Problem Solving

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

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

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


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.

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

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


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

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.

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:


Cases 2 and 3:



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.

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:


Cases 2 and 3:



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