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

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01 Mar 2016, 04:51

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

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

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10 Jun 2016, 13:05

2

This post was BOOKMARKED

AbdurRakib wrote:

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.

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

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11 Jun 2016, 01:04

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.

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

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01 Aug 2017, 17:37

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.

Attachments

gmat circle problem.png [ 217.29 KiB | Viewed 438 times ]

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

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01 Aug 2017, 21:50

∠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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