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GMAT Prep - Triangle inscribed in circle question

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by ssy » Sat Oct 27, 2007 6:00 am
In the figure attached, equilateral triangle ABC is inscribed in the circle. If the length of arc ABC=24, what is the approximate diameter of the circle?

5
8
11
15
19

11
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by Suyog » Sat Oct 27, 2007 6:42 am
I assumed the perimeter of the triangle to be 24; which gives
AB = BC = AC = 8

Image attached!

Draw a perpendicular from A so that is bisects B & C at point D and where it touches the circle as point E.

BD = CD = 4
as AB = 8, BD = 4,
Using pythagoras, AD approx is 7

Lets take center of the circle as O,
OD = DE = 3.5

AE = AD + DE
= 7 + 3.5
= 10.5 Approx 11

Please comment.
Ans C.
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Last edited by Suyog on Sat Oct 27, 2007 10:15 am, edited 1 time in total.
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by ri2007 » Sat Oct 27, 2007 6:56 am
Here is what I did - its quite simple - u only need to know the approx value of pi = 3.14

So now u know Arc ABC = 24 so circumference = 24+12 =36

36 = pi*d
so d is approx 11
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by Suyog » Sat Oct 27, 2007 7:27 am
Thanks ri...
you made it look so simple!
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by ssy » Sat Oct 27, 2007 9:09 pm
Thanks a lot Suyong and Ri 2007.

Ri2007 - How do you know that " Arc ABC = 24 so circumference = 24+12 =36"

Is it because the circle is inscribed with an equilateral triangle..and arc ABC "covers" 2 sides out of 3 sides of the triangle...which means Arc ABC =2/3 of circumference?
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by ri2007 » Sun Oct 28, 2007 8:47 am
ssy, u got it.
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by gmatguy16 » Mon Oct 29, 2007 3:38 pm
i agree with ri,not too sure about suyog,
the question says length of arc ,which is lot different than perimeter of triangle..so maybe you got same answer which was a coincidence..guys please let me know i am thinking okay?
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