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In the figure shown, line segments QS and RT are diameters of the circ

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In the figure shown, line segments QS and RT are diameters of the circ [#permalink]

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New post 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) \(64 \pi\)
[Reveal] Spoiler: OA

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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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New post 12 Nov 2010, 23:19
Expert's post
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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New post 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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New post 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 45-45-90 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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New post 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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New post 15 Apr 2016, 03:59
Expert's post
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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

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New post 18 May 2016, 11:14
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. :)
Re: In the figure shown, line segments QS and RT are diameters of the circ   [#permalink] 18 May 2016, 11:14
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