Triangle inscribed in circle

[ssuarezo]

Please help with this.
My exam is on Friday

Silvia

[asamaverick]

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.

[Testluv]

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

[ssuarezo]

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

[selango]

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