A quadrilateral with the maximum area, given a specific perimeter, is a square. Since  and a square has four equal sides, the max area is

Let denote the length of one side of a square.  This is also the top of the trapezoid. Let denote the bottom of the trapezoid. Finally, let be the height of the trapezoid.  The area of the trapezoid is then while the area of the square is .

We then have the total area as 100, so:

Now we know that the red line has length 15.  is the region of this line that is in the trapezoid. What we notice, however, is that the remainder is precisely the length of one side of a square.  So or

Rewriting the previous equation:

This is now an equation of 2 variables and we can easily cross out answers by plugging in possible values.  What we find is that for , respectively.  For we get , which is too small ( must be greater than ). For we get .

The area of a regular polygon is equal to one-half the product of its apothem - the perpendicular distance from the center to a side - and its perimeter. 

The perimeter of the octagon is 

From the diagram below, the apothem of the octagon is . 

 is one half of the sidelength, or 5.  can be seen to be the length of a leg of a  triangle with hypotenuse 10, or

This makes the apothem . 

The area is therefore 

A regular hexagon can be seen as a composite of six equilateral triangles, each of whose sidelength is the sidelength of the hexagon:

Each of the triangles has area

Substitute  to get 

Multiply this by 6: , the area of the hexagon.

This figure can be seen as a composite of two simple shapes: the rectangle with vertices , and the triangle with vertices .

The rectangle has length  and height , so its area is the product of these dimensions, or .

The triangle has as its base the length of the horizontal segment connecting  and , which is ; its height is the vertical distance from the other vertex to this segment, which is . The area of this triangle is half the product of the base and the height, which is .

Add the areas of the rectangle and the triangle to get the total area:

This figure can be seen as a smaller rectangle cut out of a larger one; refer to the diagram below.

We can fill in the missing sidelengths using the fact that a rectangle has congruent opposite sides. Once this is done, we can multiply length times height of both rectangles to get the area of each, and subtract areas:

 square feet

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

.

The garden area (EFGH) is equal to the wall area .

So

,

therefore 

.

Let  be the common measure of , , , and 

Then 

The sum of the measures of the angles of a pentagon is  degrees; this translates to the equation

or 

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

 and  are exterior angles of the regular polygons. Also, the measures of the exterior angles of any polygon, one at each vertex, total . Therefore, 

Add the measures of the angles to get :

The sum of the measures of the exterior angles of any polygon, one per vertex, is . In a regular polygon of  sides , then all  of these exterior angles are congruent, each measuring .

If  is the measure of one of these angles, then , or, equivalently, . Therefore, for  to be a possible measure of an exterior angle, it must divide evenly into 360. We divide each in turn:

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.