The circle with center O has a circumference of 6π

Question

https://gmatclub.com/forum/download/file.php?id=26991 The circle with center O has a circumference of 6\pi{\sqrt{3}} . If AC is a diameter of the circle, what is the length of line segment BC? (A) \frac{3}{\sqrt{2}} (B) 3 (C) 3\sqrt{3} (D) 9 (E) 9\sqrt{3} Kudos for a correct solution . 2015-06-16_1428.png

GMATinsight · Answer

Circumference of Circle =  6\pi{\sqrt{3}}  =  2\pi{r}  where r is the radius of circle

i.e.  r = 3{\sqrt{3}}  = AC/2

Since, Side AC of Triangle is equal to the Diameter therefore Angle ABC must be 90 degree

CONCEPT: Angle drawn in a semicircle on the circumference is always 90 degrees

i.e. Triangle ABC is 30-60-90 Traingle with ratio of the sides  x:x{\sqrt{3}}:2x

and AC =  2x = 2r = 6{\sqrt{3}}

therefore, BC =  x{\sqrt{3}} = 3{\sqrt{3}}*{\sqrt{3}}

i.e. BC = 9

Answer: Option  D

ritigl · Answer

Circumference of a circle = 2 \pi r = 6 \pi\sqrt{3} 
This means r(radius)= 3 \sqrt{3} 
AC= diameter= 2r = 6 \sqrt{3}

In triangle ABC, 
cos 30 =  \frac{BC}{AC} 
 \sqrt{3} /2 = BC/6 \sqrt{3} 
BC= 9

Option D

BuggerinOn · Answer

This is a 30-60-90 triangle and we are told that the circumference is  6\pi{\sqrt{3}} . Set  6\pi{\sqrt{3}}  =  2(\pi)r  giving you r =  3{\sqrt{3}}

You know the diameter, line AC equal 2r or  6{\sqrt{3}}

If you remember your 30-60-90 angles, you can calculate the 60 degree angle,  3{\sqrt{3}} * {\sqrt{3}}  BC = 9

balamoon · Answer

The circle with center O has a circumference of  6\pi{\sqrt{3}} . If AC is a diameter of the circle, what is the length of line segment BC?

Solution -

Angle B is always 90 degrees. So C=30 and A=60.

Triangle ABC is 30-60-90 Triangle with ratio of the sides x:x√3:2x.

2 \pi r= 6\pi{\sqrt{3}}  ->Diameter(2r) = AC = 2x = 6√3 ->x=3√3.

BC = x√3 = 9. ANS D.

Thanks,

Kudos Please.

reto · Answer

Inscribed triangle ABC is a 30-60-90 triangle. Angle ABC = 90 because the opposite line is the diameter of the Circle. Angle CAB is 60 to add up to a Total of 180 Degrees.

AC = Diameter of the Circle =  6\sqrt{3}  which equals 2x in the special right triangle. Solve for x by dividing by 2:  3\sqrt{3}  = x. Finally BC =  x\sqrt{3}  =  3\sqrt{3}*\sqrt{3} =9

Answer Choice D

Bunuel · Answer

MANHATTAN GMAT OFFICIAL SOLUTION: Some intuitive recollection of geometry rules and a picture drawn to scale can help us determine reasonable answer choices. If AC is a diameter of the circle, then triangle ABC is a right triangle, with angle ABC = 90 degrees. The shortest side of a triangle is across from its smallest angle, and the longest side of a triangle is across from its largest angle. Therefore, AC > BC > AB. The circumference of the circle = \pi{d}=6\pi{\sqrt{3}} , so d=6\sqrt{3}\approx{6*1.7}=10.2 Thus, AC ≈ 10.2 and BC < 10.2. But we can clearly see from our picture drawn to scale that BC is longer than half the diameter, so we conservatively determine that BC > 5.1. https://gmatclub.com/forum/download/file.php?id=27038 2015-06-22_1748.png

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

The circle with center O has a circumference of 6π

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

The circle with center O has a circumference of 6π

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards