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Praetorian
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[#27] 2 Min. Challenge: Sheep 19 Mar 2004, 10:52
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1. Time yourself
2. Solve as fast as you can
3. write your solution here and your time please.

A sheep is tethered to the corner of a square house of side equal to 30 m with a rope of length 60 m. what is the maximum area that can be grazed by it?

A. (3/4)*pi*(60)^2 + (1/2)*pi*(30)^2
B. (1/4)*pi*(60)^2 + (2/3)*pi*(30)^2
C. (3/4)*pi*(30)^2 + (1/2)*pi*(60)^2
D. (2/3)*pi*(30)^2 + (4/5)*pi*(60)^2
E. (1/4)*pi*(60)^2 + (1/2)*pi*(30)^2



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[#27] 2 Min. Challenge: Sheep 19 Mar 2004, 10:59
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Ans: A. (3/4)*pi*(60)^2 + (1/2)*pi*(30)^2

Another good one! thanks a lot.

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lvb9th
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[#27] 2 Min. Challenge: Sheep 19 Mar 2004, 15:04
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A as well.

The entire 3/4 of a circle at r=60, plus two quarters of a circle with r=30

<3 min
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[#27] 2 Min. Challenge: Sheep 19 Mar 2004, 23:57
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praetorian123
1. Time yourself
2. Solve as fast as you can
3. write your solution here and your time please.

A sheep is tethered to the corner of a square house of side equal to 30 m with a rope of length 60 m. what is the maximum area that can be grazed by it?

A. (3/4)*pi*(60)^2 + (1/2)*pi*(30)^2
B. (1/4)*pi*(60)^2 + (2/3)*pi*(30)^2
C. (3/4)*pi*(30)^2 + (1/2)*pi*(60)^2
D. (2/3)*pi*(30)^2 + (4/5)*pi*(60)^2
E. (1/4)*pi*(60)^2 + (1/2)*pi*(30)^2


yes, the answer is A.
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[#27] 2 Min. Challenge: Sheep 20 Mar 2004, 06:29
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Could you please explain how do you arrive at this answer? I am not quiet clear with the explanations given here.
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[#27] 2 Min. Challenge: Sheep 22 Mar 2004, 07:13
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Yes, if someone could explain with a diagram.........it would be highly appreciated!!
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[#27] 2 Min. Challenge: Sheep 22 Mar 2004, 10:04
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cbrf3
Yes, if someone could explain with a diagram.........it would be highly appreciated!!

Sorry, I don't have any software to draw a diagram, However, I'll try to
capture this question in words.

First, assume that there is no house. A goat tied to a pole and the rope
length is 60 meters. Find the area it can cover.

Second, assume that we have a house that sides are more than 60 meters
and the pole where the goat is tied is infact one of the house carner.

In this case, the goat can cover only 3/4 of the circle.


Third, now consider the house has 30 meter sides.
I goat can cover not only 3/4 of the circle, but also some more
area.

The question is how much more area can it cover?

A --------------- B
| |
| |
| |
| |
C----------------D

Assume that the goat is tied to point C. It walks down to point D.
From point D, it can cover 1/4 of circle with radius 30= (60 - 30).

Similarly, if the goat walks to point A. it can cover another 1/4 of a
circle with radius 30.

Thus, adding 3/4 of cicle with radius 60 and 2 times 1/4 area of a circle
with radius 30 yields the answer.



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