# Math Help - More Creative thinking and Circles

1. ## More Creative thinking and Circles

If there is a goat tied to a rectangular barn on a 50 foot lead and the barn is 20 feet by 20 feet (floor), what is the maximum grazing area? If there are regions you can't find the area of, provide as good an estimate as you can. Assume the goat is tied to a corner outside the barn, cannot get in, and that the barn is not grazing area. (Remember, this will be based on parts of circles, no other shapes...the goat's rope will only get shorter when he tries to go around the barn...)

How much of the 50 foot circle can the goat reach without getting interrupted by the barn? What is that area?
When the rope goes around the barn, what is the new radius? How much of a circle can it make without hitting the barn or overlapping area you've already found? What is that area?

When the rope goes around the barn the other way, what is the new radius? How much of a circle can it make without hitting the barn or overlapping area you've already found? What is that area?

The areas you found in 7 and 8 overlap each other. How much do they overlap? What *approximate* shape do they make? What is that area?
What is the total grazing area the goat can reach?

Formula's will be very much appreciated. Have no clue how to picture all of this..

2. Hi!!!

Area of a circle is $\pi r^2$

I have labeled your figure(see next post).

Note that region AOCB is $\frac{3}{4}$ of the full circle with centre at O and radius OA. Hence, area of region OACB is $\frac{3}{4}$ of the full circle with centre at O and radius OA.

Try to find area of other regions in the same way.

3. Forgot to attach figure in the last post.

4. Does this mean that I multiply 25 by pi to get the initial area?

Originally Posted by malaygoel
Forgot to attach figure in the last post.

5. I've attached a sketch of the situation a little bit more to scale so you can pick up the radii in question.

If the goat is going counterclockwise around the center it is covering 3 different areas:

1. The largest area the goat can reach is $\frac34$ of the outer circle:

$a_1=\dfrac34 \cdot \pi \cdot 50^2$

2. Then a quarter circle with radius r = 30 follows:

$a_2= \dfrac14 \cdot \pi \cdot 30^2$

3. At last the smallest quarter circle with radius r = 10:

$a_3=\dfrac14 \cdot \pi \cdot 10^2$

4. The trickiest part of the question is to calculate the overlapping areas. But according to the wording of the problem you are asked to use an approximate shape (maybe a circle(?)) to get the total area.

6. Still having trouble with the last two, especially but with problem in entirety. Thank you. Your help has allowed me to produce this work. However, my work is unfortunately wrong. Please tell me what is my flaw. Thanks for the help from all you mathematicians

6)a1=3/4(pi)(50^2)
3/4(pi)(50^2)
3/4(pi)(2500)
2500/4=625*3=1875 sqrt of 1875 = 43.301

7)a^2=(1/4)(pi)(900)= 900/4= 225 sqrt of 225 = 15

8)a^3=(1/4)(pi)(r^2)
a^3=(1/4)(pi)(10^2)= 100/4= 25 sqrt of 25= 5

9)They overlap 10 degrees. They make a pennant.

10) 15 * 5 = 75 total grazing

Originally Posted by earboth
I've attached a sketch of the situation a little bit more to scale so you can pick up the radii in question.

If the goat is going counterclockwise around the center it is covering 3 different areas:

1. The largest area the goat can reach is $\frac34$ of the outer circle:

$a_1=\dfrac34 \cdot \pi \cdot 50^2$

2. Then a quarter circle with radius r = 30 follows:

$a_2= \dfrac14 \cdot \pi \cdot 30^2$

3. At last the smallest quarter circle with radius r = 10:

$a_3=\dfrac14 \cdot \pi \cdot 10^2$

4. The trickiest part of the question is to calculate the overlapping areas. But according to the wording of the problem you are asked to use an approximate shape (maybe a circle(?)) to get the total area.

7. Originally Posted by KevinVM20
Still having trouble with the last two, especially but with problem in entirety. Thank you. Your help has allowed me to produce this work. However, my work is unfortunately wrong. Please tell me what is my flaw. Thanks for the help from all you mathematicians

6)a1=3/4(pi)(50^2)
3/4(pi)(50^2)
3/4(pi)(2500)
2500/4=625*3=1875 sqrt of 1875 = 43.301 ...... <<<<< What do you want to calculate by this?

...
1. Even though you write (pi) in your calculations the value of $\pi$ doesn't occur ... hmmmm?

2. The area
$a_1 = \dfrac34 \cdot \pi \cdot 50^2 =1875 \cdot \pi \approx 5890.5 \ (ft)^2$

3. To calculate the area outside of the $\dfrac34$-circle I've attached an enlarged drawing.

4. The area in question consists of a sector and a triangle. To calculate the distance PF use the right triangle CPF:

$(20+x)^2+x^2 = 30^2~\implies~2x^2+40x-500=0~\implies~x=\dfrac{-40 \pm \sqrt{40^2-4 \cdot 2 \cdot (-500)}}{2 \cdot 2}$

and therefore $x = -10+5\sqrt{14}\approx 8.7083$ and the distance $d(\overline{PF}) = 20+x = 10+5\sqrt{14} \approx 28.7083$

5. Use the Sine rule to calculate the angle $\alpha$:

$\sin(\alpha) = \dfrac{20+x}{30} \approx \dfrac{28.7083}{30}~\implies~ \alpha \approx 73.1256^\circ$

Now you can calculate the area of the sector (light blue):

$a_2=\dfrac{73.1256^\circ}{360^\circ} \cdot \pi \cdot 30^2 \approx 574.3\ (ft)^2$

6. The angle $\angle(QCP)=90^\circ - 73.1256^\circ \approx 16.8744^\circ$

Since you know that $|\overline{QC}|= 20$ and $|\overline{CP}|=30$ you can calculate the area of the triangle (dark blue):

$a_3 = \dfrac12 \cdot 20 \cdot 30 \cdot \sin(16.8744^\circ) \approx 87.1\ (ft)^2$

7. The total area which provides the goat with green, green grass is consequently:

$A = a_1 + 2 \cdot (a_2 + a_3) = 7213.3\ (ft)^2$

8. earboth, you are a champ!!! thanks a bundle!

Originally Posted by earboth
1. Even though you write (pi) in your calculations the value of $\pi$ doesn't occur ... hmmmm?

2. The area
$a_1 = \dfrac34 \cdot \pi \cdot 50^2 =1875 \cdot \pi \approx 5890.5 \ (ft)^2$

3. To calculate the area outside of the $\dfrac34$-circle I've attached an enlarged drawing.

4. The area in question consists of a sector and a triangle. To calculate the distance PF use the right triangle CPF:

$(20+x)^2+x^2 = 30^2~\implies~2x^2+40x-500=0~\implies~x=\dfrac{-40 \pm \sqrt{40^2-4 \cdot 2 \cdot (-500)}}{2 \cdot 2}$

and therefore $x = -10+5\sqrt{14}\approx 8.7083$ and the distance $d(\overline{PF}) = 20+x = 10+5\sqrt{14} \approx 28.7083$

5. Use the Sine rule to calculate the angle $\alpha$:

$\sin(\alpha) = \dfrac{20+x}{30} \approx \dfrac{28.7083}{30}~\implies~ \alpha \approx 73.1256^\circ$

Now you can calculate the area of the sector (light blue):

$a_2=\dfrac{73.1256^\circ}{360^\circ} \cdot \pi \cdot 30^2 \approx 574.3\ (ft)^2$

6. The angle $\angle(QCP)=90^\circ - 73.1256^\circ \approx 16.8744^\circ$

Since you know that $|\overline{QC}|= 20$ and $|\overline{CP}|=30$ you can calculate the area of the triangle (dark blue):

$a_3 = \dfrac12 \cdot 20 \cdot 30 \cdot \sin(16.8744^\circ) \approx 87.1\ (ft)^2$

7. The total area which provides the goat with green, green grass is consequently:

$A = a_1 + 2 \cdot (a_2 + a_3) = 7213.3\ (ft)^2$

9. NVM, I am still having problems with one problem.

I am still having problems with this part of the problem.

When the rope goes around the barn the other way, what is the new radius? How much of a circle can it make without hitting the barn or overlapping area you've already found? What is that area?

earboth, gave me a formula that does not seem to be working for me.

Here is the formula:
1/4 x pi x 10^2

Here is all my work :
(1/4)(pi)(r^2)
(1/4)(pi)(10^2)= 100/4(pi)

That did not work so I tried this..

(1/4)(pi)(r^2)
a^3=(1/4)(pi)(20^2)
(1/4)(pi)(400)/4= (100)(pi)= 314
They are both incorrect. I am using the formula, aren't I?

Thanks for any potential help. I am getting highly frustrated with the problem..

Originally Posted by KevinVM20
earboth, you are a champ!!! thanks a bundle!

10. Originally Posted by KevinVM20
...

How much of the 50 foot circle can the goat reach without getting interrupted by the barn? What is that area?

It's 3/4 of the large circle. I've painted it orange. The value of this area is calculated as $\bold{\color{blue}a_1}$

When the rope goes around the barn, what is the new radius?
The new radius is r = 50' - sidelength of barn = 30'. One center of the circle in question is for instance the point C.
How much of a circle can it make without hitting the barn or overlapping area you've already found? What is that area?

This is the sector which I've painted in lightblue. First you must calculate the central angle of the sector and afterwards the value of the area. Compare the steps #4 and #5 of my previous post.

When the rope goes around the barn the other way, what is the new radius? How much of a circle can it make without hitting the barn or overlapping area you've already found? What is that area?

There are only 2 small triangles left where the goat can graze too. One triangle's area is calculated as $\bold{\color{blue}a_3}$

The areas you found in 7 and 8 <<<<< not sure what you mean by that (I didn't find any 7 and 8)

overlap each other. How much do they overlap? What *approximate* shape do they make? What is that area?
What is the total grazing area the goat can reach?

Formula's will be very much appreciated. Have no clue how to picture all of this..
....

11. I understand that I am supposed to calculate the light blue. Earboth, can you walk me through what is the radius when the goat goes the opposite way of the barn? this part is very annoying. I understand what you are saying to me in the general geometric sense, yet I can't seem to find the radius of the opposite way. I really need to come up with an answer soon or I'll get a terrible grade.

Thank you

12. Originally Posted by KevinVM20
I understand that I am supposed to calculate the light blue. Earboth, can you walk me through what is the radius when the goat goes the opposite way of the barn? <<<<<< what exactly do you mean by this?
this part is very annoying. I understand what you are saying to me in the general geometric sense, yet I can't seem to find the radius of the opposite way. I really need to come up with an answer soon or I'll get a terrible grade.

Thank you
I've attached a sketch where all radii in question are marked in different colors. Maybe this helps to clarify my calculations.

I've added an enlarged sketch of the region where the overlapping happens. Since the original question asks for an approximate solution I've made 2 suggestions:
1. Use $\frac34$ of a circle with r = 10'. Then the darkbrown area is missing, that means this approximation is too small.
2. Add a square with sidelength of 10' to the 2 quartercircles with r = 10'. Then the darkbrown area was added, that means this approximation is too large.

If this doesn't help you any further I think it would be best, if you can give us the unabrigded and unedited text of the problem.

13. " When the rope goes around the barn, what is the new radius? How much of a circle can it make without hitting the barn or overlapping area you've already found? What is that area? "

(1/4)(pi)(900)= 900/4= 225 x 3.14 = 706.5 = This is correct.

"When the rope goes around the barn the other way, what is the new radius? How much of a circle can it make without hitting the barn or overlapping area you've already found? What is that area? "

I don't understand how to calculate the other way...