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B
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Difficulty:
35%
(medium)
Question Stats:
67% (01:19) correct
33%
(01:58)
wrong
based on 105
sessions
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Equilateral triangle is inscribed in a circle. If length of arc ABC is 2pi, what is the radius of the circle?
(A) 1
(B) 3/2
(C) 2
(D) 5/3
(E) 3
(A) 1
(B) 3/2
(C) 2
(D) 5/3
(E) 3
05 Mar 2011, 09:16
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banksy
Arc ABC is \(\frac{2}{3}\) of the circumference (as ABC is equilateral triangle then (arc AB)=(arc BC)=(arc AC), so (arc AB)+(arc BC)=(arc ABC)=2/3 of circumference, so \(arc \ ABC=2\pi\) basically means that the circumference is \(3\pi\) --> \(circumference=2\pi{r}=3\pi\) --> \(r=\frac{3}{2}\).
Answer: B.
Similar question: geometry-circle-triangle-from-mba-com-97393.html
Yep, thanks Bunuel.
But I don't understand why it is not possible to use another method to solve it....
it drives me crazy cos I don't understand what is wrong with it...please look..
I used formula:
Circumference of sector= (angle in front of the sector/ 360) * 2*pi*R
so if we know that ABC equilateral, so all angles=60 and all circumferences in front of A ,B and C are same. so if circumference in front of 2 angles = 2pi, so circumference in front of one angle is pi.
so
pi= 60/360*2pi*R
so pi=1/6*2pi*R
so pi=1/3pi*R
R=3....
i hope it is clear what I wrote here.
so what is wrong with it?
But I don't understand why it is not possible to use another method to solve it....
it drives me crazy cos I don't understand what is wrong with it...please look..
I used formula:
Circumference of sector= (angle in front of the sector/ 360) * 2*pi*R
so if we know that ABC equilateral, so all angles=60 and all circumferences in front of A ,B and C are same. so if circumference in front of 2 angles = 2pi, so circumference in front of one angle is pi.
so
pi= 60/360*2pi*R
so pi=1/6*2pi*R
so pi=1/3pi*R
R=3....
i hope it is clear what I wrote here.
so what is wrong with it?
