#### Answer

120$cm^{2}$

#### Work Step by Step

Given, Although not all trapezoids are cyclic, one with bases of lengths 12 cm and 28 cm and both legs of length 10 cm would be cyclic.
We need to find the area of this isosceles trapezoid.
By Brahmagupta's formula, the area of cyclic quadrilateral with sides of length a,b,c and d is
A = $\sqrt (s-a)(s-b)(s-c)(s-d)$
s = $\frac{a+b+c +d}{2}$
AB =a = 12 cm, BC = b = 10 cm
CD = c =28 cm, DA = d = 10cm
s = $\frac{12+10+28+10}{2}$
=30 cm
A = $\sqrt (30 - 12)(30 - 10)(30 -28)(30 - 10)$
=$\sqrt 18 * 20 * 2 * 20$
=120$cm^{2}$
Therefore the area of the given isosceles trapezoid = 120$cm^{2}$.