The inner circle is the largest one that can be drawn inside the square. The outer circle is the smallest one that can be drawn with the square inside it. Prove that the shaded area between the 2 circles is the same as the area enclosed by inner circle?

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The inner circle is the largest one that can be drawn inside the square. The outer circle is the smallest one that can be drawn with the square inside it. Prove that the shaded area between the 2 circles is the same as the area enclosed by inner circle?

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Answer 1 · 2017-10-14T15:52:33

If we call the radius of the smaller circle #r#, we see that #A = r^2pi#. Since the diameter measures #10#, the radius measures #5# and hence the area is #25pi#.

The diameter of the larger circle is given by pythagoras, because it can be found by drawing a diagonal through the square.

#R^2 = 10^2 + 10^2 = 200#

#R = sqrt(200) = sqrt(100 * 2) = 10sqrt(2)#

Then the area of the larger circle is #A = (10sqrt(2)/2)^2pi = 50pi#

Because the area of the inner circle is #25pi#, the area of the larger circle is

#50pi - 25pi = 25pi#

Or the same as the inner circle.

So we've proved that this is the case.

Hopefully this helps!

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The inner circle is the largest one that can be drawn inside the square. The outer circle is the smallest one that can be drawn with the square inside it. Prove that the shaded area between the 2 circles is the same as the area enclosed by inner circle?

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