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If a square is inscribed in a circle that, in turn, is inscribed in a

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If a square is inscribed in a circle that, in turn, is inscribed in a  [#permalink]

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New post 08 Dec 2010, 19:04
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If a square is inscribed in a circle that, in turn, is inscribed in a larger square, what is the ratio of the perimeter of the larger square to that of the smaller square?

A. \(\frac{1}{2}\)
B. \(\frac{1}{\sqrt{2}}\)
C. \(\sqrt{2}\)
D. \(2\)
E. \(3.14\)

M17-23

Ok, I understand this explanation from sportyrizwan:

Let "a" be the diameter of the circle
Therefore the side of the bigger square will be a and the perimeter 4a

Also a will be the diagonal of the smaller square
Therefore the side of the smaller square will be a/sqrt2 and perimeter (2a)sqrt2

Therefore the ratio of the bigger square to the smaller square is sqrt2

http://gmatclub.com/forum/square-inscribed-in-a-circle-inscribed-in-a-square-55210.html

***********

However, when I try to plug values, it doesn't work..

Let say larger square
one side x=2, so P=8

Smaller square
Diagonal = hypotenus = 2
So x^2+x^2=2^2
2x^2=4
x^2=2
x=sqrt2
P=4sqrt2

L/S = 8/4sqrt2 = 2 ???? It doesn't work and I can't find my mistake..

PLEASE help me to find my mistake!!

Thank you very much for your help!!
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Re: If a square is inscribed in a circle that, in turn, is inscribed in a  [#permalink]

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New post 09 Dec 2010, 01:46
2
3
maive wrote:
GMAT CLUB Tests, M17 q 23

A square is inscribed in a circle whic, in turn, is inscribed in a bigger square. What is the ratio of the perimeter of the bigger square to that of the smaller square?

A. 1/2
1/rad 2
sqrt2
2
pi


Look at the diagram:
Attachment:
SquareCircleSquare.jpg
SquareCircleSquare.jpg [ 10.9 KiB | Viewed 4391 times ]

You can see that a diameter of the circle equals to a diagonal of the smaller square and a side of the bigger square: \(diameter=diagonal_{small}=side_{big}\);

Now, if the side of the bigger square is \(a\) then its perimeter will be \(P=4a\);
Next, a side of a smaller square will be \(\frac{a}{\sqrt{2}}\) and its perimeter \(p=\frac{4a}{\sqrt{2}}\);

\(\frac{P}{p}=\sqrt{2}\)

Answer: C.

Or as soon as you realize that the sides of bigger and smaller squares are \(a\) and \(\frac{a}{\sqrt{2}}\) respectively, so their ration is \(\sqrt{2}\), then as P=4*side then the ratio of the sides will be the same as the ratio of the perimeters, so the same \(\sqrt{2}\).
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Re: If a square is inscribed in a circle that, in turn, is inscribed in a  [#permalink]

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New post 08 Dec 2010, 19:14
1
I think your math is off when you are trying to find a side of the smaller square.

So Large square side =2. So diameter of circle =2 - therefore 2=hypotenuse of smaller square.

Since the square is two isosceles triangles - hypotenuse=x^2 + x^2 or in other words in our case 2/sqrt2 to get one of the sides.

so perimeter of larger square = 2*4= 8 perimeter of smaller square = 2/sqrt2 * 4 = 8/sqrt2.

when you then take the ratio of larger square perimeter to smaller square it looks like this:

8/(8/sqrt2) or 8 * sqrt2/8 so the 8's cancel out and your are left with sqrt 2 - the answer.

hope it helps.
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Re: If a square is inscribed in a circle that, in turn, is inscribed in a  [#permalink]

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New post 09 Dec 2010, 04:46
1
maive wrote:

L/S = 8/4sqrt2 = 2 ???? It doesn't work and I can't find my mistake..

PLEASE help me to find my mistake!!

Thank you very much for your help!!


You got \(\frac{L}{S} = \frac{8}{4\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}\)

Sometimes, it is the silliest things that get you!
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Re: If a square is inscribed in a circle that, in turn, is inscribed in a  [#permalink]

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New post 09 Dec 2010, 07:14
I got it!!

Thanks a lot for your help, I really appreciate it!!
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Re: If a square is inscribed in a circle that, in turn, is inscribed in a  [#permalink]

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New post 24 Sep 2014, 23:16
1
maive wrote:
If a square is inscribed in a circle that, in turn, is inscribed in a larger square, what is the ratio of the perimeter of the larger square to that of the smaller square?

A. \(\frac{1}{2}\)
B. \(\frac{1}{\sqrt{2}}\)
C. \(\sqrt{2}\)
D. \(2\)
E. \(3.14\)

M17-23

Ok, I understand this explanation from sportyrizwan:

Let "a" be the diameter of the circle
Therefore the side of the bigger square will be a and the perimeter 4a

Also a will be the diagonal of the smaller square
Therefore the side of the smaller square will be a/sqrt2 and perimeter (2a)sqrt2

Therefore the ratio of the bigger square to the smaller square is sqrt2

http://gmatclub.com/forum/square-inscribed-in-a-circle-inscribed-in-a-square-55210.html

***********

However, when I try to plug values, it doesn't work..

Let say larger square
one side x=2, so P=8

Smaller square
Diagonal = hypotenus = 2
So x^2+x^2=2^2
2x^2=4
x^2=2
x=sqrt2
P=4sqrt2

L/S = 8/4sqrt2 = 2 ???? It doesn't work and I can't find my mistake..

PLEASE help me to find my mistake!!

Thank you very much for your help!!


Consider the diagram below:

Attachment:
m17-23.png
m17-23.png [ 8.46 KiB | Viewed 3542 times ]


The side of a large square is \(a\), thus its perimeter is \(4a\);

The side of a small square is \(\sqrt{(\frac{a}{2})^2+(\frac{a}{2})^2}=\frac{a}{\sqrt{2}}\), thus its perimeter is \(\frac{4a}{\sqrt{2}}\);

Hence the ratio is \(\frac{(4a)}{(\frac{4a}{\sqrt{2}})}=\sqrt{2}\).


Answer: C
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Re: If a square is inscribed in a circle that, in turn, is inscribed in a  [#permalink]

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New post 28 Oct 2016, 01:34
Bunuel : Need your help here. Why is the diagonal of the small square = a/sqrt 2?
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Re: If a square is inscribed in a circle that, in turn, is inscribed in a  [#permalink]

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New post 28 Oct 2016, 01:44
Vishvesh88 wrote:
Bunuel : Need your help here. Why is the diagonal of the small square = a/sqrt 2?


The diagonal of small square is a, not \(\frac{a}{\sqrt{2}}\). \(\frac{a}{\sqrt{2}}\) is side of a small square. Check here: if-a-square-is-inscribed-in-a-circle-that-in-turn-is-inscribed-in-a-105993.html#p1419806
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Re: If a square is inscribed in a circle that, in turn, is inscribed in a  [#permalink]

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New post 28 Oct 2016, 02:53
Bunuel : Got it mate! thanks.
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Re: If a square is inscribed in a circle that, in turn, is inscribed in a  [#permalink]

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New post 19 Nov 2016, 03:29
Assume the side of the larger square to be 8, perimeter of larger square = 32.

Diameter of the circle inscribed within a square = Side of the square , therefore radius = 4. Furthermore radius of the circle is half the diagonal length of smaller square, hence 4= √2a/2 , solving for 'a', we get 4√2. Therefore, perimeter of smaller square = 16√2. Finally, 32/16√2 = √2. [C]
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Re: If a square is inscribed in a circle that, in turn, is inscribed in a  [#permalink]

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New post 17 Dec 2018, 08:54
maive wrote:
If a square is inscribed in a circle that, in turn, is inscribed in a larger square, what is the ratio of the perimeter of the larger square to that of the smaller square?

A. \(\frac{1}{2}\)
B. \(\frac{1}{\sqrt{2}}\)
C. \(\sqrt{2}\)
D. \(2\)
E. \(3.14\)


Let the side of small square be \(= 2\)

Therefore the diagonal of small square \(= 2\sqrt{2}\)

Diagonal of small square is \(=\) diameter of the circle

Diameter of the circle is \(=\) side of the large square

Therefore side of large square \(= 2\sqrt{2}\)

Perimeter of small square \(= 4a = 4*2 = 8\)

Perimeter of large square \(= 4a = 4*2\sqrt{2} = 8\sqrt{2}\)

Ratio of perimeter of large square to perimeter of small square \(= \frac{8\sqrt{2}}{8} = \sqrt{2}\)

Answer C
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Re: If a square is inscribed in a circle that, in turn, is inscribed in a &nbs [#permalink] 17 Dec 2018, 08:54
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