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Triangle inscribed in circle

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by asamaverick » Wed Jun 16, 2010 5:49 pm
Since the triangle is an equilateral triangle we can infer that the arcs AB, BC & AC are of same length. Using this arc ABC = 2/3 perimeter of triangle = 2(pi)r * 2/3

Given Arc ABC = 24.

This leads us to 24 = 2(pi)r * 2/3
=> r = 18/pi = 18/3.14 = approx (5.7)
Hence diameter = 2r = 11.4 (approx)

C is the closest one to this hence it is the right answer.
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by Testluv » Wed Jun 16, 2010 6:25 pm
Using this arc ABC = 2/3 perimeter of triangle = 2(pi)r * 2/3
This should be: 24 = (2/3)*circumference of circle = (2/3)*2*pi*r

and we want 2r, so:

2r = 36/pi = approx. 11
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by ssuarezo » Wed Jun 16, 2010 7:10 pm
Testluv wrote:
Using this arc ABC = 2/3 perimeter of triangle = 2(pi)r * 2/3
This should be: 24 = (2/3)*circumference of circle = (2/3)*2*pi*r

and we want 2r, so:

2r = 36/pi = approx. 11
Thank u guys,
I understood what Asamaverick wanted to say, but as testLuv noted, it s always good to use the exact terminology.
The answer is more than clear
Thanks.
Silvia
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by selango » Wed Jun 16, 2010 11:30 pm
I solved this with different approach.

Length of arc ABC=Length of arc AB+Length of arc BC

The inscribed angle in center is 120.

L arc AB=120/360*pi*d

L arc BC=120/360*pi*d

2/3*pi*d=24

d=11
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