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
BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
Target Test Prep GMAT OnDemand
Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.
- 715+ score guarantee — highest in the industry (99th percentile)
- 52 chapters · 1,500+ lessons · 4,000+ practice questions
- 400 hours of video · 1,500+ instructor-led HD smartboard lessons
- 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
- 24/7 live support + weekly Zoom office hours with GMAT instructors
- TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
- 1,200+ flashcards + AI-powered study assistant & daily calendar
- OnDemand, LiveTeach & GMAT Bootcamp formats available
- Also: GRE, SAT Math & Executive Assessment courses
- MBA Admissions Consulting now available
- 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
- 200,000+ students served
- 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★
5.0
(559 reviews)
130-pt guarantee
$0 to start
then $127/mo
equialteral triangel
Source: Beat The GMAT — Problem Solving |
hi truplayer256,
Could you please tell me how did you decide 240 degrees as angle. .please explain
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
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
The powers of two are bloody impolite!!
Sorry for telling out of turnhi truplayer256,
Could you please tell me how did you decide 240 degrees as angle. .please explain
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
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...
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...
- Attachments
-
Anyone from Hyderabad or Telugu speaking community.
Searching for a serious study partner from Hyderabad or the one who work for same Company.
Searching for a serious study partner from Hyderabad or the one who work for same Company.
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
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
Ananda Chakraborty
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.
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.
Anyone from Hyderabad or Telugu speaking community.
Searching for a serious study partner from Hyderabad or the one who work for same Company.
Searching for a serious study partner from Hyderabad or the one who work for same Company.
















