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# Square A is inscribed in circle X. Square B is inscribed in

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Square A is inscribed in circle X. Square B is inscribed in [#permalink]

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06 Jun 2012, 03:25
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Square A is inscribed in circle X. Square B is inscribed in circle Y. The perimeter of square B is twice the perimeter of square A. The area of circle Y is equal to k multiplied by the area of circle X. What is the value of k ?

A. 2
B. 3
C. 4
D. 6
E. 8
[Reveal] Spoiler: OA

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Re: Square A is inscribed in circle X. Square B is inscribed in [#permalink]

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06 Jun 2012, 04:23
macjas wrote:
Square A is inscribed in circle X. Square B is inscribed in circle Y. The perimeter of square B is twice the perimeter of square A. The area of circle Y is equal to k multiplied by the area of circle X. What is the value of k ?

A. 2
B. 3
C. 4
D. 6
E. 8

If Peri(sqrB)=2 x Per(SqrA)
=> side(sqrB)= 2 x side(SqrA)
=> diagonal(sqrB)= 2 x daigonal(SqrA)
=> Radius of circle B= 2 x Radius of circle B [ as diagonal of square is daimeter of circle]
=> Area of Circle B = 4 x area of cicle a [As area =pie x (radius)^2]
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Re: Square A is inscribed in circle X. Square B is inscribed in [#permalink]

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07 Jun 2012, 02:50
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Let a = side of square A;
Let b = side of square B;

Perimeter of B = 4b
Perimeter of A = 4a

Perimeter of B = 2 * Perimeter of A
4b = 2 * 4a;
b = 2a;

From Pythagoras rule, length of a diagonal equals \sqrt{2} * side of the square:
Diagonal of square A = $$\sqrt{2} * a$$
Diagonal of square B = $$\sqrt{2} * b = \sqrt{2} * 2a = \sqrt{8} * a$$

Radius of circle X = Half of Diagonal of square A, since square A is inscribed in circle X = $$\sqrt{2} * a/2;$$
Radius of circle Y = Half of Diagonal of square B, since square B is inscribed in circle Y =$$\sqrt{8} * a/2;$$

Area of circle X = $$Pi * r^2 = Pi * (\sqrt{2} * a/2)^2 = Pi * a/2;$$
Area of circle Y = $$Pi * r^2 = Pi * (\sqrt{8} * a/2)^2 = Pi * 2a;$$

Area of circle Y = $$k *$$Area of circle X
$$Pi * 2a = k * Pi * a/2$$
$$2 = k * 1/2$$
$$k = 4;$$
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Last edited by Squish on 08 Jun 2012, 01:56, edited 1 time in total.

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Re: Square A is inscribed in circle X. Square B is inscribed in [#permalink]

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07 Jun 2012, 20:08
Brotha wrote:
Let a = side of square A;
Let b = side of square B;

Perimeter of B = 2b
Perimeter of A = 2a

Perimeter of B = 2 * Perimeter of A
2b = 2 * 2a;
b = 2a;

From Pythagoras rule, length of a diagonal equals \sqrt{2} * side of the square:
Diagonal of square A = $$\sqrt{2} * a$$
Diagonal of square B = $$\sqrt{2} * b = \sqrt{2} * 2a = \sqrt{8} * a$$

Radius of circle X = Half of Diagonal of square A, since square A is inscribed in circle X = $$\sqrt{2} * a/2;$$
Radius of circle Y = Half of Diagonal of square B, since square B is inscribed in circle Y =$$\sqrt{8} * a/2;$$

Area of circle X = $$Pi * r^2 = Pi * (\sqrt{2} * a/2)^2 = Pi * a/2;$$
Area of circle Y = $$Pi * r^2 = Pi * (\sqrt{8} * a/2)^2 = Pi * 2a;$$

Area of circle Y = $$k *$$Area of circle X
$$Pi * 2a = k * Pi * a/2$$
$$2 = k * 1/2$$
$$k = 4;$$

If one side is a then the perimeter should be 4a, right ?Why u wrote 2a?Can you explain?

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Re: Square A is inscribed in circle X. Square B is inscribed in [#permalink]

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08 Jun 2012, 01:57
Aaack! Sorry - my mistake! Have updated it - it should be 4a.
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Re: Square A is inscribed in circle X. Square B is inscribed in [#permalink]

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08 Jun 2012, 05:58
Circle X
let side of square = 1
radius = $$\sqrt{2}$$
area = $$2*(\sqrt{2})^2$$ = $$4$$

Circle y
let side of square = 2
radius = $$\sqrt{8}$$
area = $$2*(\sqrt{8})^2$$ = $$16$$

area of y / area of x = k
16 / 4 = 4

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Re: Square A is inscribed in circle X. Square B is inscribed in [#permalink]

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08 Jun 2012, 07:23
[quote="pike"]Circle X
let side of square = 1
radius = $$\sqrt{2}$$
area = $$2*(\sqrt{2})^2$$ = $$4$$

Circle y
let side of square = 2
radius = $$\sqrt{8}$$
area = $$2*(\sqrt{8})^2$$ = $$16$$

area of y / area of x = k
16 / 4 = 4[/quote

I want to mention one thing...u said the radius is root 2 for the circle X but actually it's the diagonal of the square not the radius and same for the circle Y..correct me if I am wrong..

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Re: Square A is inscribed in circle X. Square B is inscribed in [#permalink]

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25 Sep 2014, 13:28
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Square A is inscribed in circle X. Square B is inscribed in [#permalink]

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29 Sep 2014, 02:32
pike wrote:

Circle y
let side of square = 2
radius = $$\sqrt{8}$$
area = $$2*(\sqrt{8})^2$$ = $$16$$

area of y / area of x = k
16 / 4 = 4

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Re: Square A is inscribed in circle X. Square B is inscribed in [#permalink]

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29 Sep 2014, 02:34
farukqmul wrote:
I want to mention one thing...u said the radius is root 2 for the circle X but actually it's the diagonal of the square not the radius and same for the circle Y..correct me if I am wrong..

As square is inscribed, "Diagonal Of Square" = "Diameter of Circle"
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Re: Square A is inscribed in circle X. Square B is inscribed in   [#permalink] 29 Sep 2014, 02:34
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