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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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19 Jul 2018, 20:55
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Difficulty:

15% (low)

Question Stats:

84% (00:48) correct 16% (01:21) wrong based on 31 sessions

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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 [ 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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19 Jul 2018, 21:08
1
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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19 Jul 2018, 22:28
Bunuel wrote:

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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19 Jul 2018, 22:32
Bunuel wrote:

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.

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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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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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