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gmat prep - equilateral triangle inscribed in a circle

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gmat prep - equilateral triangle inscribed in a circle

by yankees660 » Tue Apr 21, 2009 9:41 am
"in the figure above, equilaterial triangle ABC is inscribed in the circle. if the length of arc ABC is 24, what is the approximate diameter of the circle"

the drawing is a circle with a regular equilateral triangle inscribed, A to the left, B at the top, C to the right

a - 5
b - 8
c- 11
d- 15
e - 19

OA = C

thanks for the help!
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by maihuna » Tue Apr 21, 2009 9:57 am
this one is a dirty tricky question : ABC here means length of arc AB+BC. The arc is part of chord with equal chord AB BC CA. So total arc length = 36

Perimeter = 2IIr = 36 => 2r= dia = 36/2II = 36/3.14 = appro 11.
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by mike22629 » Tue Apr 21, 2009 10:14 am
Circum. = 36

Circum. = pie(d)

hence:

d= 36/ pie
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by Thouraya » Sat Jun 19, 2010 4:36 am
Hi,

Can someone please explain this in detail? I didn't get it..


THANKS!:)
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by Patrick_GMATFix » Sat Jun 19, 2010 5:22 am
yankees660 wrote:"in the figure above, equilaterial triangle ABC is inscribed in the circle. if the length of arc ABC is 24, what is the approximate diameter of the circle"

the drawing is a circle with a regular equilateral triangle inscribed, A to the left, B at the top, C to the right

a - 5
b - 8
c- 11
d- 15
e - 19

OA = C

thanks for the help!
The solution relies on 4 properties and 1 formula:

1) Each angle in an equilateral is 60 degrees
2) The measure of an arc is always twice the measure of the remote angle that forms the arc.
3) An entire circle is 360 degrees
4) Length of an arc is proportional to its degree measure
5) circumference = 2*pi*diameter.

Since angle B=60, minor arc AC must be 120 (1/3 of the circle), so the remainder of the circle (arc ABC) must be 2/3 of the circle. Since this arc is a length of 24, 2/3 of the circle is 24, so the circumference must be 36.

With the circumference, it's easy to find the diameter. The answer is C

You can see a detailed solution with graphs as well as a video solution at GMATPrep Question 1304. If you struggle with this type of question, use the Drill Engine to generate timed drills and set topic='geometry' and difficulty='400-500 AND 500-600'

Good Luck
-Patrick
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by Thouraya » Sun Jun 20, 2010 1:56 am
Amazing Patrick! Thanks a million for the detailed replies!

Just to double check, in your calculation, you're assuming that the pie is approx. 3.14, right?

Cuz following your steps, I am reaching a point where I have: 4*pie*r=72
so 4*pie= 12.56, ie: 13
then, 72/13=5.sth, so the diameter is approx 11, right?


Thanks again:)
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by Patrick_GMATFix » Sun Jun 20, 2010 3:59 am
You are right Thouraya, though on the test I use '3' to approximate 'pi'. It's just so much faster to work with 3 than it is to work with 3.14.

If I need a more exact answer, I use 22/7 (the exact value of pi) or in some cases 3.14.

-Patrick

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by atg212 » Fri Jul 09, 2010 12:42 pm
Please explain how arc ABC pertains to the arc AB & BC combined. I find it misleading...I thought it was referring to the arc AC as in the arc that refers to angle B?

just trying to make sure I don't make the same preventable mistake again on the actual exam.

Thanks.
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by Patrick_GMATFix » Fri Jul 09, 2010 12:47 pm
Hi atg212,

The arc name corresponds to the trail of letters around the circumference of the circle. So arc ABC is found by trailing from point A through B, to C.

If the question had simply referred to arc AC, there would be no way to know whether it would refer to the arc that goes from A to B to C, or the arc that goes from A to C directly. In such cases, the question will refer to "minor" arc AC (shorter distance) or "major" arc AC (longer distance)

Hope that helps,
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by luiscarlos59 » Sun Apr 17, 2011 11:41 am
Patrick

with no angles did u assumed based on the picture it was a 60,60,60 angle?

how do we know when to assume that or for example why u didnt assumed it was a 45,45,90
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by Patrick_GMATFix » Sun Apr 17, 2011 12:11 pm
Hi Luis Carlos,

Good questions. Assuming is a bad idea. In this case we don't have to because the question stem describes the triangle as "equilateral", so it must be 60-60-60.

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