__Problem Statement__

Farmer Jake has a cow, some rope, and a ten foot by ten foot square barn. Farmer Jake ties his cow to one corner of the outside of the barn using exactly 100ft of rope. What is the total grazing area of Farmer Jake's cow?

__Process__

The Cow Problem asked us to find the total grazing area of a cow tied to the outside corner of a ten foot by ten foot square barn by a rope that is 100ft long. My first step was to draw a diagram of the situation so that I could better visualize what exactly the problem was asking. Here is the diagram that I came up with:

By drawing this diagram, I learned a number of things:

1) The point at which Farmer Jake's cow is tied to the barn is basically the center of a circle.

2) The rope by which the cow is constrained represents the radius of the circle.

3) The rope is uninhibited for exactly three-quarters of a revolution around the center.

4) As the rope is pulled past the adjacent outside corner of the barn, it begins to bend around that corner, and not the corner to which it was originally fixed.

1) The point at which Farmer Jake's cow is tied to the barn is basically the center of a circle.

2) The rope by which the cow is constrained represents the radius of the circle.

3) The rope is uninhibited for exactly three-quarters of a revolution around the center.

4) As the rope is pulled past the adjacent outside corner of the barn, it begins to bend around that corner, and not the corner to which it was originally fixed.

I figured that the majority of the cow's grazing area could be calculated by finding the area of three-quarters of the area of a 100ft radius circle:

A=0.75(pi*100^2)

A=0.75(pi*10,000)

A=0.75(31,415.93)

A=23,561.95

A=0.75(pi*100^2)

A=0.75(pi*10,000)

A=0.75(31,415.93)

A=23,561.95

Now, the area of the cow's three-quarter revolution was

**23,561.95 cubic feet**. The main problem was then finding the area of the region beyond the rope's three-quarter revolution. I was stumped.Mr. B finally showed us a way to split the diagram into several manageable geometric shapes. Here's what we did:

As you can see, the length of the rope essentially decreases as it begins to bend around the outside corner of the barn. Because of this, the grazing area converges at a single point, exactly forty-five degrees opposite the barn. We drew two lines, which both began at this convergence point, but ended at either corner of the barn. We then drew a line connecting the first two lines, which split the square barn into two right triangles. These three lines produced three geometric shapes which we could use to find the total grazing area of Farmer Jake's cow:

1) An isosceles triangle

2) Two circle sectors

First, we solved for the area of the isosceles triangle.

1) An isosceles triangle

2) Two circle sectors

First, we solved for the area of the isosceles triangle.

The formula for the area of a triangle is "base times height divided by two". To find the area of this triangle, we would need to find both the base (b) and the height (h). We knew that each side of the barn is ten feet long. Using the right triangle created by the diagonal line that we drew earlier, we plugged the values for (a) and (b) into the pythagorean theorem equation. In this situation, the function looked like this:

(10^2)+(10^2)=c^2

100+100=c^2

200=c^2

14.14=c

(10^2)+(10^2)=c^2

100+100=c^2

200=c^2

14.14=c

Now that we had solved for the length of the base of the isosceles triangle, we were able to solve for the height. By splitting the triangle in half, we created two identical right triangles, with base lengths of 7.07. At this point, we could utilize the pythagorean theorem to solve for the height:

(7.07^2)+(h^2)=90^2

49.98+h^2=8100

h^2=8050.02

h=89.72

(7.07^2)+(h^2)=90^2

49.98+h^2=8100

h^2=8050.02

h=89.72

Now that we had the base and the height of our isosceles triangle, we were capable of finding it's total area:

A=(14.14*89.72)/2

A=(1268.64)/2

A=634.32

A=(14.14*89.72)/2

A=(1268.64)/2

A=634.32

Now we had found the area of the isosceles that we created. Still, this was not our final step. We still needed to subtract the area of the barn which overlapped the triangle, because the cow's grazing area is outside of the barn. Since the overlap consisted of exactly half of the area of the barn, we simply found the area of the barn and divided by two:

A=(10*10)/2

A=(100)/2

A=50

A=(10*10)/2

A=(100)/2

A=50

So, the area of the isosceles triangle minus the overlapping space equals

**584.32 cubic feet.**Next, we found the area of the two circle sectors. The formula for the area of a circle sector is "(theta/360)pi*radius^2)". In this situation, we needed to find theta (the angle of the circle sector).

To find theta, we would need to find the angle (n) of our isosceles triangle. To do this, we plugged the base and height lengths of the triangle that we had previously found into the inverse sine function.

-SIN(89.72/90)=n

-SIN(0.9968)=n

85.48=n

-SIN(89.72/90)=n

-SIN(0.9968)=n

85.48=n

Now that we had calculated the angle measure of the isosceles triangle, we could solve for the measure of theta. Because a straight line has an angle measure of 180 degrees, then:

(45)+(85.48)+(theta)=180

130.48+theta=180

theta=49.52

(45)+(85.48)+(theta)=180

130.48+theta=180

theta=49.52

Now that we had found the value of theta, we could plug our values into the circle sector formula and find the total area:

A=(49.52/360)pi*(90^2)

A=(0.14)pi*8100

A=(0.14)*25,446.9

A=3,562.57

A=(49.52/360)pi*(90^2)

A=(0.14)pi*8100

A=(0.14)*25,446.9

A=3,562.57

We had found the area of one circle sector, but we still needed to consider the second circle sector created by the rope wrapping around the other end of the barn. To do this, we simply multiplied the area of one circle sector by two:

A=3,562.57*2

A=7125.14

A=3,562.57*2

A=7125.14

The combined area of the circle sectors was

**7125.14 cubic feet.**.Now all we had left to do was find the sum of our three area totals, including the three-quarter revolution, the isosceles triangle minus the overlap, and the two circle sectors:

Total Grazing Area=(23,561.95)+(584.32)+(7125.14)

Total Grazing Area=31,271.41

Total Grazing Area=(23,561.95)+(584.32)+(7125.14)

Total Grazing Area=31,271.41

**The Total Grazing Area of Farmer Jake's Cow was equal to 31,271.41 cubic feet.**

For me, splitting the diagram up into manageable pieces was very helpful because I was able to deal with geometric shapes that I could actually calculate the areas of. In addition, being able to use formulas and functions such as the pythagorean theorem and the sine function aided me in finding the angle measurements without calculating it by hand. It was much easier to simply plug my values into my calculator, as opposed to writing it out by hand.

__Reflection__

The most useful things that I learned during the cow problem were the trigonometric functions such as sine, cosine, tangent, and the pythagorean theorem. They were a brand new mathematical concept for me and they really helped me solve the cow problem. At first, I had a hard time wrapping my head around using SIN, COS, and TAN. It didn't help that I couldn't actually find the inverse SIN, COS, and TAN buttons on my calculator. But now I know, and I'm glad I learned how to do it. If I were to grade myself on this unit, I would give myself a, eighty-two percent (B-) because I did learn a lot but I didn't do so well on the test, which shows that I didn't study enough.