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655-705 (Hard)|   Geometry|      
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rohan2345
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Can a circular table close cover a rectangular table completely? 26 Aug 2019, 00:58
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E
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65% (hard)

Question Stats:

38% (01:06) correct 62% (01:23) wrong based on 77 sessions

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Can a circular table close cover a rectangular table completely?

(1) The diameter of the close is greater than the length of the table.

(2) The area of the cloth is greater than the area of the table.
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E
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ccooley
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Can a circular table close cover a rectangular table completely? 26 Aug 2019, 13:29
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Here's a quick drawing illustrating why the answer is E: even if the diameter of the circle is greater than the length of the rectangle, and the area of the circle is much greater than the length of the rectangle, it's possible that the corners of the rectangle will still stick out:

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Can a circular table close cover a rectangular table completely? 21 Sep 2019, 22:42
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ccooley
Here's a quick drawing illustrating why the answer is E: even if the diameter of the circle is greater than the length of the rectangle, and the area of the circle is much greater than the length of the rectangle, it's possible that the corners of the rectangle will still stick out:

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Just to make sure I am thinking straight, if the diameter of the circle would have been equal to the diagonal of the rectangle, then it would be sufficient. right?
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Fdambro294
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Can a circular table close cover a rectangular table completely? Updated on: 26 Aug 2021, 01:32
Originally posted by Fdambro294 on 26 Aug 2021, 01:22.
Last edited by Fdambro294 on 26 Aug 2021, 01:32, edited 2 times in total.
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You can use the coordinate plane just to verify that the circle can meet both conditions in the statements and still not cover the rectangle.


Let the rectangle dimensions be 6 by 8

Put 1 vertex at (0 ,0)

2nd vertex at (0 , 6)

3rd vertex at (8 , 0)

4th vertex at (8 , 6)


Now can we have a circle NOT cover the area of the rectangle and:

-the diameter be greater than the Length of the rectangle (call L = 6)

And

-the area of circle > area of rectangle (6 * 8 = 48)

????


To maximize the chance that the Circle can cover the rectangle we will want to match the center of the circle with the symmetrical center of the rectangle. This will be at point (4 , 3)

Let the Diameter of the Circle be = 8

Which means the radius = 4

Area of Circle = (4)^2 * (pi) = 16 * (a little greater than 3) = a value a little greater than the rectangle’s area of 48 ———> statement 2 is satisfied

Also, if the length of the rectangle = 6 ———> statement 1 is satisfied as the Diameter of 8 is greater

To determine whether the circle will cover the rectangle we can ask the following question:

Will the length from the center of the circle (4 , 3) to the Origin Vertex of the rectangle(0 , 0) be LESS than the radius of the circle ——- such that the Circle will cover the vertex at (0 , 0) (and all the other vertices)???

The answer is NO, the circle will not cover the vertex at (0 , 0) (and the other 3 vertices of the rectangle)

Using the Pythagorean Triplet 3-4-5 to our advantage, you can see that the length from the center of the circle to the vertex is = 5

However, the radius of the Circle is only = 4

Thus there will be a small triangular-like area outside of the circle but inside the rectangle that will NOT be covered by the circle.

This gives us a NO with both statements satisfied.

We can easily find a YES by making the diameter of the circle = 20.

-E-

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Can a circular table close cover a rectangular table completely? 26 Aug 2021, 01:24
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Yes, in such a case we can inscribe the rectangle inside the circle.

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ccooley
Here's a quick drawing illustrating why the answer is E: even if the diameter of the circle is greater than the length of the rectangle, and the area of the circle is much greater than the length of the rectangle, it's possible that the corners of the rectangle will still stick out:


Just to make sure I am thinking straight, if the diameter of the circle would have been equal to the diagonal of the rectangle, then it would be sufficient. right?
Show more

Posted from my mobile device
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