ }$

Add the measures of the angles to get $\displaystyle m\angle AVB$:

$\displaystyle m\angle AVB = m\angle 1 + m\angle 2 = 72 + 60 = 132^{\circ }$

The sum of the measures of the exterior angles of any polygon, one per vertex, is $\displaystyle 360^{\circ }$. In a regular polygon of $\displaystyle N$ sides , then all $\displaystyle N$ of these exterior angles are congruent, each measuring $\displaystyle \frac{360^{\circ }}{N}$.

If $\displaystyle x$ is the measure of one of these angles, then $\displaystyle x = \frac{360^{\circ }}{N}$, or, equivalently, $\displaystyle N = \frac{360^{\circ }}{x}$. Therefore, for $\displaystyle x$ to be a possible measure of an exterior angle, it must divide evenly into 360. We divide each in turn:

$\displaystyle 360 \div 6 = 60$

$\displaystyle 360 \div 9 = 40$

$\displaystyle 360 \div 15 = 24$

$\displaystyle 360 \div 16 = 22.5$

Since 16 is the only one of the choices that does not divide evenly into 360, it cannot be the measure of an exterior angle of a regular polygon.

Each angle of a square measures $\displaystyle 90^{\circ }$; each angle of a regular pentagon measures $\displaystyle 180 \cdot3 \div 5 = 108^{\circ }$. To get $\displaystyle m\angle 1$, subtract:

$\displaystyle 108 - 90 = 18^{\circ }$.

Call $\displaystyle x$ the measure of $\displaystyle \angle A$

$\displaystyle m \angle A = m \angle C = m \angle E = x$

$\displaystyle m \angle A + 6^{\circ } = m \angle B$, and $\displaystyle \angle B \cong \angle D \cong \angle F$

so 

$\displaystyle m \angle B = m \angle D = m \angle F = x + 6$

The sum of the measures of the angles of a hexagon is $\displaystyle 180 (6 - 2) = 720 ^{ \circ }$, so 

$\displaystyle m \angle A + m \angle B + m \angle C + m \angle D + m \angle E + m \angle F = 720$

$\displaystyle x+ (x+6) +x + (x+6) + x + (x+6) = 720$

$\displaystyle 6x + 18 = 720$

$\displaystyle \ 6x = 702$

$\displaystyle \ x = 702 \div 6 = 117$, which is the measure of $\displaystyle \angle A$.

The sum of the degree measures of any polygon is $\displaystyle 360^{\circ }$. A regular polygon with $\displaystyle N$ sides has exterior angles of degree measure $\displaystyle \left (\frac{360}{N} \right )^{\circ }$. For this to be an integer, 360 must be divisible by $\displaystyle N$. 

We can test each of our choices to see which one fails this test.

$\displaystyle 360 \div 24 = 15$

$\displaystyle 360 \div 30 = 12$

$\displaystyle 360 \div 45 = 8$

$\displaystyle 360 \div 80 = 4.5$

$\displaystyle 360 \div 90 = 4$

Only the eighty-sided regular polygon fails this test, making this the correct choice.

The measure of each interior angle of a regular pentagon is 

$\displaystyle \frac{180 (5-2)}{5} = \frac{180 (3)}{5} = 108^{\circ }$

The measure of each interior angle of a regular hexagon is 

$\displaystyle \frac{180 (6-2)}{6} = \frac{180 (4)}{6} = 120^{\circ }$

The measure of $\displaystyle \angle ABC$ is the difference of the two, or $\displaystyle 12 ^{\circ }$.

The sum of the measures of the nine angles of any nonagon is calculated as follows:

 $\displaystyle 180 \cdot \left ( 9-2\right ) =180 \cdot 7= 1,260^{\circ }$

Divide this number by nine to get the arithmetic mean of the measures:

$\displaystyle 1,260\div 9 = 140 ^{\circ }$

The mean of the measures of the four angles of the quadrilateral and the five angles of of the pentagon is their sum divided by 9.

The sum of the measures of the interior angles of any quadrilateral is $\displaystyle 180 (4-2) = 360^{\circ }$. The sum of the measures of the interior angles of any pentagon is $\displaystyle 180 (5-2) = 540^{\circ }$.

The sum of the measures of the interior angles of both polygons is therefore $\displaystyle 360 + 540 = 900^{\circ }$. Divide by 9:

$\displaystyle 900 \div 9 = 100^{\circ }$