Problem Solving

i got it right, educated guess.... but what is the right approach?

There are a total of 240 degrees in arc ABC. We're told that Arc ABC is equal to 24.

pi*d*240/360=24

pi*d*2/3=24

36/pi=d

36/3.14 ~ 11 C

you gotta be careful posting GMAT Prep questions due to copyright issues.

hi truplayer256,


Could you please tell me how did you decide 240 degrees as angle. .please explain

by symmetry,
the three arcs have to be of same length. so if arc ABC is 24, the smaller arc will be 12
2*pi*r=36
2r=36*7/22~ 11

hi truplayer256,


Could you please tell me how did you decide 240 degrees as angle. .please explain

Sorry for telling out of turn
Angle of an equilateral trianagle LB=60
angle subtended at center by the arc of the circle AC at center=120 by properties of circle

Thus angle subtended by ABC at center=360-120=240=4n/3

Rahul

Thanks for explanation

Hi,

Find the circle shown in the attachment,

The problem can be solved in the similar way but with help of no specific formula to find angle at the center...

<A= <B =<C=60.

AO, BO , CO are the angular bisectors and they meet at the center of the circle as per the properties of equilateral triangle.

Hence <OAC= <OCA = 30 ===> <AOC = 120.

Hence angle made by the arc ABC at center is 360-120= 240.

Hence subsitute the angle and find the answer..

Hope it helps...

Hi,
Thanks for the explanation. But would the same theory be valid if it were an isosceles triangle instead of equilateral? Let me explain, here angle B = 60 degree, and the angle at center (angle AOC) is 120 degree. If ABC was an isosceles triangle with angle B = 50 degree (say). Then would the angle at center (angle AOC) would be 100 degree? It would be of great help if you can explain the theorem involved as well.


Thanks & Regards,
Ananda

It cannot be vaild for an isosceles triangle, as I used 2 properties very specific to equilateral triangle to solve this problem.

1) Each Angle is 60.
2) Angular bisectors will meet at Center of circle.

There would be another strategy to solve for isosceles triangle, if I find I will let you know.