Hi
This has been discussed previously
https://www.beatthegmat.com/largest-area ... html#50963
Cheers
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GMAT Prep - Triangle Area
Source: Beat The GMAT — Problem Solving |
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.
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.
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
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Well going to this by a simpler approach : Since the 2 sides of a triangle are of same length we have an isoceles triangle.
The isoceles triangle will have the maximum area when it is a right angled isoceles , where the two sides having the same length are at 90 deg.
So the area will 1/2 * r * r = 1/2 * 1 * 1 = 0.5. (r is the radius of the circle).
The isoceles triangle will have the maximum area when it is a right angled isoceles , where the two sides having the same length are at 90 deg.
So the area will 1/2 * r * r = 1/2 * 1 * 1 = 0.5. (r is the radius of the circle).
What happens to the vertex on the center???? That fact is taken nowhere into account! am i missing the point?? can someone help?spanlength wrote:Well going to this by a simpler approach : Since the 2 sides of a triangle are of same length we have an isoceles triangle.
The isoceles triangle will have the maximum area when it is a right angled isoceles , where the two sides having the same length are at 90 deg.
So the area will 1/2 * r * r = 1/2 * 1 * 1 = 0.5. (r is the radius of the circle).
The vertex is still in the center. It's the corner. It's where the right angle is formed? Base =1. Height = 1. Connect the dots of the two points.KICKGMATASS123 wrote:What happens to the vertex on the center???? That fact is taken nowhere into account! am i missing the point?? can someone help?spanlength wrote:Well going to this by a simpler approach : Since the 2 sides of a triangle are of same length we have an isoceles triangle.
The isoceles triangle will have the maximum area when it is a right angled isoceles , where the two sides having the same length are at 90 deg.
So the area will 1/2 * r * r = 1/2 * 1 * 1 = 0.5. (r is the radius of the circle).
use the Sine formulae for area of triangle
Area = (1 / 2)* OB * OA * sin(O)
max value of sine = 90 degree
therefore area= 1/2 *1*1*1 = .5
Area = (1 / 2)* OB * OA * sin(O)
max value of sine = 90 degree
therefore area= 1/2 *1*1*1 = .5
