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The ratio of the area of the larger square to the area of th

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The ratio of the area of the larger square to the area of th [#permalink] New post 16 Sep 2017, 10:31
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Question Stats:

40% (02:09) correct 60% (02:02) wrong based on 10 sessions



Attachment:
squareinsidecircle.jpg
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Quantity A
Quantity B
The ratio of the area of the larger square to the
area of the smaller square
Twice the ratio of the area of the smaller circle
to the area of the larger circle


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: The ratio of the area of the larger square to the area of th [#permalink] New post 05 Oct 2017, 02:14
Even though it can be easier to solve it by plugging in numbers as suggest on Manhattan Prep, I think I find a way to solve it generally. Here is my try.

Quantity A is \(\frac{2r_1^2}{2r_2^2}\), where \(r_1\) is the radius of the larger circle and \(r_2\) the radius of the smaller one.

Quantity B is \(2*\frac{\pi*r_2^2}{\pi*r_1^2}\), keeping the same notation.

Now, let's try to get some information from the figure to simplify those ratios.

It is easy to notice that the diagonal of the smaller square is the diameter of the larger circle. That's could be a nice relationship to use.
In formula, this means that \(2r_1 = 2r_2\sqrt(2)\).

Substituting in the quantities, they become

Q1: \(\frac{(2r_2\sqrt(2))^2}{2r_2^2}\)

Q2: \(2*\frac{\pi*r_2^2}{\pi(r_2\sqrt(2))^2}\)

Simplifying we get

Q1: \(\frac{(8r_2^2}{4r_2^2}\)

Q2: \(2*\frac{r_2^2}{2r_2^2}\)

Simplifyng once more,

Q1: \(2\)

Q2: \(1\)

Thus, Q1 is larger than Q2. Answer A!
Re: The ratio of the area of the larger square to the area of th   [#permalink] 05 Oct 2017, 02:14
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