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 →

GMAT Prep: Geometry Explain

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 120
Joined: Thu Sep 02, 2010 8:56 am
Location: India
Thanked: 7 times
GMAT Score:730
by euro » Sat Oct 30, 2010 3:58 am
What is the greatest possible are of a triangular region with one vertex at the center of a circle of radius 1 and the other two vertices on the circle?

(A) (Sq.Root 3)/ 4

(B) 1/2

(C) Pi/4

(D) 1

(E) (Sq. Root 2)

[spoiler]OA is (B)[/spoiler]
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
Master | Next Rank: 500 Posts
Posts: 286
Joined: Tue Sep 21, 2010 5:36 pmFollowed by: 5 members
Location: Kolkata, India
Thanked: 11 times

by pesfunk » Sat Oct 30, 2010 5:17 am
The triangle of largest area of all those inscribed in a given circle is equilateral;

Since the equilateral triangle has sides of length 1, the OA should be A

Could someone please confirm ??
euro wrote:What is the greatest possible are of a triangular region with one vertex at the center of a circle of radius 1 and the other two vertices on the circle?

(A) (Sq.Root 3)/ 4

(B) 1/2

(C) Pi/4

(D) 1

(E) (Sq. Root 2)

[spoiler]OA is (B)[/spoiler]
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 120
Joined: Thu Sep 02, 2010 8:56 am
Location: India
Thanked: 7 times
GMAT Score:730

by euro » Sat Oct 30, 2010 5:24 am
@pesfunk: The OA is already mentioned. Read the question carefully; it doesn't ask for the area of the largest triangle that can be inscribed in the circle. If you have a different answer, then please share your explanation (how you arrived at the answer). It will be helpful to others also.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 286
Joined: Tue Sep 21, 2010 5:36 pmFollowed by: 5 members
Location: Kolkata, India
Thanked: 11 times

by pesfunk » Sat Oct 30, 2010 5:27 am
"greatest possible are" ??
euro wrote:@pesfunk: The OA is already mentioned. Read the question carefully; it doesn't ask for the area of the largest triangle that can be inscribed in the circle. If you have a different answer, then please share your explanation (how you arrived at the answer). It will be helpful to others also.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members
Location: Bangalore, India
Thanked: 146 times

by shovan85 » Sat Oct 30, 2010 5:32 am
euro wrote:What is the greatest possible are of a triangular region with one vertex at the center of a circle of radius 1 and the other two vertices on the circle?

(A) (Sq.Root 3)/ 4

(B) 1/2

(C) Pi/4

(D) 1

(E) (Sq. Root 2)

[spoiler]OA is (B)[/spoiler]
This is asking for AREA ;)

Keep this as a RULE:
"Isosceles triangle will have MAXIMUM area when the traingle is a Right Angle Triangle".

This is because of Trigonomatry rules as Sine starts from 0(0) and goes to 1(90) and then back to 0 (180). Dont bother. Keep the rule :)

Thus two sides 1 and 1 each. you have height and base. Thus area: 1/2
If the problem is Easy Respect it, if the problem is tough Attack it
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1119
Joined: Fri May 07, 2010 8:50 amFollowed by: 3 members
Thanked: 29 times

by diebeatsthegmat » Mon Nov 01, 2010 9:27 am
shovan85 wrote:
euro wrote:What is the greatest possible are of a triangular region with one vertex at the center of a circle of radius 1 and the other two vertices on the circle?

(A) (Sq.Root 3)/ 4

(B) 1/2

(C) Pi/4

(D) 1

(E) (Sq. Root 2)

[spoiler]OA is (B)[/spoiler]
This is asking for AREA ;)

Keep this as a RULE:
"Isosceles triangle will have MAXIMUM area when the traingle is a Right Angle Triangle".

This is because of Trigonomatry rules as Sine starts from 0(0) and goes to 1(90) and then back to 0 (180). Dont bother. Keep the rule :)

Thus two sides 1 and 1 each. you have height and base. Thus area: 1/2
but the question is about how to find the greatest possible of a triangular region...its not about how to find the greatest area...

i am thinking to solve the problem by the following way... dunno if i am right...
with 3 points, they can make a trangle or a straight line which is demeter, so there will be only 2 possibility... one is triangle, one is a straight line... so B makes sense
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Mon Nov 01, 2010 9:48 am
euro wrote:What is the greatest possible are of a triangular region with one vertex at the center of a circle of radius 1 and the other two vertices on the circle?

(A) (Sq.Root 3)/ 4

(B) 1/2

(C) Pi/4

(D) 1

(E) (Sq. Root 2)

[spoiler]OA is (B)[/spoiler]
If we are given two sides of a triangle, the greatest possible area will be achieved if the two sides form a right angle so that one of the sides is the base and the other side is the height.

For an illustration of why this rule holds true, please check my second post in the following thread:

https://www.beatthegmat.com/kaplan-trian ... 66820.html

In the problem above r=1, so the 2 given sides each have a length of 1.
Thus, the greatest possible area will be achieved if these two sides form a right angle so that b=1 and h=1.
A = 1/2 * 1 * 1 = 1/2.

The correct answer is B.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1