Given AB || CD and AB = CD = 2.

Let O be the center of the cirle and EF be perpendicular from AB to CD thru O. EF = 2.

With this we have, AE = 1(Since OE bisects AB), OE = 1. <AEO = 90 ==> OA = sqrt(2). By Pythogorous theorem.

Hence

Radius of the circle = sqrt(2)Now Area of the region formed by ABCD = Area of the sectors OCA + Area of the sector OBD + Area of the triangle OAB + Area of the triangle OCD.

Area of the sector OCA = 90 / 360 * Pi * 2 = pi / 2. -----> i

Area of the sector OBD = 90 / 360 * Pi * 2 = pi / 2. -----> ii

Area of the triangle OAB = 1/2 * 1 * 2 = 1 -----> iii

Area of the triangle OCD = 1/2 * 1 * 2 = 1 -----> iv

ie.

Total area of the region formed by ABCD = Pi + 2 (i + ii + iii + iv)

My apologies for not having a diagram. Not sure how to soft copy diagram.