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
Area of triangle in circle
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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.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
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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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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