Area of triangle in circle

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khizarj: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Tue Jun 14, 2011 2:35 pm

by khizarj » Fri Jun 17, 2011 7:32 am

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

OA

Spoilers not working for some reason

OA: 1/2

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Source: Beat The GMAT — Problem Solving | Fri Jun 17, 2011 7:32 am

Frankenstein: Legendary Member; Posts: 1448; Joined: Tue May 17, 2011 9:55 amFollowed by: 53 members; Location: India; Thanked: 375 times

by Frankenstein » Fri Jun 17, 2011 7:40 am

Hi,
If A is the angle subtended by the base of the triangle at the center, then area of triangle is (1/2)(r)(r)sin A
r=1, so area is (1/2)sin A
For area to be maximum sin A should be maximum
max(sin A) =1 when A is 90 degrees.
So, (area)max = 1/2

Cheers!

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bajwa2307: Senior | Next Rank: 100 Posts; Posts: 40; Joined: Tue Jan 18, 2011 5:14 amFollowed by: 2 members; Location: Bangalore; Thanked: 3 times

by bajwa2307 » Fri Jun 17, 2011 7:41 am

refer to this link for the solution

https://www.beatthegmat.com/largest-area ... html#50963

Verbal is testing my patience

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Fri Jun 17, 2011 7:43 am

khizarj wrote:3) What is the greatest possible area of a triangular region with one vertex at the center of a circle of radius 1 and the other two vertices on the circle

Easiest way to solve this problem is to use trigonometric formula for the area of the triangle : Area of a triangle with two sides of length a and b is given by (1/2)*ab*sin(x), where x is the angle between side a and b.

Here, area of the triangle = (1/2)*1*1*sin(x) = sin(x)/2

Therefore, when the sine of the angle between the two sides of the triangle originating from the center of the circle will be maximum, the area of the triangle will be maximum. Sine of any angle will be maximum, i.e. 1 when the measure of the angle will be 90 degrees.

Hence, maximum possible area of the triangle = 1*(1/2) = 1/2

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Fri Jun 17, 2011 8:30 am

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)

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 possible area will be achieved when a right angle is placed between them. The result is that one of the sides becomes the base, the other side becomes the height.

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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