Begin with the trapezoid. The area of a trapezoid is `A=1/2(b_1+b_2)h` where `b_1,b_2` are the lengths of the parallel sides (the bases) and `h` is the distance between the parallel sides (the height.)

So `20=1/2(b_1+b_2)4` (Noting that the height is made up of 2 radii of length 2)

Then `b_1+b_2=10`

Now draw the trapezoid circumscribed about a circle. Let the trapezoid be called ABCD where `bar(AB)` is the shorter base. Let the circle intersect the trapezoid at E,F,G,H where E is on `bar(AB)` , F is on `bar(BC)` (one of the congruent sides), G is on the other base and H on the other congruent side.

If the center of the circle is O; note that EBFO is a kite -- 2 pairs of consecutive sides are congruent. OE=OF since they are radii, and BE=BF since they are tangents to a circle drawn from a point. Likewise CGOF is a kite.

Thus one leg of the trapezoid is BF+FC. We can substitute so that the length of the leg is BE+CG. But this is 1/2 of AB+CD=10.

**Therefore the lengths of the legs (the congruent sides) is 5.**

** We can use the Pythagorean theorem to verify the results. Drop a perpendicular from B to `bar(CD)` forming a right triangle. The legs of the triangle have lengths 3 and4, thus the hypotenuse, which is the leg of the trapezoid, is 5.**

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