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# A rectangle is inscribed in a circle whose radius equals R.

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A rectangle is inscribed in a circle whose radius equals R. [#permalink]

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17 Sep 2004, 12:56
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A rectangle is inscribed in a circle whose radius equals R. It if is known that this rectangle is not a square, what could be the perimeter of that rectangle?

(A) 4*sqrt(2)*R
(B) 3*R
(C) 5*R
(D) 6*R
(E) pi*R
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17 Sep 2004, 13:06
I am getting 2R+2 sqrt(3) R

This value is closer to C.
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17 Sep 2004, 14:32
Would go with B......if correct, will post the solution!!
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17 Sep 2004, 23:09
sorry guys, I don't have an OA (the question is from a PDF without answers), but please do publish an explenation
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18 Sep 2004, 02:22
I am not sure we can solve this.
Answer A is the square solution
E is impossible
between the 3 remaining answers how could we choose knowing that we do not know anything about diagonals angles ...
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18 Sep 2004, 02:41
I pick D.
The question is "what COULD be the perimeter".

The solution for a square is 4*sqrt(2)*R .
The perimeter of a rectangle should be greater than the perimeter of a square , so the only answer left is D.
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18 Sep 2004, 03:08
hmmm... ..took R as the dia wheras it shud be 2R!!

Last edited by cbrf3 on 18 Sep 2004, 04:52, edited 1 time in total.
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18 Sep 2004, 04:04
Correct reasoning Dookie.

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18 Sep 2004, 07:51
Dookie,

I do not understand your opinion

Circle perimeter= 2.pi.R > 6.R (your suggestion)

Besides there is no reason why the rectangle perimeter should be greater than the circle perimeter. IMO it is exactly the opposite way : circle arc is greater than a line inscribed in it...

If x is the diagonal angle, rectangle perimeter = 4R(sinx+cosx)

so min = 4R when rectangle has a nearly 0 width and max = 4Rsqrt(2) which is the square perimeter...
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18 Sep 2004, 10:40
twixt wrote:
Dookie,Besides there is no reason why the rectangle perimeter should be greater than the circle perimeter.

twixt, I think Dookie is comparing an inscribed square perimeter (not circle perimeter) with inscribed rectangle perimeter.

Dookie wrote:
I pick D.
The question is "what COULD be the perimeter".
The solution for a square is 4*sqrt(2)*R .
The perimeter of a rectangle should be greater than the perimeter of a square , so the only answer left is D.

Dookie, do you have a proof for, Perimeter of an inscribed rectangle > Perimeter of an inscribed square?

I drew a circle of R=1 and inscribed a square (a = 1.414, Per = 5.6) and a rectangle (l = 1.87, b = 0.7, Per = 5.14) and it shows the opposite.
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18 Sep 2004, 22:19
On second thought, my earlier post is wrong. The answer should be C.
The Maximal permieter seems to be in the Square case. (4*sqrt(2)*R)
And the minimal is a "flat" rectangle with a perimeter of about 4*R.
So the correct choice will be between those values and the only answer is C.
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19 Sep 2004, 08:24
The answer to this question is shd be A

The diameter of the circle which happens to be the diagonal of the rectangle is 2R,
Now apply the angle 45: 45: 90 rule, lenght of lines in such triangle are in the ratio x: x: xsqrt(2).
For this question
x sqrt(2) = 2R
x= 2R/sqrt(2) = R*sqrt(2)

Perimeter= 4x = 4R*sqrt(2)
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20 Sep 2004, 02:30
So what's the final answer ?
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03 Oct 2004, 12:57
Perhaps here
http://www.gmatclub.com/phpbb/viewtopic.php?t=10391
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Franky
http://franky4gmat.blogspot.com

03 Oct 2004, 12:57
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# A rectangle is inscribed in a circle whose radius equals R.

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