I need to find a maximum area of trapezoid inscribed in circle. The base is diameter of the circle, (1,3) is center of the circle and radius is 4.

Answer 1

Location of center not play role. If center combain with wertices of trapezoid we get tree triangles with angles ∝, 180° - 2∝ and ∝ and sides containing this angles is 4(radius of circle). So area of trapezoid a is equal sum of areas of this 3 triangles:

A = 1/2·4·4sin∝ + 1/2·4·4sin(180° - 2∝) + 1/2·4·4sin∝= 8sin∝ + 8sin2∝ + 8sin∝ = 16sin∝ + 8sin2∝.

Now A' = 16cos∝ + 16cos2∝ = 0, cos∝ + 2cos²∝ - 1 = 0; (2cos∝ - 1)(cos∝ + 1) = 0, from here cos∝ = 1/2, ∝ = 60° or cos ∝ = - 1, ∝ = 180°. Now A'' = - 16sin∝ - 32sin2∝ = -16sin60° - 32sin120° = -24sqrt(3) < 0. So we have max at ∝ = 60°. A_max = 16sin60° + 8sin120° = 12(3)^1/2