find the lengths of the sides of the triangle of maximum area that can be drawn inside a circle of radius r with one vertex of the triangle at the center of the circle, and the other two vertices on the circle.

In a two-dimensional plane, a triangle's area is the area that it completely encloses.
A triangle is a closed form with three sides and three vertices, as is common knowledge.
The entire area occupied by a triangle's three sides is referred to as its area. Half of the product of the triangle's base and height provides the standard formula for calculating the area of the triangle.

With one of the triangle's vertices in the circle's centre and the other two on the circumference, we must determine the lengths of the sides of the triangle with the largest possible area that can be drawn within.

Let,

b = base of triangle

h = height of triangle

R = Radius of circle

Area of triangle = bh/2

The area of a triangle is bh/2, but since b/2 is shared by a triangle, the maximum area is constrained by the height h.

This would be the perpendicular height to the base, which is the radius R

The base is two radii b=2R and height is h= R

Therefore, the lengths of the sides of a triangle are as follows:

base(b) = 2R and height(h) = R

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