text: Where is the angle Θ?  Is it at the origin, or is it one of the sides of the trapezoid?\n\nEdit: from your description, it sounds like Θ is at the origin.  And the point at the upper right of the trapezoid would be (x,y).  This point is also (sin(Θ), cos(Θ))\n\nYou have an isosceles trapezoid formed by the points (1,0), (x, y), (-x, y), (-1, 0).  And x = cos(Θ), y = sin(Θ).\n\nThe area of a trapezoid is given by the formula:\n(b1 + b2)/2 * h\n\nIn your case:\nBase 1 goes from from (-1,0) to (1,0) for a length of 2.\nBase 2 goes from (-x,y) to (x,y) for a length of 2x or 2 cos(Θ).\n\nSo the average of these two:\n (2 + 2 cos(Θ))/2\n= 1 + cos(Θ)\n\nAnd the height is y or just sin(Θ).\n\nSo putting it all together, a formula for the area of your trapezoid is:\n\nf(Θ) = sin(Θ) * (1 + cos(Θ) )\n\nAs requested, there is part (a)...\n\nFor part (b) ... take the derivative and set it to zero.  You will get a Θ between 0° and 90° (π/2 radians)... I suspect that the maximum area will be at Θ = 60° (π/3 radians).  Just thinking about it visually, that would be the top half of a regular hexagon...