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Re: The ratio of the area of the larger square to the area of th [#permalink]
Assume the radius of the small circle = x. Then the side of the small square = 2x.
Area of small circle = pi*x^2
Area of small square = (2x)^2 = 4x^2

Next, the diameter of the bigger circle makes a 45-45-90 triangle with the side of the small square, so the diameter of the larger circle = 2*sqrt(2)*x = the side of the large square. And then the radius of the larger circle is half that = sqrt(2)*x

Area of large circle = pi*(sqrt(2)*x)^2 = 2*pi*x^2
Area of large square = (2*sqrt(2)*x)^2 = 4*2*x^2 = 8x^2

Now let's look at the ratios in the problem.
Ratio A = large square / small square = (8x^2) / (4x^2) = 2
Ratio B = 2 * small circle / large circle = 2 * (pi*x^2) / (2*pi*x^2) = (2*pi*x^2) / (2*pi*x^2) = 1

So A is bigger.
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Re: The ratio of the area of the larger square to the area of th [#permalink]
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