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The ratio of the area of the larger square to the area of th
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16 Sep 2017, 10:31
16 Sep 2017, 10:31
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Question Stats:
51% (02:33) correct
48% (01:52) wrong based on 194 sessions
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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.
Re: The ratio of the area of the larger square to the area of th
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05 Oct 2017, 02:14
05 Oct 2017, 02:14
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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!
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
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20 Aug 2019, 20:57
20 Aug 2019, 20:57
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This problem is based on two theories of geometry,
1) If a circle (Cs)is inscribed in a square which is in turn inscribed in another circle (Cb), then the area of the bigger circle (Cb) is twice as the area of the smaller circle(Cs)
2) If a square(Ss) is inscribed in a circle which is in turn inscribed in another square (Sb), then the area of the bigger square (Sb) is twice as the area of the smaller square (Ss)
As per the theories mentioned above , we get
Value of A = area of Sb/area of Ss = 2
Value of B = 2 * ( area of Cs/area of Cb) = 2 * 1/2 = 1
hence the answer is option A.
1) If a circle (Cs)is inscribed in a square which is in turn inscribed in another circle (Cb), then the area of the bigger circle (Cb) is twice as the area of the smaller circle(Cs)
2) If a square(Ss) is inscribed in a circle which is in turn inscribed in another square (Sb), then the area of the bigger square (Sb) is twice as the area of the smaller square (Ss)
As per the theories mentioned above , we get
Value of A = area of Sb/area of Ss = 2
Value of B = 2 * ( area of Cs/area of Cb) = 2 * 1/2 = 1
hence the answer is option A.
