My answer is B.
Since the radius is 1. We have a triangle with two sides equal to 1. To get the maximum area the triangle should be a right triangle should be a right triangle.
Therefore the angle at the center should be 90 degree. Which gives us the base and height of the triangle with 1 units each and therefore the area is 1/2*1*1 --> 1/2.
largest area of of a triangle GMAT prep prob?
Source: Beat The GMAT — Problem Solving |
-
subha_sri8
- Senior | Next Rank: 100 Posts
- Posts: 34
- Joined: Tue Jan 08, 2008 11:13 am
- Thanked: 2 times
Good one Subha_sri8! 
Didnt know that to get maximum area the triangle must be a rt triangle. Makes this problem simple.
Another approach:
Area of a triangle = 1/2 b *h
If the area = 1 then either b or h would have to be 2. H cannot be 2 since all three vertices lie on the circle. If b were 2, then the diameter would be the base and that is not possible for a triangle.
So eliminate 1 and all choices with values greater than 1 i.e. root2.
Eliminate Pi/4 since we are not concerned with the arc of the circle for a triangle inscribed in a circle. 1/2 *b*h for line segments cannot contain Pi.
You are left with 1/2 and root3/4. 1/2 is the greater value. So try to analyse it first. 1/2*b*h can be 1/2 only if b=h. This is possible if the central angle is 90. Hence 1/2 is the answer.
Could you please let us know if the answer is correct? Thanks.
Didnt know that to get maximum area the triangle must be a rt triangle. Makes this problem simple.Another approach:
Area of a triangle = 1/2 b *h
If the area = 1 then either b or h would have to be 2. H cannot be 2 since all three vertices lie on the circle. If b were 2, then the diameter would be the base and that is not possible for a triangle.
So eliminate 1 and all choices with values greater than 1 i.e. root2.
Eliminate Pi/4 since we are not concerned with the arc of the circle for a triangle inscribed in a circle. 1/2 *b*h for line segments cannot contain Pi.
You are left with 1/2 and root3/4. 1/2 is the greater value. So try to analyse it first. 1/2*b*h can be 1/2 only if b=h. This is possible if the central angle is 90. Hence 1/2 is the answer.
Could you please let us know if the answer is correct? Thanks.
Clarification for the part:
H cannot be 2 since all three vertices lie on the circle. If b were 2, then the diameter would be the base and that is not possible for a triangle.
For a triangle with the center of a circle as one vertex, the diameter cannot be the base of the triangle (since all three vertices cannot lie in a straight line).
H cannot be 2 since all three vertices lie on the circle. If b were 2, then the diameter would be the base and that is not possible for a triangle.
For a triangle with the center of a circle as one vertex, the diameter cannot be the base of the triangle (since all three vertices cannot lie in a straight line).
-
durgesh79
- Master | Next Rank: 500 Posts
- Posts: 167
- Joined: Tue Apr 22, 2008 12:48 am
- Thanked: 15 times
a little extra knowledge of trigonometry will help here.
Area of a any trangle is 1/2 * side1 * side2 * sin(angle between side1 and side2)
Now in this case, since the two other points are on the circle.
side 1 = side 2 = radius of the circle = 1
Area = 1/2*1*1*sinX
the highest value of SinX is 1 when X = 90 degree.
Area = 1/2
Area of a any trangle is 1/2 * side1 * side2 * sin(angle between side1 and side2)
Now in this case, since the two other points are on the circle.
side 1 = side 2 = radius of the circle = 1
Area = 1/2*1*1*sinX
the highest value of SinX is 1 when X = 90 degree.
Area = 1/2
GMAT/MBA Expert
-
Ian Stewart
- GMAT Instructor
- Posts: 2623
- Joined: Mon Jun 02, 2008 3:17 am
- Location: Montreal
- Thanked: 1090 times
- Followed by:355 members
- GMAT Score:780
Imagine the circle is in the co-ordinate plane, centre O at (0,0). You might as well let one of the points A be at (1,0) (you can rotate the circle to get it there if you need to). Consider OA to be the base of our triangle: b=1.
Now, if (c,d) is the third point in the triangle, then the height will be |d|. To get the largest area we need the largest height, and that clearly happens when (c,d) is (0,1) or (0.-1). So the maximum area is 1*1/2 = 1/2.
Now, if (c,d) is the third point in the triangle, then the height will be |d|. To get the largest area we need the largest height, and that clearly happens when (c,d) is (0,1) or (0.-1). So the maximum area is 1*1/2 = 1/2.
-
beeparoo
- Master | Next Rank: 500 Posts
- Posts: 200
- Joined: Sun Jun 17, 2007 10:46 am
- Location: Canada
- Thanked: 9 times
That was brilliant! I am a visual learner so this trick helped me by leaps and bounds. Thanks a lot!Ian Stewart wrote:Imagine the circle is in the co-ordinate plane, centre O at (0,0). You might as well let one of the points A be at (1,0) (you can rotate the circle to get it there if you need to). Consider OA to be the base of our triangle: b=1.
Now, if (c,d) is the third point in the triangle, then the height will be |d|. To get the largest area we need the largest height, and that clearly happens when (c,d) is (0,1) or (0.-1). So the maximum area is 1*1/2 = 1/2.
Cheers,
Sandra
-
gaurav_846
- Newbie | Next Rank: 10 Posts
- Posts: 1
- Joined: Wed Jun 04, 2008 8:09 pm
- Location: US
I think for the maximum area of any triangle , the triangle should be equilateral. Therefore the area of a equilateral traingle is (3)^1/2 / 4 .Also the values of sides = 1.vishubn wrote:Can anyone figure out wat is the approach plzzzz
cheers
Live Life!!
-
onlyGmat2008
- Newbie | Next Rank: 10 Posts
- Posts: 9
- Joined: Wed May 21, 2008 9:53 am
-
AleksandrM
- Legendary Member
- Posts: 566
- Joined: Fri Jan 04, 2008 11:01 am
- Location: Philadelphia
- Thanked: 31 times
- GMAT Score:640
Why couldn't I make the base go all the way across the circle? Is it because that would make it "on the circle" instead of just "at the center" of the circle?
Yes, that's right. The question states that one of the vertices lies at the center of the circle. If the base were to lie across the center of the circle, it would be impossible for the origin to be a vertice of the triangle.AleksandrM wrote:Why couldn't I make the base go all the way across the circle? Is it because that would make it "on the circle" instead of just "at the center" of the circle?
-
umaa
- Legendary Member
- Posts: 727
- Joined: Sun Jun 08, 2008 9:32 pm
- Thanked: 8 times
- Followed by:1 members
What is the greatest possible area of a triangle region with one vertex at the center of a circle of radius 1 and other two vertices on the circle?
Pls find the attachment for the rough draw.
Assume a square is lying on a circle and the diagonal of the square in the circle's diameter. "The greatest possible area of a triangle region with one vertex at the center of a circle" means, its the diagonal of the triangle. If radius is 1, then diameter(diagonal) is, 2.
If the diagonal is going at the center of the circle, then the would definitely be the right angle triangle. Other two vertices are lying on the circle. (It doesn't mean that for right angle triangle the two sides should be equal.)
Both the vertices wouldn't go above the diagonal area. It should be below 2.
so, if i take both the vertices as 2 *(considering diagonal) 1/2 * 2 * 2 = 2, the area wouldn't go above 2 and wouldn't go below 1. so, it should be between 1 and 2. With this, we can eliminate A, B and C.
Take 1/2 * b * h = 1;
b=2/h;
b^2 + h^2 = 4
4/h^2 + h^2 = 4
4+ h ^ 4 - 4 h^2=0
h^4-4h^2+4=0 (take h^2 = m)
m^2 - 4m + 4=0
m=2
so, h = Sqroot(2).
b=2/Sqroot(2)
which is, Sqroot(2);
Guys, let me know if i'm wrong.[/img]
Pls find the attachment for the rough draw.
Assume a square is lying on a circle and the diagonal of the square in the circle's diameter. "The greatest possible area of a triangle region with one vertex at the center of a circle" means, its the diagonal of the triangle. If radius is 1, then diameter(diagonal) is, 2.
If the diagonal is going at the center of the circle, then the would definitely be the right angle triangle. Other two vertices are lying on the circle. (It doesn't mean that for right angle triangle the two sides should be equal.)
Both the vertices wouldn't go above the diagonal area. It should be below 2.
so, if i take both the vertices as 2 *(considering diagonal) 1/2 * 2 * 2 = 2, the area wouldn't go above 2 and wouldn't go below 1. so, it should be between 1 and 2. With this, we can eliminate A, B and C.
Take 1/2 * b * h = 1;
b=2/h;
b^2 + h^2 = 4
4/h^2 + h^2 = 4
4+ h ^ 4 - 4 h^2=0
h^4-4h^2+4=0 (take h^2 = m)
m^2 - 4m + 4=0
m=2
so, h = Sqroot(2).
b=2/Sqroot(2)
which is, Sqroot(2);
Guys, let me know if i'm wrong.[/img]
- Attachments
-
umaa, u have made two mistakes.
1) ur drawing shows that u haven't taken centre as vertex for one side. The triangle line is maybe passing thru centre but not is not a vertex. A vertex at centre means that it's a meeting point for two different sides of triangle.
2) you have already assumed the area to be 1. that's what we want to find out.
we know radius is one so to sides have length of 1. After that u can use any of the approach give above e.g Ian's approach to get the answer
1) ur drawing shows that u haven't taken centre as vertex for one side. The triangle line is maybe passing thru centre but not is not a vertex. A vertex at centre means that it's a meeting point for two different sides of triangle.
2) you have already assumed the area to be 1. that's what we want to find out.
we know radius is one so to sides have length of 1. After that u can use any of the approach give above e.g Ian's approach to get the answer
