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Into a square with side K is inscribed a circle with radius

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Director
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Into a square with side K is inscribed a circle with radius [#permalink] New post 18 Oct 2007, 12:47
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

62% (02:26) correct 37% (02:04) wrong based on 29 sessions
Into a square with side K is inscribed a circle with radius r. If the ratio of area of square to the area of circle is P and the ratio of perimeter of the square to that of the circle is Q. Which of the following must be true?

(A) P/Q > 1
(B) P/Q = 1
(C) 1 > P/Q > 1/2
(D) P/Q = 1/2
(E) P/Q < 1/2
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Feb 2014, 09:19, edited 1 time in total.
Edited the question and added the OA.
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 [#permalink] New post 18 Oct 2007, 13:14
(B) for me :)

We have:
o k= 2*r
o Area of the circle = pi * r^2
o Area of the square = k^2 = 4 * r^2
o P = k^2 / (pi * r^2) = 4 / pi
o Perimeter of the circle = 2*pi*r = pi * k
o Perimeter of the square = 4*k
o Q = 4*k / pi * k = 4 / pi = P

Finally,
P = Q
<=> P/Q = 1
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 [#permalink] New post 18 Oct 2007, 13:17
Fig wrote:
(B) for me :)

We have:
o k= 2*r
o Area of the circle = pi * r^2
o Area of the square = k^2 = 4 * r^2
o P = k^2 / (pi * r^2) = 4 / pi
o Perimeter of the circle = 2*pi*r = pi * k
o Perimeter of the square = 4*k
o Q = 4*k / pi * k = 4 / pi = P

Finally,
P = Q
<=> P/Q = 1


oops...OA is not unconvincing anymore..:)
I missed the equation k = 2r. Thanks.
OA is B indeed.
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Re: Into a square with side K is inscribed a circle with radius [#permalink] New post 25 Feb 2014, 09:13
Bunuel,

Can you look into this solution.

Inscribed means circle is touching all the 2 sides of a square?

OR

2R<K

According to me this should be the condition:

2R<=K



Final solution will be P/Q >=1 NOT P/Q = 1
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Re: Into a square with side K is inscribed a circle with radius [#permalink] New post 25 Feb 2014, 09:40
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honchos wrote:
Into a square with side K is inscribed a circle with radius r. If the ratio of area of square to the area of circle is P and the ratio of perimeter of the square to that of the circle is Q. Which of the following must be true?

(A) P/Q > 1
(B) P/Q = 1
(C) 1 > P/Q > 1/2
(D) P/Q = 1/2
(E) P/Q < 1/2

Bunuel,

Can you look into this solution.

Inscribed means circle is touching all the 2 sides of a square?

OR

2R<K

According to me this should be the condition:

2R<=K



Final solution will be P/Q >=1 NOT P/Q = 1


When a circle is inscribed in a square it touches all 4 sides of the square:
Attachment:
Untitled.png
Untitled.png [ 3.85 KiB | Viewed 247 times ]

Thus when a circle is inscribed in a square, the diameter of the circle is equal to the side length of the square.

Into a square with side K is inscribed a circle with radius r. If the ratio of area of square to the area of circle is P and the ratio of perimeter of the square to that of the circle is Q. Which of the following must be true?

(A) P/Q > 1
(B) P/Q = 1
(C) 1 > P/Q > 1/2
(D) P/Q = 1/2
(E) P/Q < 1/2

Into a square with side K is inscribed a circle with radius r --> k = 2r --> ;

The ratio of area of square to the area of circle is P --> \frac{k^2}{\pi{r^2}}=\frac{(2r)^2}{\pi{r^2}}=\frac{4}{\pi}=P.

The ratio of perimeter of the square to that of the circle is Q --> \frac{4k}{2\pi{r}}=\frac{4(2r)}{2\pi{r}}=\frac{4}{\pi}=Q.

Thus we have that P=Q --> P/Q=1.

Answer: B.

Hope it's clear.
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Re: Into a square with side K is inscribed a circle with radius [#permalink] New post 25 Feb 2014, 21:17
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honchos wrote:
Bunuel,

Can you look into this solution.

Inscribed means circle is touching all the 2 sides of a square?

OR

2R<K

According to me this should be the condition:

2R<=K



Final solution will be P/Q >=1 NOT P/Q = 1


Also, check out the following posts on regular polygons inscribed in circles and circles inscribed in regular polygons:

http://www.veritasprep.com/blog/2013/07 ... relations/
http://www.veritasprep.com/blog/2013/07 ... other-way/
http://www.veritasprep.com/blog/2013/07 ... n-circles/
http://www.veritasprep.com/blog/2013/07 ... -polygons/
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Re: Into a square with side K is inscribed a circle with radius   [#permalink] 25 Feb 2014, 21:17
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