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In the figure, vertex Q of square OPQR is on circle with centre O. If

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In the figure, vertex Q of square OPQR is on circle with centre O. If  [#permalink]

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New post 19 Jul 2018, 20:55
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In the figure, vertex Q of square OPQR is on circle with centre O. If the area of the square is 8, what is the area of the circle?


A. \(8 \pi\)

B. \(8 \sqrt{2} \pi\)

C. \(16 \pi\)

D. \(32 \pi\)

E. \(64 \pi\)

Attachment:
figure 11.jpg
figure 11.jpg [ 14.68 KiB | Viewed 448 times ]

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In the figure, vertex Q of square OPQR is on circle with centre O. If  [#permalink]

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New post 19 Jul 2018, 21:08
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vertex Q of square OPQR is on circle with centre O

=> radius of the circle = diagonal of the square

area of square is \(a^2\) = 8

=> a = \(2\sqrt{2}\)

diagonal of the square = \(\sqrt{a^2 + a^2}\)

=> diagonal = \(\sqrt{8 + 8}\)

=> diagonal = \(\sqrt{16}\) = 4

=> radius of the circle = 4

=> Area of the circle = \(4^2 \pi\) = \(16 \pi\)

Hence option C
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Re: In the figure, vertex Q of square OPQR is on circle with centre O. If  [#permalink]

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New post 19 Jul 2018, 22:28
Bunuel wrote:
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In the figure, vertex Q of square OPQR is on circle with centre O. If the area of the square is 8, what is the area of the circle?


A. \(8 \pi\)

B. \(8 \sqrt{2} \pi\)

C. \(16 \pi\)

D. \(32 \pi\)

E. \(64 \pi\)

Attachment:
figure 11.jpg


OA:C
As per figure,Diagonal (\(diagonal_{square}\)) of Square is equal to Radius of circle.
Area of Square = \(\frac{1}{2}*Diagonal_{square}^2=8\)
\(Diagonal_{square}^2=16\);
As \(Diagonal_{square}\) = Radius of circle
\({Radius}^2 =16\)
Area of circle= \(\pi *\) \({Radius}^2=\pi*16\)
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Re: In the figure, vertex Q of square OPQR is on circle with centre O. If  [#permalink]

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New post 19 Jul 2018, 22:32
Bunuel wrote:
Image
In the figure, vertex Q of square OPQR is on circle with centre O. If the area of the square is 8, what is the area of the circle?


A. \(8 \pi\)

B. \(8 \sqrt{2} \pi\)

C. \(16 \pi\)

D. \(32 \pi\)

E. \(64 \pi\)

Attachment:
figure 11.jpg



Note :

Area of the square = 1/2 \((d)^2\)

Given

1/2*\((d)^2\) = 8

d = 4.

Now diagonal has become the radius of the of circle.

Area of the circle = pie * \(r^2\)

= 16 pie.

The best answer is C.
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Re: In the figure, vertex Q of square OPQR is on circle with centre O. If  [#permalink]

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New post 20 Jul 2018, 01:13
+1 for C.

Area of the Square = L^2 = 8
L = 2 root 2

Diagonal of the Square = D
D^2 = (2root2)^2 + (2root2)^2
D^2 = 16
D = 4 = Radius of the circle

Area of the Circle = (Radius)^2*pi = (4^2)pi = 16 pi

Hence, C.
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Re: In the figure, vertex Q of square OPQR is on circle with centre O. If &nbs [#permalink] 20 Jul 2018, 01:13
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