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Equilateral triangle inscribed in a circle

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Source: — Problem Solving |

by theCodeToGMAT » Sun Dec 15, 2013 9:58 pm
[spoiler]{C}[/spoiler]
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by Stuart@KaplanGMAT » Sun Dec 15, 2013 11:43 pm
josh80 wrote:In the figure above (absent here), equilateral triangle ABC is inscribed in a circle. If the length of the arc ABC is 24, what is approximate diameter of the circle?

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Hi!

There's an excellent chance that on test day you'll see a question that involves the circle sector formula - that's exactly what we have here.

Here's the formula:

length of arc/circumference = angle of arc/360 = area of sector/area of circle

In other words, every aspect of the sector has the same ratio to the relevant aspect of the circle.

In this question, we're given arc length ABC. Since we have an equilateral triangle inscribed in the circle, the 3 vertices divides the circumference into 3 parts. ABC covers 2/3 of the circumference, so we know that 24 = 2/3 the circumference.

Now we can solve the formula:

24 = (2/3)pi*diameter
(3/2)24 = pi*diameter
36=pi*diameter
36/pi = diameter

Since pi is a bit more than 3, 36/pi will be a bit less than 12. Choose (C)!
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Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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