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Largest possible area of triangle inside circle

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Hello,

Can you please help with this:

What is the largest possible area of a triangle inside a circle of radius 1 if one
vertex of the triangle is on the center of the circle and the other two vertices are
on the circumference?

A) 1/3
B) 1/2
C) 1
D) pi/2
E) pi

OA: B


I came up with the following triangle. To find the largest possible area is it necessary that the triangle should be equilateral? However, I am still getting the wrong answer. Can you please assist?

Thanks,
Sri
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by GMATGuruNY » Sun Mar 16, 2014 4:50 am
gmattesttaker2 wrote:Hello,

Can you please help with this:

What is the largest possible area of a triangle inside a circle of radius 1 if one
vertex of the triangle is on the center of the circle and the other two vertices are
on the circumference?

A) 1/3
B) 1/2
C) 1
D) pi/2
E) pi

OA: B
To MAXIMIZE the area of a triangle when given 2 sides:
Place a RIGHT ANGLE between the two given sides so that one of the sides becomes the BASE, while the other side becomes the HEIGHT.
In the problem above, since the radius of the circle is 1, two sides of the inscribed triangle must each have a length of 1.

Case 1: The two given sides form a right angle
Image
Here:
AC=b=1
AB=h=1
A = (1/2)bh = (1/2)(1)(1) = 1/2.

Cases 2 and 3: The two given sides do NOT form a right angle
Image
Image
In Cases 2 and 3:
AC=b=1.
Since AB=1, h<1.
Since b=1 and h<1, the area is LESS than 1/2.

Thus, Case 1 -- where a right angle is placed between the two given sides -- offers the maximum possible aren: 1/2.

The correct answer is B.
Mitch Hunt
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by gmattesttaker2 » Mon Mar 17, 2014 8:40 pm
GMATGuruNY wrote:
gmattesttaker2 wrote:Hello,

Can you please help with this:

What is the largest possible area of a triangle inside a circle of radius 1 if one
vertex of the triangle is on the center of the circle and the other two vertices are
on the circumference?

A) 1/3
B) 1/2
C) 1
D) pi/2
E) pi

OA: B
To MAXIMIZE the area of a triangle when given 2 sides:
Place a RIGHT ANGLE between the two given sides so that one of the sides becomes the BASE, while the other side becomes the HEIGHT.
In the problem above, since the radius of the circle is 1, two sides of the inscribed triangle must each have a length of 1.

Case 1: The two given sides form a right angle
Image
Here:
AC=b=1
AB=h=1
A = (1/2)bh = (1/2)(1)(1) = 1/2.

Cases 2 and 3: The two given sides do NOT form a right angle
Image
Image
In Cases 2 and 3:
AC=b=1.
Since AB=1, h<1.
Since b=1 and h<1, the area is LESS than 1/2.

Thus, Case 1 -- where a right angle is placed between the two given sides -- offers the maximum possible aren: 1/2.

The correct answer is B.
Hello Mitch,

Thanks a lot for the explanation. I was just wondering how we know for Case 2 and 3 that h is less than 1 i.e. the following:
Since AB=1, h<1.


Is this a property of the triangle? Thanks for your help.

Best Regards,
Sri

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by Matt@VeritasPrep » Mon Mar 17, 2014 11:28 pm
Sri, the easiest way to see this is to take a triangle with legs of 1 and watch what happens to the height. Your base will always be 1, but the height will begin to shrink: you can see this in GMATGuruNY's diagrams, where the height of the big triangle is also the leg of a smaller right triangle that has a hypotenuse of 1. (For instance, in the second diagram, the triangle creating by joining A, B, and the point at which the height meets AC.) Since the legs of a right triangle are always less than the hypotenuse, the height must therefore be less than 1.

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by GMATGuruNY » Tue Mar 18, 2014 5:27 am
gmattesttaker2 wrote:.
Hello Mitch,

Thanks a lot for the explanation. I was just wondering how we know for Case 2 and 3 that h is less than 1 i.e. the following:
Since AB=1, h<1.


Is this a property of the triangle? Thanks for your help.

Best Regards,
Sri
The SHORTEST distance between a point and a line is the PERPENDICULAR distance, as shown below:
Image

Cases 2 and 3:
Image
Image

In each case, h = the perpendicular distance between point B and line AC.
Thus, h = the shortest distance between point B and line AC.
In each case, AB is a NON-PERPENDICULAR distance between point B and line AC.
Thus, AB is NOT the shortest distance between point B and line AC.
Since h = the shortest distance, and AB is not the shortest distance, h < AB.
Since h < AB and AB=1, we get:
h < 1.
Mitch Hunt
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GMATGuruNY@gmail.com

If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon.

Available for tutoring in NYC and long-distance.
For more information, please email me at GMATGuruNY@gmail.com.
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by gmattesttaker2 » Wed Mar 19, 2014 7:22 pm
GMATGuruNY wrote:
gmattesttaker2 wrote:.
Hello Mitch,

Thanks a lot for the explanation. I was just wondering how we know for Case 2 and 3 that h is less than 1 i.e. the following:
Since AB=1, h<1.


Is this a property of the triangle? Thanks for your help.

Best Regards,
Sri
The SHORTEST distance between a point and a line is the PERPENDICULAR distance, as shown below:
Image

Cases 2 and 3:
Image
Image

In each case, h = the perpendicular distance between point B and line AC.
Thus, h = the shortest distance between point B and line AC.
In each case, AB is a NON-PERPENDICULAR distance between point B and line AC.
Thus, AB is NOT the shortest distance between point B and line AC.
Since h = the shortest distance, and AB is not the shortest distance, h < AB.
Since h < AB and AB=1, we get:
h < 1.
Great. Thanks a lot Mitch for the excellent explanation with the diagrams.

Best Regards,
Sri