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GMATprep - triangle inside circle

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jayhawk2001: Community Manager; Posts: 789; Joined: Sun Jan 28, 2007 3:51 pmFollowed by: 1 member; Location: Silicon valley, California; Thanked: 30 times

by jayhawk2001 » Sat Jun 16, 2007 2:34 pm

OA after a few reply. Please provide explanations.

What is the greatest possible area 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. sqrt(3) / 4
B. 1/2
C. pi / 4
D. 1
E sqrt(2)

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Source: Beat The GMAT — Problem Solving | Sat Jun 16, 2007 2:34 pm

AMalik: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Fri Dec 15, 2006 4:25 am

by AMalik » Sat Jun 16, 2007 10:40 pm

Hi,
For greatest possible area, triangle should be an isosceles right angled triangle.
With one vertex at centre of the circle n other two on the circle.

Area will be=1/2*B*H

Base and Height are 1 (radius of the circle)

So, area should be 1/2*1*1 = 1/2

So, it should be B

Amit

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zozo123: Senior | Next Rank: 100 Posts; Posts: 56; Joined: Tue Feb 13, 2007 6:19 am

by zozo123 » Sun Jun 17, 2007 4:18 am

Hi,
For greatest possible area, triangle should be an equilateral triangle!!

its area is 1/2*B*H with B = 1 and H = sqrt(3)/2

The result is : A = sqrt(3)/4

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jayhawk2001: Community Manager; Posts: 789; Joined: Sun Jan 28, 2007 3:51 pmFollowed by: 1 member; Location: Silicon valley, California; Thanked: 30 times

by jayhawk2001 » Sun Jun 17, 2007 7:56 am

Good call. OA is B.

An isosceles right angled triangle is the isosceles triangle with the
greatest area.

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simplythebest: Junior | Next Rank: 30 Posts; Posts: 22; Joined: Mon Sep 18, 2006 10:17 am

by simplythebest » Mon Jun 18, 2007 2:10 am

I would Still go with Option A.

An Equilateral triangle will have Maximum area.

Jay, I hope we can draw an Equi triangle with the above mentioned constraints in the Problem.

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f2001290: Master | Next Rank: 500 Posts; Posts: 400; Joined: Sat Mar 10, 2007 4:04 amFollowed by: 1 member; Thanked: 1 times

by f2001290 » Mon Jun 18, 2007 7:24 am

I got a formula from Trigonometry to solve this:

Area of a Triangle = (1/2)ab sin(x)

where a,b are the sides of the triangle and x is the included angle between these two sides.

We have maximum area when x is 90, because sin(90) is 1.
whereas sin(60) is sqrt(3)/2

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wongee: Senior | Next Rank: 100 Posts; Posts: 76; Joined: Sat Jul 28, 2007 1:55 pm; Thanked: 1 times

by wongee » Mon Nov 19, 2007 12:16 pm

How can one assume that the angle formed at the center of the circle is 90 degrees?

Thanks!

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