See the diagram.
Draw the trapezium as shown. As it is an isosceles trapezium, CB = DA also. The adjacent angles are same : ie., angle D = angle C and angle A = angle B.
Draw the circle inscribing the trapezium ABCD. It touches the trapezium at E, F, G and H. As AB || DC, the line joining EF is the diameter of the circle. EF is perpendicular to both AB and DC which are parallel tangents. Parallel tangents can only be at the ends of a diameter to the circle.
Now, from symmetry, E and F are midpoints of AB and CD. Hence AE = EB = 5 cm. Also, CF = DF = 15 cm.
As DG and DF are tangents to the circle from a point D, they are equal. Hence, DG = 15 cm. Similarly, CH = 15 cm too.
As AG, and AE are two tangents drawn from a point A to the circle, they are equal. Hence, AG = AE = 5 cm.
The sides AD and CB are = DG + AG = 15 + 5 = 20 cm
Draw a line AI, from A, perpendicular to CD meeting at I. Now , AIFE is a rectangle. Hence, IF = AE = 5 cm.
Thus, DI = DF - IF = 15 - 5 = 10 cm
The triangle DAI is a right angle triangle. Thus apply Pythagoras theorem.
AD² = DI² + IA²
IA² = 15² - 10² = 125 cm²
IA = 5√5 cm
IA = the diameter EF of the circle , as IAEF is a rectangle.
Area of the circle = π (5√5/2)² cm²
= 125 π / 4 cm²
Area of the trapezium --- if you want = 1/2 *AB+CD) EF
= 1/2 * (10+30) 5√5 = 100√5 cm²