As a matter of fact, when I had solved this very same question from the Math Video Practice set, I did have a doubt. Why is it that you solve this in a roundabout way when a more direct solution presents itself as the area of the circle – area of the square? Also, since the diagonal of the square measures 5 units, it would seem that the circle would intersect the square at the point diagonally opposite to its center. Shouldn’t the solution be 25pi – 12?

The problem is a little different from the way you are interpreting it. If you look above, our cow is restricted in how far he can move. Once he has reached the corner of the shed the rope will be limited to two inches or one inch long depending on which corner he is standing. Once he tries to go around he can only move the length of a quarter circle. The radius for the first circle is 1 (hence 1/4pi) and for where the rope stretches out to 2 (pi). From there we can get the answer be subtracting it from the area covered by where the rope has a radius of 5 (which is a large 3/4 circle).

While it is true that a complete circle would intersect the shack at the far side, that point is irrelevant to the question, because our cow cannot walk that far – he is limited by the length of the rope to a 3/4 circle as seen below.

Well I did understand what you meant Chris but wouldn’t your stated solution imply that the center of the tether has been shifted to that point say from which the quarter circle radius becomes 2 inches? For clarifying my point, lets assume that the shed is now absent. Wouldn’t the area accessible to the cow be the area of the circle with radius 5 inches? And therefore, after having included the shed equation into the question, wouldn’t the area presently accessible be the difference of the two areas? Assuming the cow is standing 5 inches from the center directly below it. In that situation the length of rope accessible to it would be limited to 2 inches. But how can you state that it will always be 2 inches in that area? Shouldn’t it be 5 inches – (length of whatever line segment of the radius line which is inside the shed)?

As Manhantesh noted, the shack is limiting the cow’s range of motion. When the cow walks to the corner of the shack where the width is 3 inches, there is only 2 inches of rope left. Now when the cow attempts to move to the other side of the shack, it can stray no farther than 2 inches from the corner. That range of motion is essentially a small circle with radius 2. (In the diagram this area is represented by a pink quarter circle). That small circle does not cover the entire range of the original circle (radius 5) minus the shed.

Magoosh blog comment policy: To create the best experience for our readers, we will only approve comments that are relevant to the article, general enough to be helpful to other students, concise, and well-written! 😄 Due to the high volume of comments across all of our blogs, we cannot promise that all comments will receive responses from our instructors.

We highly encourage students to help each other out and respond to other students' comments if you can!

If you are a Premium Magoosh student and would like more personalized service from our instructors, you can use the Help tab on the Magoosh dashboard. Thanks!

When it can cover the distance with a diagonal length why it cannot cover 25 pi

Got the solution as shed was not allowing to move beyond walls !!!!!!

thanks

As a matter of fact, when I had solved this very same question from the Math Video Practice set, I did have a doubt. Why is it that you solve this in a roundabout way when a more direct solution presents itself as the area of the circle – area of the square? Also, since the diagonal of the square measures 5 units, it would seem that the circle would intersect the square at the point diagonally opposite to its center. Shouldn’t the solution be 25pi – 12?

Hi Adrian,

The problem is a little different from the way you are interpreting it. If you look above, our cow is restricted in how far he can move. Once he has reached the corner of the shed the rope will be limited to two inches or one inch long depending on which corner he is standing. Once he tries to go around he can only move the length of a quarter circle. The radius for the first circle is 1 (hence 1/4pi) and for where the rope stretches out to 2 (pi). From there we can get the answer be subtracting it from the area covered by where the rope has a radius of 5 (which is a large 3/4 circle).

While it is true that a complete circle would intersect the shack at the far side, that point is irrelevant to the question, because our cow cannot walk that far – he is limited by the length of the rope to a 3/4 circle as seen below.

Hope that helps!

Well I did understand what you meant Chris but wouldn’t your stated solution imply that the center of the tether has been shifted to that point say from which the quarter circle radius becomes 2 inches? For clarifying my point, lets assume that the shed is now absent. Wouldn’t the area accessible to the cow be the area of the circle with radius 5 inches? And therefore, after having included the shed equation into the question, wouldn’t the area presently accessible be the difference of the two areas? Assuming the cow is standing 5 inches from the center directly below it. In that situation the length of rope accessible to it would be limited to 2 inches. But how can you state that it will always be 2 inches in that area? Shouldn’t it be 5 inches – (length of whatever line segment of the radius line which is inside the shed)?

As Manhantesh noted, the shack is limiting the cow’s range of motion. When the cow walks to the corner of the shack where the width is 3 inches, there is only 2 inches of rope left. Now when the cow attempts to move to the other side of the shack, it can stray no farther than 2 inches from the corner. That range of motion is essentially a small circle with radius 2. (In the diagram this area is represented by a pink quarter circle). That small circle does not cover the entire range of the original circle (radius 5) minus the shed.