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GRE Geometry: Practice Question of the Week #31 Answer

Here are the answer and explanation to yesterday’s practice question! Let us know if you have any questions about it.

 

 

Video Explanation:

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Margarette Jung graduated from UC Berkeley as a double major in English Literature and Linguistics. She has been teaching for large companies and tutoring privately for more than 5 years. Follow her on Google+!

6 Responses to GRE Geometry: Practice Question of the Week #31 Answer

  1. Mahantesh December 29, 2011 at 4:31 am #

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

    • Mahantesh December 29, 2011 at 4:50 am #

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

  2. Adrian December 28, 2011 at 1:27 am #

    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?

    • Chris Lele
      Chris December 28, 2011 at 1:51 pm #

      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!

      • Adrian December 29, 2011 at 12:04 am #

        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)?

        • Chris Lele
          Chris December 29, 2011 at 11:30 am #

          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.


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