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In the figure shown, line segments QS and RT are diameters of the circ [#permalink]
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12 Nov 2010, 22:44
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In the figure shown, line segments QS and RT are diameters of the circle. If the distance between Q and R is \(\frac{8}{\sqrt{2}}\), what is the area of the circle? (A) \(4 \pi\) (B) \(8 \pi\) (C) \(16 \pi\) (D) \(32 \pi\) (E) \(64 \pi\)
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Last edited by Engr2012 on 15 Apr 2016, 04:30, edited 2 times in total.
Reformatted the question



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Re: In the figure shown, line segments QS and RT are diameters of the circ [#permalink]
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12 Nov 2010, 23:19



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Re: In the figure shown, line segments QS and RT are diameters of the circ [#permalink]
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14 Nov 2010, 16:12
Line joining Q n R makes right isoscelleous(45, 45, 90) triangle where hypotenus = 8/sqrt2
Thus radius = 8/(sqrt2 * sqrt2) = 4
Area = pie*r^2 = 16 pie
Answer: C



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Re: In the figure shown, line segments QS and RT are diameters of the circ [#permalink]
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21 Mar 2016, 18:23
please use the math formulas to remove all ambiguities... we can draw a line from Q to R, with a length of
\(\frac{8}{\sqrt{2}}\)
we know that both legs are radii. applying the 454590 triangle rules, we can identify the radius, but it's easier to write the pythagorean formula:
\((\frac{8}{\sqrt{2}})^2 = \frac{64}{2} = 2r^2\) \(32 = 2r^2\) \(r^2 = 16.\)
Area of a circle is: \(pi*r^2\)
C



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Re: In the figure shown, line segments QS and RT are diameters of the circ [#permalink]
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15 Apr 2016, 03:54
Hi Bunuel, COuld you please tell why do we consider the distance between Q and R not as the length of arc from Q to R ??? Bunuel wrote: monirjewel wrote: In the figure shown, line segments QS and RT are diameters of the circle. If the distance between Q and R is 8/root2, what is the area of the circle? (A) 4 pie (B) 8 Pie (C) 16 pie (D) 32 Pie (E) 64 Pie Let the center of the circle be O. Then OQ and OR will be the radii. Now as triangle OQR is right angle then QR will be its hypotenuse hence \(QR^2=(\frac{8}{\sqrt{2}})^2=r^2+r^2\) > \(2r^2=32\) > \(r^2=16\) > \(area=\pi{r^2}=16\pi\). Answer: C.



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Re: In the figure shown, line segments QS and RT are diameters of the circ [#permalink]
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15 Apr 2016, 03:59
prateek720 wrote: Hi Bunuel, COuld you please tell why do we consider the distance between Q and R not as the length of arc from Q to R ??? Bunuel wrote: monirjewel wrote: In the figure shown, line segments QS and RT are diameters of the circle. If the distance between Q and R is 8/root2, what is the area of the circle? (A) 4 pie (B) 8 Pie (C) 16 pie (D) 32 Pie (E) 64 Pie Let the center of the circle be O. Then OQ and OR will be the radii. Now as triangle OQR is right angle then QR will be its hypotenuse hence \(QR^2=(\frac{8}{\sqrt{2}})^2=r^2+r^2\) > \(2r^2=32\) > \(r^2=16\) > \(area=\pi{r^2}=16\pi\). Answer: C. The distance between two points always means the shortest distance between those two points, which is the length of a straight line between them. Hope it's clear.
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Re: In the figure shown, line segments QS and RT are diameters of the circ [#permalink]
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18 May 2016, 11:14
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Hi Bunuel, COuld you please tell why do we consider the distance between Q and R not as the length of arc from Q to R ??? Bunuel wrote: monirjewel wrote: In the figure shown, line segments QS and RT are diameters of the circle. If the distance between Q and R is 8/root2, what is the area of the circle? (A) 4 pie (B) 8 Pie (C) 16 pie (D) 32 Pie (E) 64 Pie Let the center of the circle be O. Then OQ and OR will be the radii. Now as triangle OQR is right angle then QR will be its hypotenuse hence \(QR^2=(\frac{8}{\sqrt{2}})^2=r^2+r^2\) > \(2r^2=32\) > \(r^2=16\) > \(area=\pi{r^2}=16\pi\). Answer: C. [/quote] The distance between two points always means the shortest distance between those two points, which is the length of a straight line between them. Hope it's clear.[/quote] Hi Bunuel, How can I make such assumption on the GMAT, unless otherwise the information that you provided above is mentioned. I am not sure whether I would call this a common sense to assume that the problem is talking about the "shortest distance" and not the arc length. If possible, please share some similar questions or relevant material, in which certain assumptions like the one stated above is considered a fundamental knowledge. Also, I must say that your contribution to the success of vary many GMAT takers is immeasurable. Thanks a lot for everything you have done, and for everything you are doing.



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Re: In the figure shown, line segments QS and RT are diameters of the circ [#permalink]
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14 Jul 2016, 14:30
herein wrote: Hi Bunuel, How can I make such assumption on the GMAT, unless otherwise the information that you provided above is mentioned. I am not sure whether I would call this a common sense to assume that the problem is talking about the "shortest distance" and not the arc length. If possible, please share some similar questions or relevant material, in which certain assumptions like the one stated above is considered a fundamental knowledge. Also, I must say that your contribution to the success of vary many GMAT takers is immeasurable. Thanks a lot for everything you have done, and for everything you are doing. hey herein, I almost made the same mistake when attempted this question. As far as I know, the GMAT always specifically tells you whenever the arch length shall be used. There is no reference to arch length in this question, so the shortest distance shall be used, as suggested by Bunuel.
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Re: In the figure shown, line segments QS and RT are diameters of the circ [#permalink]
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12 Jun 2017, 15:30
mvictor wrote: please use the math formulas to remove all ambiguities... we can draw a line from Q to R, with a length of
\(\frac{8}{\sqrt{2}}\)
we know that both legs are radii. applying the 454590 triangle rules, we can identify the radius, but it's easier to write the pythagorean formula:
\((\frac{8}{\sqrt{2}})^2 = \frac{64}{2} = 2r^2\) \(32 = 2r^2\) \(r^2 = 16.\)
Area of a circle is: \(pi*r^2\)
C How did you know that you needed to square 8 / \sqrt{2}?




Re: In the figure shown, line segments QS and RT are diameters of the circ
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12 Jun 2017, 15:30








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