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Senior Manager  Joined: 11 Jun 2007
Posts: 478
Into a square with side K is inscribed a circle with radius  [#permalink]

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9 00:00

Difficulty:   45% (medium)

Question Stats: 70% (02:10) correct 30% (02:39) wrong based on 397 sessions

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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

Originally posted by eyunni on 18 Oct 2007, 13:47.
Last edited by Bunuel on 25 Feb 2014, 10:19, edited 1 time in total.
Edited the question and added the OA.
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Re: Into a square with side K is inscribed a circle with radius  [#permalink]

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4
1
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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SVP  Joined: 01 May 2006
Posts: 1520

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1
(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
Senior Manager  Joined: 11 Jun 2007
Posts: 478

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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]

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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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Math Expert V
Joined: 02 Sep 2009
Posts: 58427
Re: Into a square with side K is inscribed a circle with radius  [#permalink]

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2
1
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 [ 3.85 KiB | Viewed 5563 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.

Hope it's clear.
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Re: Into a square with side K is inscribed a circle with radius  [#permalink]

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eyunni 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

Area of square=K^2
Radius of Circle=K/2(as shown in fig.)
Area of circle=pi*(K/2)^2---->pi*K^2/4
Ratio(P)=K^2/(pi*K^2/4)----->4/pi
Again Perimeter of square=4K
Perimeter of circle=2*pi*K/2---->pi*K
Ratio (Q)= 4K/pi*K-------4/pi
So P/Q=(4/pi)/(4/pi)=1
Ans B
Attachments Untitled (1).png [ 7.81 KiB | Viewed 2958 times ]

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GMAT 1: 750 Q49 V43 Re: Into a square with side K is inscribed a circle with radius  [#permalink]

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eyunni 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

Area of the square = $$k^2$$
Area of the circle = $$Pi * r^2$$; Diameter of the circle is k ; so the radius $$r=\frac{k}{2}$$
Area of the circle = $$pi * \frac{k^2}{4}$$

Ratio $$P = \frac{K^2}{Pi * k^2/4} = \frac{4}{Pi}$$

Perimeter of square = $$4k$$
Perimeter of circle = $$2pi*r; 2pi$$ is diameter $$= k$$
Perimeter of the circle = $$pi* k$$

Ratio $$Q= \frac{4k}{pi*k} = \frac{4}{Pi}$$

Now P/Q become $$\frac{4}{pi}$$ divided by $$\frac{4}{pi}$$ = 1
B
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Joined: 07 Sep 2016
Posts: 29
Re: Into a square with side K is inscribed a circle with radius  [#permalink]

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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

Inscribed means(In geometry) :draw (a figure) within another so that their boundaries touch but do not intersect.
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Posts: 13204
Re: Into a square with side K is inscribed a circle with radius  [#permalink]

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