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GMAT Prep: Geometry Explain

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GMAT Prep: Geometry Explain

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]
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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]
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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.
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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.
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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
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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
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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.
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