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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. # The ratio of the area of the larger square to the area of th  Question banks Downloads My Bookmarks Reviews Important topics
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Moderator  Joined: 18 Apr 2015
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The ratio of the area of the larger square to the area of th [#permalink]
Expert's post 00:00

Question Stats: 53% (02:40) correct 46% (02:02) wrong based on 13 sessions

Attachment: squareinsidecircle.jpg [ 17.64 KiB | Viewed 1037 times ]

 Quantity A Quantity B The ratio of the area of the larger square to thearea of the smaller square Twice the ratio of the area of the smaller circleto 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]
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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