A farmer has a cow tethered to an outside corner of a 10' x 10' barn with 100' of rope. He only wants to plant grass in places that the cow can reach, no more and no less. Calculate how much square feet of grass the farmer needs to plant.

My first attempt at this problem was that I drew how the problem looked. At the beginning, I wasn't really sure what I was supposed to be solving for, so I drew what the problem looked liked so that I would have a better understanding of what exactly I was trying to solvr.

This is my final diagram. We knew that the cow could eat 3/4 of the circle with no problem. So the only area that we had to figure out was 1/4 of the circle. It does sound easy, but it wasn't. The 1/4 of the circle had a dimple so we had to use different shapes and methods to solve that area. The different methods we used were:
Area of a circle = πr²Area of a triangle = b x h/2 Angle of a triangle = SOHCAHTOA **SIN: 0pposite/Hypotenuse****COS: Adjacent/Hypotenuse****TAN: Opposite/Adjacent**
Side of a triangle = Pythagorean Theorem (a²+b²=c²)Based on these methods, I was able to split the 1/4 of the circle into different shapes and find the area of those shapes and add them all together. The shapes we broke down were the two pizza slices, the two triangles, and 3/4 of the circle. |

The total area of the cow problem is
31,148.5. I got this area by finding the area of each different shape and adding those areas all together. Each piece was essential in finding the area of the cow problem because since there was a dimple in the other 1/4 of the circle, we had to use shapes that we already knew how to solve the area for and add those all together to find the total area. |

The solution to find the total area of the Cow Problem is: **(area of the 3/4 of the circle) + (area of triangle in between two sectors) + 2(area of sector) - (area of 1/2 of the barn) = total area the cow can graze**

1. Area of a circle = πr^2

For the 3/4 of a circle, we used the formula for the area of a circle, which is*A=πr^2*, and multiplied it by 3/4.

The radius was 100 because that's how long the rope was. When we plugged it into the formula we got 10,000π. 10,000π was the area of a whole circle but the cow could only graze 3/4 of a circle with the radius 100. That is why we multiplied it by 3/4. This got us with the area of**23,561.94 ft**.

**2. The Pythagorean Theorem: a^2 + b^2 = c^2**

In order to find the area of the triangle indicated below, we first need to figure out the diagonal length going across the barn. We can find this using the Pythagorean theorem.

The diagonal length is*c *in the equation because it is opposite the right angle in the triangle, making it the hypotenuse. We know that *a *is equal to 10 and *b *is equal to 10. We learned about radicals to help us with this step. When we solve the equation out we get c=10√2.

Now that we have the base of the triangle's length, we need to find the height. If we split the triangle straight down the middle from it's highest point to the base we get two symmetrical right triangles. Now that we have a right angle we can use the Pythagorean Theorem again!

If we are looking at just the left right triangle, we know that the**hypotenuse **in this triangle is 90 ft. because from the bottom left corner of the barn anywhere the rope goes is equal to 90 ft. Once the cow goes past the upper left corner of the barn the rope becomes 80 ft.

We know the base of the right triangle is 5√2 because the whole base was equal to 10√2. When we spilt the triangle in half, we split the base in half making it 5√2.

We have two lengths in the equation so we can solve for the third one. This time we have to be careful with what numbers we have and what number we're trying to solve for. We have the hypotenuse and the base. That means we have*c *and *b *of the equation. When we solve it out, we get the height of the triangle equals 89.72 ft.

**3. Area of a triangle = 1/2bh**

a) Now we have all the numbers we can find the area of the triangle. The height is 89.72 and the base is 10√2. When we solve it out we get the area of the triangle equals** 634.42 ft**.

b) There is one more catch to this. We have been solving for this triangle but a part of the triangle is in the barn. The cow is attached to the outside of the barn and the barn has no doors. The actual shape we are trying to calculate looks like the one in the picture shown to the right. This means we have to subtract the part of the barn that we have been including in our calculations to find the real area of the weird shape.

c) We have been including exactly half of the barn in our calculations which is the shape of a triangle. We can use the formula for the area of a triangle again to find the area we need to subtract. The height of the barn is 10 and the base of the barn is 10. After we plug it into the formula we get the area of the triangle equals**50 ft**.

**Inverse Trigonometry**

All we have left to do is find the area of the sectors. Inverse trigonometry is very important in helping us find the area of them.

a) Our goal is to find the angle x in the diagram below. In order to do this we need to use the information we already know. We know that a straight line is equal to 180°. We know the very left angle is 45° because it's exactly half of the square barn which is a right angle. We can solve for the middle angle using inverse trig!

b) Because we have all three side lengths of the triangle we can use any function (sine, cosine, or tangent) to find the angle (y). Let's use sine. Sine is opposite divided by hypotenuse. The opposite of angle*y *is 89.72 and the hypotenuse is 90. After plugging it into the calculator, you get angle *y *is equal to 85.47.

c) We have three things that equal 180°. We have figured out two of those values so now we can solve for the third. Angle x + angle y + 45°**=** 180°

After solving this equation we got the angle of the sector was equal to 49.53°.

**5. Area of a sector = πr^2 x degree of sector/360**

a) The second to last step is to solve for the area of the sector. A sector is essentially a part of a circle. The degree of straight line is 180° and the degree of a right angle is 90°. If we think about a quarter or a half of a circle, those angles are 90 and 180° respectively. If a quarter of a circle is equal to 90°, that makes sense because a full circle is 360°. 90/360 equals a quarter. The same goes for half of a circle. 180/360 equals one half. This means that whatever degree a sector is, if we put it over 360 that gets us the portion of the circle. If we multiply the area of a circle times the portion that we want (the sector), we get the area of just the sector.

b) We know the angle of our sector is 49.53° and the radius of our circle is 90 ft. We can plug in these numbers to our equation and we get the area of the sector equals**3,501.07 ft.**

**6. Add everything together!**

Going back to the formula we wrote before, **(area of the 3/4 of the circle) + (area of triangle in between two sectors) + 2(area of sector) - (area of 1/2 of the barn) = total area the cow can graze**, now we have all the values we need to solve it out!

Area of the 3/4 of the circle**= **23,561.94 ft.

Area of triangle in between two sectors**= **634.42 ft.

Area of sector**= **3,501.07 ft.

Area of 1/2 of the barn**= **50 ft.

23,561.94 + 634.42 + 2(3,501.07) - 50 = 31,148.5 ft.

**The cow can graze an area of 31,148.5 ft.**

1. Area of a circle = πr^2

For the 3/4 of a circle, we used the formula for the area of a circle, which is

The radius was 100 because that's how long the rope was. When we plugged it into the formula we got 10,000π. 10,000π was the area of a whole circle but the cow could only graze 3/4 of a circle with the radius 100. That is why we multiplied it by 3/4. This got us with the area of

In order to find the area of the triangle indicated below, we first need to figure out the diagonal length going across the barn. We can find this using the Pythagorean theorem.

The diagonal length is

Now that we have the base of the triangle's length, we need to find the height. If we split the triangle straight down the middle from it's highest point to the base we get two symmetrical right triangles. Now that we have a right angle we can use the Pythagorean Theorem again!

If we are looking at just the left right triangle, we know that the

We know the base of the right triangle is 5√2 because the whole base was equal to 10√2. When we spilt the triangle in half, we split the base in half making it 5√2.

We have two lengths in the equation so we can solve for the third one. This time we have to be careful with what numbers we have and what number we're trying to solve for. We have the hypotenuse and the base. That means we have

b) There is one more catch to this. We have been solving for this triangle but a part of the triangle is in the barn. The cow is attached to the outside of the barn and the barn has no doors. The actual shape we are trying to calculate looks like the one in the picture shown to the right. This means we have to subtract the part of the barn that we have been including in our calculations to find the real area of the weird shape.

c) We have been including exactly half of the barn in our calculations which is the shape of a triangle. We can use the formula for the area of a triangle again to find the area we need to subtract. The height of the barn is 10 and the base of the barn is 10. After we plug it into the formula we get the area of the triangle equals

All we have left to do is find the area of the sectors. Inverse trigonometry is very important in helping us find the area of them.

a) Our goal is to find the angle x in the diagram below. In order to do this we need to use the information we already know. We know that a straight line is equal to 180°. We know the very left angle is 45° because it's exactly half of the square barn which is a right angle. We can solve for the middle angle using inverse trig!

b) Because we have all three side lengths of the triangle we can use any function (sine, cosine, or tangent) to find the angle (y). Let's use sine. Sine is opposite divided by hypotenuse. The opposite of angle

c) We have three things that equal 180°. We have figured out two of those values so now we can solve for the third. Angle x + angle y + 45°

After solving this equation we got the angle of the sector was equal to 49.53°.

a) The second to last step is to solve for the area of the sector. A sector is essentially a part of a circle. The degree of straight line is 180° and the degree of a right angle is 90°. If we think about a quarter or a half of a circle, those angles are 90 and 180° respectively. If a quarter of a circle is equal to 90°, that makes sense because a full circle is 360°. 90/360 equals a quarter. The same goes for half of a circle. 180/360 equals one half. This means that whatever degree a sector is, if we put it over 360 that gets us the portion of the circle. If we multiply the area of a circle times the portion that we want (the sector), we get the area of just the sector.

b) We know the angle of our sector is 49.53° and the radius of our circle is 90 ft. We can plug in these numbers to our equation and we get the area of the sector equals

Area of the 3/4 of the circle

Area of triangle in between two sectors

Area of sector

Area of 1/2 of the barn

23,561.94 + 634.42 + 2(3,501.07) - 50 = 31,148.5 ft.

If I were to grade myself on this unit I would give myself a B. My understanding of the cow problem is still sort of confusing so that is why I gave myself a B. What pushed my thinking was when we had the group test. I think that the group test really helped me understand the problem more and I was able to contribute to my group because I understood the different methods that were being used and how to use them for this problem. I think a lot of the methods for solving for area/angle was more of a refresher for me. Last year and even in freshman year, we had worked with those methods so I kinda already knew how the methods worked and that was easy for me.