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555-605 (Medium)|   Geometry|      
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Asifpirlo
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What is the area of circle O above? Updated on: 25 Aug 2013, 23:20
Originally posted by Asifpirlo on 25 Aug 2013, 17:45.
Last edited by Bunuel on 25 Aug 2013, 23:20, edited 1 time in total.
Edited the question.
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

25% (medium)

Question Stats:

73% (01:27) correct 27% (02:04) wrong based on 107 sessions

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Attachment:
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circle 2.png [ 4.3 KiB | Viewed 5847 times ]
What is the area of circle O above?

(A) 24π
(B) 36π
(C) 48π
(D) 64π
(E) 72π

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a real-different sort of problem
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B
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Bunuel
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What is the area of circle O above? 25 Aug 2013, 23:23
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What is the area of circle O above?

(A) 24π
(B) 36π
(C) 48π
(D) 64π
(E) 72π

Notice that ABCO is a rectangle --> diagonal AC = diagonal OB = 6.

Since OB is the radius of the circle, then the area is \(\pi{r^2}=36\pi\).

Answer: B.
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What is the area of circle O above? 26 Aug 2013, 00:04
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In the Quadilateral(OABC), since Ang(O),Ang(B), and Ang(C) are Right angles, Ang(A) also must be right angle.
(Since sum of interior angles of a Quadilateral = (n-2) * 180 where n = number of sides)


Hence we could say, Quadilateral(OABC) is a Rectangle.

Using the properties of Rectangle, AC = OB = 6

Hence Area of the Circle = Pi* OB^2=
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Pi * 6^2=36Pi
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What is the area of circle O above? 26 Aug 2013, 01:13
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Smallwonder
In the Quadilateral(OABC), since Ang(O),Ang(B), and Ang(C) are Right angles, Ang(A) also must be right angle.
(Since sum of interior angles of a Quadilateral = (n-2) * 180 where n = number of sides)


Hence we could say, Quadilateral(OABC) is a Rectangle.

Using the properties of Rectangle, AC = OB = 6

Hence Area of the Circle = Pi* OB^2=
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Pi * 6^2=36Pi
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it's not locked brother. you can comment here as you did.
Already this post received expert advice. it's not locked...

however , nice solution :) :)
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What is the area of circle O above? 22 Feb 2018, 20:42
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What is the area of circle O above?

(A) \(24\pi\)

(B) \(36\pi\)

(C) \(48\pi\)

(D) \(64\pi\)

(E) \(72\pi\)

Show Spoiler
Attachment:
2018-02-23_0740.png
2018-02-23_0740.png [ 15.69 KiB | Viewed 3617 times ]
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What is the area of circle O above? 22 Feb 2018, 22:31
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AC=OB, OB=6, area = pi*r2 , area = pi*6square , area = 36pi
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What is the area of circle O above? 23 Feb 2018, 02:02
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Bunuel

What is the area of circle O above?

(A) \(24\pi\)

(B) \(36\pi\)

(C) \(48\pi\)

(D) \(64\pi\)

(E) \(72\pi\)

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similar triangles ac=ob= 6

area = pi r^2 = 36 pi

(B) imo
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What is the area of circle O above? 23 Feb 2018, 13:42
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Bunuel

What is the area of circle O above?

(A) \(24\pi\)

(B) \(36\pi\)

(C) \(48\pi\)

(D) \(64\pi\)

(E) \(72\pi\)

Show Spoiler
Attachment:
2018-02-23_0740.png
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Quick solution: OABC is either a rectangle or a square.* Either way, diagonal lengths are equal.

Mentally draw another diagonal, from O to B.
AC = OB = radius = 6

Area of circle = \(\pi r^2 = 36\pi\)

Answer B

*Three right angles = two sets of parallel and perpendicular lines with 4 right angles => the figure is a rectangle or a square. OR:

The figure OABC is a parallelogram with four right angles.
Three right angles are indicated. Their sum = 270°.
Quadrilateral's internal angles sum to 360°, and (360°-270°) = 90° = fourth angle. Four right angles in a quadrilateral create two sets of parallel and perpendicular lines. The figure is a square or a rectangle.
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What is the area of circle O above? 23 Feb 2018, 14:41
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didn't think much
suppose AC is r. then the area is 36pi. C,D, E are out right away.
even if the OC was the r, then OC would be equal to 3sqrt(3), and area would be 27. A is out
B is left
I didn't consider the OB to be equal to AC...i guess I just missed it, and it seemed easier for me to go with my POE method.
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What is the area of circle O above? 12 May 2019, 05:08
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