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

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Into a square with side K is inscribed a circle with radius  [#permalink]

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New post Updated on: 25 Feb 2014, 10:19
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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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New post 25 Feb 2014, 22:17
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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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New post 18 Oct 2007, 14:14
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
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  [#permalink]

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New post 18 Oct 2007, 14:17
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]

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New post 25 Feb 2014, 10:13
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]

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New post 25 Feb 2014, 10:40
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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
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.

Answer: B.

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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New post 21 Jul 2016, 10:58
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
Untitled (1).png [ 7.81 KiB | Viewed 2958 times ]

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Re: Into a square with side K is inscribed a circle with radius  [#permalink]

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New post 29 Jul 2016, 22:58
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
Answer is 1
B
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Re: Into a square with side K is inscribed a circle with radius  [#permalink]

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New post 29 Jun 2017, 00:39
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



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

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Re: Into a square with side K is inscribed a circle with radius   [#permalink] 01 Oct 2018, 12:00
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