BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

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
Start Free 5-Day Trial Read 559 Reviews →

Length of an arc - central angle confusion

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 421
Joined: Sun Apr 17, 2011 4:27 amFollowed by: 2 members
Location: India
Thanked: 6 times
GMAT Score:620
by vinni.k » Sat May 18, 2013 11:21 am
Triangle ABC is inscribed in a circle, such that AC is a diameter of the circle and angle BAC is 45. If the area of triangle ABC is 84.5 square units, what is the length of arc BC ?

[spoiler](13√2π)/4[/spoiler]

I will really appreciate if anyone can explain this question. This one is from MGMAT 4th edition. The explanation says to assume central angle but the drawn diameter doesn't allow for central angle. How can central angle be drawn in this case ?

Thanks & Regards
Vinni
Attachments
Inscribed triangle.jpg
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 423
Joined: Fri Jun 11, 2010 7:59 amFollowed by: 2 members
Location: Seattle, WA
Thanked: 86 times

by srcc25anu » Sat May 18, 2013 3:36 pm
AC is diameter. therefore Triangle CAB is right angled at B and angle B = 90 degrees. angle CAB = 45 therefore angle ACB must also be 45 degrees. So BC = AB = lets say X
Now area of ABC = 1/2 * base * Height or (1/2 * CB * BA) = 84.5
Since AB = BC therefore each of AB and BC = 1/2 * x^2 = 84.5 and x^2 = 169 or x = 13
ABC has sides in the ratio of 1:1:√2
so AC = 13√2 which is the diameter. hence radius = 13√2/2
CAB = 45 degrees so the central angle made by CB at centre O will be 90 degrees.
Arc CB = 2 * π * R * 90 / 360
Arc CB = [2 * π * (13√2 / 2)]/4 = (13√2π)/4
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1