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In a circle centered on point O with radius 8, diameter AC and poi

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In a circle centered on point O with radius 8, diameter AC and poi  [#permalink]

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New post 28 Feb 2016, 07:21
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A
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D
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  45% (medium)

Question Stats:

64% (00:58) correct 36% (01:02) wrong based on 353 sessions

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In a circle centered on point O with radius 8, diameter AC and point B on the circle, pictured above, what is the length of line segment (AB)?

(1) ∠AOB = 120°
(2) OB = BC

Attachment:
Geometry_Q16Image.jpg
Geometry_Q16Image.jpg [ 6.62 KiB | Viewed 7575 times ]

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Re: n a circle centered on point O with radius 8, diameter AC and poin  [#permalink]

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New post 10 Jun 2016, 12:05
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AbdurRakib wrote:
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In a circle centered on point O with radius 8, diameter AC and point B on the circle, pictured above, what is the length of line segment AB?


1.m∠AOB= 120°

2.OB=BC



1. m∠AOB= 120°. So, m∠OAB + m∠OBA = 60°. (As sum of all angles in a triangle in 180°). OA = OB = radius, so m∠OAB = m∠OBA = 30°. m∠ABC = 90° and this is a 30-60-90 triangle. As you know one side (diameter or AC), you should be able to calculate the other 2 sides. Sufficient.

2.OB=BC. BOC is an equilateral triangle. So, m∠BCO = 60 = m∠BCA. As ABC is a right triangle, you have a 30-60-90 triangle. As you know one side (diameter or AC), you should be able to calculate the other 2 sides. Sufficient.

Answer is D.
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Re: In a circle centered on point O with radius 8, diameter AC and poi  [#permalink]

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New post 01 Mar 2016, 03:51
1) tells you that the triangle on left is 120 - 30 - 30, and triangle on right is 60-60-60, and big combined triangle is 30-60-90. Can determine side lenghts with this info.
Suff

2) Side length of triangle on right gives you same info as 1) above.
Again, Suff

D.
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Re: In a circle centered on point O with radius 8, diameter AC and poi  [#permalink]

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New post 11 Jun 2016, 00:04
d

1)angle 120..from which u can find out that triangle BOC is equilateral as OB and OC are radii of circle .ABC is right angled triangle Apply pythagoras theorem.

2) Same here also.OB=BC. triangle BOC is equilateral and ABC is right angled .Apply pythagoras theorem.
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Re: In a circle centered on point O with radius 8, diameter AC and poi  [#permalink]

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New post 13 Jun 2016, 23:55
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Bunuel wrote:
Image
In a circle centered on point O with radius 8, diameter AC and point B on the circle, pictured above, what is the length of line segment (AB)?

(1) ∠AOB = 120°
(2) OB = BC

Attachment:
Geometry_Q16Image.jpg


Statement 1
∠AOB = 120°. This means ∠BOC = 60°

Also because OA = OB (radius). Triangle AOB is a isosceles triangle. Therefore ∠OAB = ∠OBA = 30°

because AC is the diameter and angle ABC is right angle, so angle BCO will be 60 degrees.

Now in Triangle BOC, angle BCO = ∠BOC = 60°. Therefore OBC will also be 60°

So the triangle BOC is equilateral. Therefore side BC = side BO = radius of circle.

We know the radius, so we know BO. We know the radius so we know diameter AC.

From Pythagorus theorem (applied on triangle ABC) we can get AB.

Statement 1 is sufficient.

Statement 2

OB = BC

We straight away get value of BC = side BO = radius of circle.

We know the radius, so we know BO. We know the radius so we know diameter AC.

From Pythagorus theorem (applied on triangle ABC) we can get AB.

Statement 2 is also sufficient.

D is the answer.
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In a circle centered on point O with radius 8, diameter AC and poi  [#permalink]

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New post 01 Aug 2017, 16:37
I think this is a little harder than Medium you need to know a lot of properties at once.

1) When you have a diameter with a triangle, the diameter is the hypotenuse. Therefore, the opposite side ∠ABC = 90
2) We also know that OB is a radius because it goes from the center to a point on a circle. Any point of the radius going to the edge of the circle is also a radius. Therefore OB=8 and OC =8
3) From this we know that Triangle OBC is Isosceles with BO and OC being equal meaning that ∠B and angle ∠C share a value

From this good start we see the diagram:
1) if ∠AOB =120 then the other side of that line ∠COB bust be 60 cuz it's a straight line and a line adds to 180. Now if we know that is 60 we know (180-60)/2 is the ∠ of B and ∠C since it is isosceles. Therefore we know now that this triangle is equilateral
2) From this we can subscribe a point of 8 on BC. From this we know it's a 90 degree triangle with one leg of 8² * X²=16² and from this we can solve

Similarly from Prompt 2
Knowing OB=BC
1) We know that OB =Radius =8 therefore BC =8 and we already knew that OC =8 So again equilateral and therefore we can solve.
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Re: In a circle centered on point O with radius 8, diameter AC and poi  [#permalink]

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New post 01 Aug 2017, 20:50
∠ABC = 90°
Per Right Triangle Altitude Theorem:
AB square = AO X AC
AO = OC = 8
AC = AO + OC = 16
AB square = 8 X 16 = 128
Hence, AB could be determined from the Q stem itself.

So, irrespective of what information is given in each Statement, AB will be determined.
Hence, answer is D even before reviewing Statement 1 and Statement 2.

Please suggest whether the aforementioned is correct or not.
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In a circle centered on point O with radius 8, diameter AC and poi  [#permalink]

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New post 13 Nov 2017, 22:48
GoodnessShallPrevail wrote:
∠ABC = 90°
Per Right Triangle Altitude Theorem:
AB square = AO X AC
AO = OC = 8
AC = AO + OC = 16
AB square = 8 X 16 = 128
Hence, AB could be determined from the Q stem itself.

So, irrespective of what information is given in each Statement, AB will be determined.
Hence, answer is D even before reviewing Statement 1 and Statement 2.

Please suggest whether the aforementioned is correct or not.


My friend sorry to break it to you but above highlighted are a couple of incorrect statement in your method.

I am not sure from where did you get this Right Triangle Altitude Theorem because, in actual Altitude Theorem, the line segment BO has to have an angle of 90 Degree with line segment AC. (that is how you can call BO an altitude of the triangle.)

Hence, you are getting a wrong calculation.

Here is how I would approach this :

We are asked to find AB.
Now, Triangle ABC is a right angle triangle.
So, AC^2 = AB^2 + BC^2
i.e AB^2 = AC^2 - BC^2

Now we know AC = 8 (twice the radius OC or OA or BO)
And if we know BC, we can find AB.
Hence, effectively, this question is to find BC.

Moving to the statements,
St-1 : Angle AOB = 120.
So, In triangle AOB, Since AO=OB (both are radius), Angle OAB = Angle OBA
So from, Angle OAB + Angle OBA + Angle AOB = 180
2*(Angle OBA) = 180-120 = 60
So angle OBA = 30
Therefore, Angle OBC = 90-30 = 60.

Hence, in the triangle, OBC, all angles are 60.
Therefore, its an equilateral triangle and all sides have length 8 (same as radius OC).
So, BC = 8.

Hence this statement is sufficient.

Statement-2 :
OB = BC
Pretty straight forward, OB = radius = 8 = BC
Hence, we got BC and so can calculate AB.

So, this statement is sufficient as well.

OPtion D is the correct answer.

P.S : If you were to calculate the value of AB, it would be 8Sqrt3. (which is NOT same as what you were calculating).
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Re: In a circle centered on point O with radius 8, diameter AC and poi  [#permalink]

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New post 04 Dec 2018, 09:31
Bunuel wrote:
Image
In a circle centered on point O with radius 8, diameter AC and point B on the circle, pictured above, what is the length of line segment (AB)?

(1) ∠AOB = 120°
(2) OB = BC

Attachment:
Geometry_Q16Image.jpg


well i see a lot op answers already posted so it might be redundent to post one more,but id like to post my way of solving the above proble.

So we know
1.that all readius are equal
2.Angle made by a triangle with diameter as one of the side will always be 90 degrees(most important point,if you know this then consider problem solved)

Basically DS points 1 and 2 are both saying the same thing...that OC and OB are eqaul and if that is true then we get two sides of a right triangle and to find AB we simply use pythagorus theorm.

Answer is C.
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Re: In a circle centered on point O with radius 8, diameter AC and poi &nbs [#permalink] 04 Dec 2018, 09:31
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