The Cow Problem (Trigonometry)

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Linty Fresh

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A cow is tethered to one corner of a square barn, 10 feet by 10 feet with a rope 100 feet long. What is the maximum grazing area for the cow?

OK, if it were just a case of marking off a sector, I know I could use the s=r*theta formula to find the area of a sector within which the cow can wander, but I'm not sure how the dimensions of the barn come into play. Do they form a triangle within the circle. Any ideas?

Thanks.

Linty
 
The area of the circle to consider is 3/4 of a full circle at a radius of 100 and for the last 1/4 of the circle you need to consider 2 more 1/4 circles of 90 radius (as the new centres become the edges along the building.

Can you work from this hint?
 
Hmmm . . . OK

OK, so using the area formula A=.5*r^2*theta, I get:

100 ft. radius
A=.5*100^2*(3*pi/2)=23,561.9

90 ft radius
A=.5*90^2*(pi/2)=6361.7 times 2, because there are two sectors.

Which gives me 36,285.3 sq ft.

Now this is too high, according to the answer in the book. Should I be using triangle area formulae for the second part of the problem with the 90 ft radius? I'm not quite sure what I'm missing.

Thanks.

Linty
 
If you draw a picture you will see that the two 90 ft parts over-lap in a triangle. You are counting that twice when you double. You have to workout the it's area and subtract it from your total area.
 
OK, so I drew a picture, and I came up with a triangle with sides 90 ft, 90 ft, and 10ft. I calculated the area to be 449.3 sq ft. Subtracting that from 23,561.9+(2*6361.7), I get 35,836 sq ft. Still a little high. Is it safe to assume that the both sides will be 90 feet?

Thanks so much for your help, btw

Linty
 
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