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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1 Answer
Oct 14, 2017

If we call the radius of the smaller circle rr, we see that A = r^2piA=r2π. Since the diameter measures 1010, the radius measures 55 and hence the area is 25pi25π.

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 = 200R2=102+102=200

R = sqrt(200) = sqrt(100 * 2) = 10sqrt(2)R=200=1002=102

Then the area of the larger circle is A = (10sqrt(2)/2)^2pi = 50piA=(1022)2π=50π

Because the area of the inner circle is 25pi25π, the area of the larger circle is

50pi - 25pi = 25pi50π25π=25π

Or the same as the inner circle.

So we've proved that this is the case.

Hopefully this helps!