Answer:
let x = shortest base of the trapezoid
A = h (x/2 + r) = âš(r² - (x/2)²) (r + x/2)
= âš(1 - (x/2)²) (1 + x/2) .... if r = 1
dA/dx = (2 - x - x²) / (2âš(4 - x²))
= 0 at x = 1 .... which is a relative maximum because d²A/dx² < 0
A = (3/4) âš3
