Can a circular table close cover a rectangular table completely?

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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Yes, in such a case we can inscribe the rectangle inside the circle.



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