This problem asks for us to solve for the perimeter of a square that has a length of  inches. The perimeter can be solved for using , where w means width and l means length. 

In this kind of problem, it's important to remember that squares have 4 equal sides. This means that their lengths equal the width - therefore, it can be misleading to think there isn't enough information. 

Therefore, the square has a perimeter of  inches. 

The problem states that Jill wants to make a pen in the shape of an equilateral triangle. This means the triangle will have three equal sides. With only  feet of fencing, she wishes to make the largest triangle possible. In order to solve for the longest possible whole number length of the triangle pen, we must first utilize some concepts associated with perimeter. 

Perimeter is the sum of all sides. This means that with  feet of fencing, the sum of the three sides cannot exceed  feet. Using this information, we can solve for the maximal side length:

if we set up an equation where all three sides are summed, we can set it equal to , keeping in mind that this is the limit. But we must also keep in mind that s cannot be a decimal number - it must be a whole number. 

Because we are restrained by  feet, we must round the value for s down. This would entail that the side of the pen can be  feet maximum. 

 is a right triangle with legs ; its area is half the product of its legs, which is

 is a right triangle with legs 

and

 ;

its area is half the product of its legs, which is

The shaded region is the former triangle removed from the latter triangle; its area is the difference of the two: .

This region is therefore

 of .

 is a right triangle with legs ; its area is half the product of its legs, which is

 is a right triangle with legs 

and

 ;

its area is half the product of its legs, which is

The shaded region is the former triangle removed from the latter triangle; its area is the difference of the two: .

A hexagon has six sides; a regular hexagon with perimeter 36 has sidelength 

.

A regular hexagon can be divided into six triangles, each of which can be easily proved equilateral, as seen in the diagram below:

Each equilateral triangle has sidelength 6, so each has area

.

The total area of the hexagon is the area of six such triangles:

A regular hexagon can be divided into six triangles, each of which can be easily proved equilateral, as seen in the diagram below:

All segments shown are congruent, and, since the diameter shown in the original diagram is 4, each sidelength is half this, or 2.

Each equilateral triangle has area

.

There are six such triangles, so the total area of the hexagon is six times this, or .

Write the area of a square.

Substitute the side into the formula.

The answer is:  

Examine the bottom figure, in which the bottom two sides have been connected. Note that the figure is now a rectangle cut out of a rectangle, and, since the opposite sides of a rectangle have the same length, we can fill in some of the side lengths as shown:

The figure is a 60-by-40 rectangle cut from a 100-by-100 square, so, to get the area of the figure, subtract the area of the former from that of the latter. The area of a rectangle is equal to the product of its dimensions, so the areas of the rectangle and the square are, respectively,

and 

,

making the area of the figure

.

Since the circle is inscribed in the square, the diameter of the circle is the same length as the length of a square.

Start by finding the area of the square.

For the given square,

Now, because the diameter of the circle is the same as the length of a side of the square, we now also know that the radius of the circle must be . Next recall how to find the area of a circle.

Plug in the found radius to find the area of the circle.

Now, the shaded area is the area left over when the area of the circle is subtracted from the area of the square. Thus, we can write the following equation to find the area of the shaded region.

Start by drawing out the square yard and the circular pool in a way that maximizes the area of the pool.

Notice that the diameter of the pool will be the same length as the side of the square. 

Since the question asks about the area that is left over after the pool is built, we can find that area by subtracting the area of the pool from the area of the square.

Start by finding the area of the square.

Next, find the area of the circular pool.

Since the diameter of the pool is , the radius of the pool must be . Recall how to find the area of a circle:

Plug in the radius of the circle.

Subtract the area of the circle from that of the square.