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by tito1545 » Thu Mar 03, 2011 12:01 pm
A step by step approach for the below please ?

Whats is the greatest possible area of a triangular region with one vertex at the centre 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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by Night reader » Thu Mar 03, 2011 12:09 pm
We need to maximize the area of triangle --> max(1/2 *base*height)
first assign that whatever the shape of our triangle the base is the radius :) then continue with defining the height, which must be a line segment perpendicular to the base it could be radius too; and i think we are done 1/2 *1*1=1/2
tito1545 wrote:A step by step approach for the below please ?

Whats is the greatest possible area of a triangular region with one vertex at the centre 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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by tito1545 » Thu Mar 03, 2011 12:20 pm
cheers m8
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by Anurag@Gurome » Thu Mar 03, 2011 6:59 pm
tito1545 wrote:A step by step approach for the below please ?

Whats is the greatest possible area of a triangular region with one vertex at the centre 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)

Solution:
Area of a triangle is ½ * (product of any two sides)*(sine of included angle between these two sides).
Let us take the sides which form the radius and let the included angle between them be x.
So area of triangle is (1/2) * 1 * 1 *sin x = ½*sin x.
Obviously area is maximum if sin x is maximum.
Maximum value of sin x is 1.
Or maximum area is ½*1 = ½.
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by tito1545 » Fri Mar 04, 2011 8:26 am
Cheers Anurag ,

A good approach , i had forgotten the formula mula A=1/2 * Sinx * product of sides.

Looks like you have an engineering background.
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by GMATGuruNY » Fri Mar 04, 2011 8:36 am
tito1545 wrote:A step by step approach for the below please ?

Whats is the greatest possible area of a triangular region with one vertex at the centre 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)

Image

The drawings above show 3 different versions of the triangle.

Leftmost drawing: b=1, h=1.
Middle drawing: b=1, h<1.
Rightmost drawing: b=1, h<1.

Notice that in each triangle b=1, but only in the leftmost triangle does h=1. In the other two triangles, h<1, resulting in a smaller area. The drawings above illustrate the following rule:

Given two sides of triangle, the greatest area will be achieved when a right angle is placed between them. This rule will hold true for any triangle in which we're given 2 sides.

Thus, the leftmost triangle above will yield the greatest area: 1/2 * 1 * 1 = 1/2.

The correct answer is B.
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