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In the figure above, the diameter of the circle is 12.
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22 Dec 2022, 10:01
22 Dec 2022, 10:01
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Question Stats:
61% (01:54) correct
38% (02:15) wrong based on 39 sessions
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GRE In the figure above, the diameter of the circle.jpg [ 22.39 KiB | Viewed 2723 times ]
In the figure above, the diameter of the circle is 12.
Quantity A |
Quantity B |
The area of rectangle PQRS |
80 |
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: In the figure above, the diameter of the circle is 12.
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04 Jan 2023, 03:30
04 Jan 2023, 03:30
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OE
To picture the smallest possible rectangle, envision one extremely short and wide, as in the figure at left, on the next page. This rectangle would have a width approaching that of the circle's diameter (12), but a height approaching o. Thus, the area would approach o. To picture the largest possible rectangle, envision a perfect square. As in the figure at right below, the diagonal PR = 12, which is also the diameter of the circle. This diameter cuts the square into two 45-45-90 triangles, which have sides in the proportion \(1:1:\sqrt{2}\) .
Therefore, the side of the square would equal \(\frac{12}{\sqrt{2}}= 6 \sqrt{2}\) , and the area would equal \((6 \sqrt{2} )^2=72\) , which is less than Quantity B (80). Therefore, the rectangle will always have an area smaller than Quantity B. Thus, Quantity B is greater.
To picture the smallest possible rectangle, envision one extremely short and wide, as in the figure at left, on the next page. This rectangle would have a width approaching that of the circle's diameter (12), but a height approaching o. Thus, the area would approach o. To picture the largest possible rectangle, envision a perfect square. As in the figure at right below, the diagonal PR = 12, which is also the diameter of the circle. This diameter cuts the square into two 45-45-90 triangles, which have sides in the proportion \(1:1:\sqrt{2}\) .
Therefore, the side of the square would equal \(\frac{12}{\sqrt{2}}= 6 \sqrt{2}\) , and the area would equal \((6 \sqrt{2} )^2=72\) , which is less than Quantity B (80). Therefore, the rectangle will always have an area smaller than Quantity B. Thus, Quantity B is greater.
Re: In the figure above, the diameter of the circle is 12.
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08 Apr 2023, 20:14
08 Apr 2023, 20:14
Didn't understand how did you assume it as square
