what is the maximum possible area of a triangle inscribed in a circle with radius R?

Answer 1

The maximum possible area would be a equilateral triangle:
A = 3*(Rsin30)(Rcos30)
A = 3*(R^2)(1/2)((√3)/2)
A = 3((√3)/4)R^2

Answer 2

I think Cynthia's answer is the perimeter, not the area.

The maximum area occurs when the triangle is an equilateral triangle.

Then the radius R forms the hypotenuse of a 30-60-90 right triangle (draw a line from the center perpendicular to a side to see it). The shorter side is R/2...and if you draw the radius up to the point, you'll see that the height of the triangle is R + R/2 = 3R/2. Then using 30-60-90 triangle relationships, the base of the equilateral triangle will be 2(R√3/2) = R√3.

So the area would be:

A = 1/2 b h

= 1/2(3R/2)(R√3) = 3R²√3/4

Answer 3

[(3√3) / 4] * [r^2]
.........or..........
[3*(r^2)*√3] / 4

whichever you prefer
__________________________________________________

The largest triangle that can be inscribed in a circle.

Finding the area of the triangle:
The highest point of the triangle to the center of the circle = r
The center of the circle to the bottom edge of the triangle = r/2 (after applying the 30-60-90 triangle side-ratios)
Thus, the height of the triangle = r + r/2 = 3r/2
The base of the triangle = r√3 (twice the size of the (r√3)/2 length

Therefore, the area of the triangle = (BH)/2 = [(r√3)(3r/2)]/2 = [3*(r^2)*√3] / 4 = [(3√3)/4]*[r^2]

Answer 4

It would be an equilateral triangle.

Let
s = length of side of maximum inscribed equilateral triangle

The central angle θ = 2π/3

sin(θ/2) = sin(π/3) = (s/2)/R = √3/2
s/R = √3
s = R√3

The area A, of an equilateral triangle with side s is:

A = (1/2)(s)(√3/2)s = (√3/4)s² = (√3/4)(R√3)² = (3√3/4)R²

Answer 5

(3R^2√3)/4

I'm sorry, but I don't know how to show work on the computer...