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In the figure below, a square is inscribed in a circle. If the area of

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bumblebee14
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In the figure below, a square is inscribed in a circle. If the area of [#permalink] New post   Updated on: 14 Jul 2019, 21:26
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In the figure below, a square is inscribed in a circle. If the area of the square region is 16, what is the area of the circular region?


(A) 2π
(B) 4π
(C) 8π
(D) 12π
(E) 16π

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Attachment:
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square-in-circle.png [ 19.04 KiB | Viewed 62474 times ]

Answer:

Area of a square = (Diagonal)^2 /2

16 = (Diagonal)^2 /2

32 = (Diagonal)^2
Diagonal = 4 (2)^(1/2)
Diagonal = Diameter in the given figure
Radius = Diameter/2
Radius = 2(2)^(1/2)
Area of a circle = πr^2
= π [2(2)^(1/2)]^2
= 8π

Answer is C
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C

Originally posted by bumblebee14 on 01 Mar 2015, 08:31.
Last edited by Bunuel on 14 Jul 2019, 21:26, edited 2 times in total.
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Re: In the figure below, a square is inscribed in a circle. If the area of [#permalink] New post   07 Jul 2019, 08:59
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bumblebee14 wrote:
In the figure below, a square is inscribed in a circle. If the area of the square region is 16, what is the area of the circular region?


(A) 2π
(B) 4π
(C) 8π
(D) 12π
(E) 16π



Let side of Square be \(= a\)

Radius of Circle \(= r\)

Area of square \(= a^2= 16\)

Side of square \(= a = 4\)

Diagonal of square \(= 4\sqrt{2}\)

Diagonal of square \(=\) Diameter of circle

\(4\sqrt{2} = 2r\)

\(r = \frac{4\sqrt{2}}{2} = 2\sqrt{2}\)

Area of circle \(= \pi r^2 = \pi (2\sqrt{2})^2 = 8\pi\)

Answer C
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Re: In the figure below, a square is inscribed in a circle. If the area of [#permalink] New post   01 Mar 2015, 16:02
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Hi bumblebee14,

It's great that you posted this question. However, by posting your explanation (and the correct answer) immediately under it, it's difficult for any of the other users to avoid looking at your answer. I'd suggest that you post all of that as a 'spoiler', so that others can attempt this question without any part of the explanation staring them right in the face.

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Re: In the figure below, a square is inscribed in a circle. If the area of [#permalink] New post   01 Mar 2015, 23:39
Also, the question says: "In the figure above"; can you please post the figure, so that readers have clarity on this.
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Re: In the figure below, a square is inscribed in a circle. If the area of [#permalink] New post   04 Mar 2018, 15:27
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Hi All,

In these types of multi-shape questions, it's usually a must to focus on the radius of the circle. Do you know it's value? Can you figure it out? How does the radius "interact" with the other shape(s)?

Here, we know that the area of the square is 16, so each side of the square = 4.

The diagonal of the square = the diameter of the circle. So….

4(root2) = diameter
2(root2) = radius

Next, plug the radius into the formula for area of a circle: pi(radius)^2

pi(2root2)^2 = 8pi

Final Answer:
Show Spoiler
C


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Re: In the figure below, a square is inscribed in a circle. If the area of [#permalink] New post   15 Feb 2019, 10:13
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I still dont understand the explanation : ((
Would someone be able to help pls?
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Re: In the figure below, a square is inscribed in a circle. If the area of [#permalink] New post   19 Feb 2019, 16:26
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hsn81960 wrote:
I still dont understand the explanation : ((
Would someone be able to help pls?


Isosceles is 1:1:root2

so diagonal(and diameter) is 4(root2)

R = 1/2 diameter means diameter is 2root2
Plug into pi r^2

First find r^2 = 2^2(root2)^2 = 2*4
2*4Pi = 8pi
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Re: In the figure below, a square is inscribed in a circle. If the area of [#permalink] New post   07 Jul 2019, 03:24
Can someone please explain why you Pythagoras theorem wouldn't work here?

4^2 + 4^2 = 32
root(32) = diameter
Root(16)=radius

pi[root(16)]^2 = Pi16.

Please help?
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Re: In the figure below, a square is inscribed in a circle. If the area of [#permalink] New post   27 Jul 2019, 14:35
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I made the same error as you did. Pythagoras does work here, but 16π is a wrong from calculation error.

\(4^2 + 4^2 = c^2\)
\(16 + 16 = c^2\) (--> THIS DOES NOT EQUATE TO C = 16!)
\(32 = c^2\)
\(c = 4 \sqrt{2}\)

c = the diameter (d) of this circle. Need the radius which is \(\frac{d}{2}\).
\(r = \frac{4 \sqrt{2}}{2}\)
\(r = 2 \sqrt{2}\)

Now plug in our radius amount into the circle area formula \(A = π r^2\)
\(A = π (2 \sqrt{2})^2\)
\(= π (4 * 2)\)
= 8π

hanutsingh wrote:
Can someone please explain why you Pythagoras theorem wouldn't work here?

4^2 + 4^2 = 32
root(32) = diameter
Root(16)=radius

pi[root(16)]^2 = Pi16.

Please help?
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In the figure below, a square is inscribed in a circle. If the area of [#permalink] New post   18 Aug 2019, 19:55
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radius of circle = 4 root 2 / 2
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Re: In the figure below, a square is inscribed in a circle. If the area of [#permalink] New post   14 Sep 2020, 08:45
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Radius of the circle is \(\frac{1}{2} * diameter\)

Here the diagonal of the square = \(a\sqrt{2} = 4\sqrt{2}\), is the diameter of the circle.

Radius of the circle is \(\frac{1}{2} * 4\sqrt{2} = 2\sqrt{2}\)

Area of the circle = \(\pi r^2 = \pi*(2\sqrt{2})^2 = 8\pi\)



Option C

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Re: In the figure below, a square is inscribed in a circle. If the area of [#permalink] New post   10 Dec 2020, 01:19
Given,
Area of square =16
So,
S^2 = 16
S = √16
=4

As we know,
Diameter of Circle = Diagonal of Square

So,
2r = 4S
2r = 4*4 [S=4]
2r = 8
r = 8÷2
= 4

So, area of square = πr^2
= π×4^2
= π×8
= π8

Ans:8π
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Re: In the figure below, a square is inscribed in a circle. If the area of [#permalink] New post   03 Feb 2023, 11:36
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Re: In the figure below, a square is inscribed in a circle. If the area of [#permalink]
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