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Intern  Joined: 18 Mar 2012
Posts: 46
GPA: 3.7
In the figure above, point O is the center of the circle  [#permalink]

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8 00:00

Difficulty:   65% (hard)

Question Stats: 72% (03:28) correct 28% (03:28) wrong based on 202 sessions

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File comment: Image of circle photo.JPG [ 175.07 KiB | Viewed 7377 times ]
In the figure above, point O is the center of the circle and line segment BD is tangent to the to the circle at point C. If BC = 4, OB = 8, and OC = CD, then what is the area of the region whose perimeter is radius OA, arc ACE, and radius OE?

A. 8π
B. 10π
C. 12π
D. 16π
E. 20π

Originally posted by alex1233 on 05 Mar 2013, 11:38.
Last edited by Bunuel on 06 Mar 2013, 02:25, edited 1 time in total.
Edited the question.
Intern  Joined: 18 Feb 2013
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GMAT 1: 710 Q49 V38 Re: In the figure above, point O is the center of the circle  [#permalink]

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4
My question is how did you know where to put point A and point E? Was the image included in the problem or did you draw it?

Angle OCD is 90, and given OC=CD, we know angle COD and CDO =45 degrees.
OB=8, CB=4, and properities of the special 30, 60, 90 triangle. x, 2x, $$x\sqrt{3}$$, we find out the radius is $$4\sqrt{3}$$.
We also know angle COB, which is opposite of x in the triangle = 30 degrees
Radius= $$4\sqrt{3}$$ Area=48pie
Angle AOE =45+30=75 degree

$$\frac{75}{360}$$*48pie=10pie
Intern  Joined: 18 Mar 2012
Posts: 46
GPA: 3.7
Re: In the figure above, point O is the center of the circle  [#permalink]

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DoItRight wrote:
My question is how did you know where to put point A and point E? Was the image included in the problem or did you draw it?

Angle OCD is 90, and given OC=CD, we know angle COD and CDO =45 degrees.
OB=8, CB=4, and properities of the special 30, 60, 90 triangle. x, 2x, $$x\sqrt{3}$$, we find out the radius is $$4\sqrt{3}$$.
We also know angle COB, which is opposite of x in the triangle = 30 degrees
Radius= $$4\sqrt{3}$$ Area=48pie
Angle AOE =45+30=75 degree

$$\frac{75}{360}$$*48pie=10pie

It was part of the question I did not draw it
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: In the figure above, point O is the center of the circle  [#permalink]

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Hi All,

The GMAT is based heavily on patterns, so building up your 'pattern-matching' skills is a valuable part of your training. When complex-looking questions appear, they are almost certainly going to be based on a series of overlapping patterns (since it's difficult to make a question complex by accident), so you should be on the lookout for "little" patterns, then think about how they 'connect' to one another.

Here, the first rule that you need to know is that lines that are TANGENT to a circle always form 90 degree angles. This means that triangles OBC and ODC are both RIGHT triangles.

With triangle OBC, we're given two of the sides: one of the legs is 4 and the hypotenuse is 8. You should be thinking....."what type of right triangle has a hypotenuse that is exactly DOUBLE one of its legs.....?" The pattern is that it's a 30/60/90 right triangle.

Next, with triangle ODC, we're told that the two legs of that right triangle are equal. What type of right triangle has two legs that are EQUAL....? The pattern is that it's an ISOSCELES right triangle, so we're dealing with a 45/45/90 right triangle.

From here, it's just a few more steps to figure out the central angle of the circle and the sector area of that piece of circle.

As you continue to study, remember that you're not expected to do every step of a question 'all at once.' Break prompts into small pieces, look for patterns and do the work on the pad.

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Re: In the figure above, point O is the center of the circle  [#permalink]

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OB = 8 = 2*4
CB = 4
Using pythagoras therom we can deduce OC = 4√3
As we can see this triangle follows 30-60-90 Right angled triangle hence angle COB = 30 degrees.
We know from the question OC = DC hence
Computing OD will be 4√6 which means triangle ODC is a 45-45-90 right triangle hence angle DOC = 45degrees.
Finally, OE=OC=OA = 4√3 as all are the radii of the circle.
The area covering OE,OC,OA arcs EC and CA is to be calculated using the proportion of central angle as follows:
[[(75°)/(360°)]* π*r^(2 ) =>[5/24]* π*(4√3)^2=>10π
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Re: In the figure above, point O is the center of the circle  [#permalink]

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alex1233 wrote:
Attachment:
photo.JPG
In the figure above, point O is the center of the circle and line segment BD is tangent to the to the circle at point C. If BC = 4, OB = 8, and OC = CD, then what is the area of the region whose perimeter is radius OA, arc ACE, and radius OE?

A. 8π
B. 10π
C. 12π
D. 16π
E. 20π

OB= 8, BC = 4

Since DB is tangent on point C, it makes 90 degrees angle at C

So OC = \sqrt{OB^2 - CB^2}= 4\sqrt{3}

Sides are in ratio 1:2:\sqrt{3}, Hence angle COB= 30 degrees

OC=DC; that means ODC is an isosceles triangle with angle DOC= 45 degrees

Angle DOB= 30+45= 75 degrees

Area= 75/360 *π \sqrt{3}^2= 10π

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Re: In the figure above, point O is the center of the circle  [#permalink]

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_________________ Re: In the figure above, point O is the center of the circle   [#permalink] 18 Aug 2018, 11:19
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