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GMAT Prep - Geometry Question

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by rickyishere » Thu Feb 10, 2011 2:32 pm
Hey guys,

Can someone explain how to solve this ? :

What is the greatest possible area of a triangular region with one vertex at the center of the circle of radius 1 and other two vertices on the circle ?

a) (sqrt(3))/4
b) 1/2
c) pie/4
d) 1
e) sqrt(2)

Thanks !!
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Source: — Data Sufficiency |
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by GMATGuruNY » Thu Feb 10, 2011 3:15 pm
rickyishere wrote:Hey guys,

Can someone explain how to solve this ? :

What is the greatest possible area of a triangular region with one vertex at the center of the circle of radius 1 and other two vertices on the circle ?

a) (sqrt(3))/4
b) 1/2
c) pie/4
d) 1
e) sqrt(2)

Thanks !!
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 a triangle, the greatest area will be achieved when a right angle is placed between them (as in the leftmost triangle).

Thus, the greatest possible area = 1/2 * 1 * 1 = 1/2.

The correct answer is B.
Last edited by GMATGuruNY on Sun Apr 22, 2012 5:14 am, edited 1 time in total.
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by rickyishere » Thu Feb 10, 2011 3:21 pm
GMATGuruNY wrote:
rickyishere wrote:Hey guys,

Can someone explain how to solve this ? :

What is the greatest possible area of a triangular region with one vertex at the center of the circle of radius 1 and other two vertices on the circle ?

a) (sqrt(3))/4
b) 1/2
c) pie/4
d) 1
e) sqrt(2)

Thanks !!
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 (as in the leftmost triangle).

Thus, the greatest possible area = 1/2 * 1 * 1 = 1/2.

The correct answer is B.
Thanks for replying, what you bolded above is that a rule which will always holds true ?
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by GMATGuruNY » Thu Feb 10, 2011 6:48 pm
rickyishere wrote:
GMATGuruNY wrote:
rickyishere wrote:Hey guys,

Can someone explain how to solve this ? :

What is the greatest possible area of a triangular region with one vertex at the center of the circle of radius 1 and other two vertices on the circle ?

a) (sqrt(3))/4
b) 1/2
c) pie/4
d) 1
e) sqrt(2)

Thanks !!
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 (as in the leftmost triangle).

Thus, the greatest possible area = 1/2 * 1 * 1 = 1/2.

The correct answer is B.
Thanks for replying, what you bolded above is that a rule which will always holds true ?
Yes, the rule applies to any triangle: given two sides, we'll achieve the greatest area by placing a right angle between the two sides so that one side becomes the base and the other side becomes the corresponding height.

Thus:
Given a side of 6 and a side of 8, the greatest possible area = 1/2*6*8 = 24.
Given a side of 10 and a side of 20, the greatest possible area = 1/2*10*20 = 100.
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As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

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by banibhusan » Sat Feb 12, 2011 10:13 am
It's a straight forward theory that you can apply. An area of a triangle is 1/2*Base*Height. So area will be maximum when its a right angle. If you try to increase or decrease the angle with the base fixed, you will see a decrease in the height, which will reduce the area of the triangle.
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by alex.gellatly » Sat Apr 21, 2012 6:53 pm
GMATGuruNY wrote:
rickyishere wrote:
GMATGuruNY wrote:
rickyishere wrote:Hey guys,

Can someone explain how to solve this ? :

What is the greatest possible area of a triangular region with one vertex at the center of the circle of radius 1 and other two vertices on the circle ?

a) (sqrt(3))/4
b) 1/2
c) pie/4
d) 1
e) sqrt(2)

Thanks !!
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 (as in the leftmost triangle).

Thus, the greatest possible area = 1/2 * 1 * 1 = 1/2.

The correct answer is B.
Thanks for replying, what you bolded above is that a rule which will always holds true ?
Yes, the rule applies to any triangle: given two sides, we'll achieve the greatest area by placing a right angle between the two sides so that one side becomes the base and the other side becomes the corresponding height.

Thus:
Given a side of 6 and a side of 8, the greatest possible area = 1/2*6*8 = 24.
Given a side of 10 and a side of 20, the greatest possible area = 1/2*10*20 = 100.
This rule holds true for all polygons right?
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by [email protected] » Sun Apr 22, 2012 5:10 am
Hey GMATguru that was a wonderful explanation given by you... I genuinely missed out on this question...

The area of the triangle is 1/2 X b X h.

From that perspective, the area of 1/2 is the highest. Hence the correct answer is B.
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by simone88 » Sun Apr 22, 2012 8:06 am
I think that the solution of GMATGuru is very smart. I want to post another solution so that if you don't have that idea you can solve this problem anyway.
In the image below you can see that the area of the triangle is (2x)*y/2=x*y. y=sqrt(1-x^2) so you have to maximize x*sqrt(1-x^2). By differentiating you get x=1/sqrt(2) and the maximum 1/2.
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