#### Answer

1764$mm^{2}$

#### Work Step by Step

Given, AB = 39 mm, BC =52mm
CD =25mm, DA = 60mm
By Brahmaguptas 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 = 39 mm, BC = b = 52mm
CD = c =25mm, DA = d = 60mm
s = $\frac{39+52+25 +60}{2}$
=88mm
A = $\sqrt (88-39)(88-52)(88-25)(88-60)$
=$\sqrt 49*36*63*28$
=1764$mm^{2}$
Therefore the area of the cyclic quadrilateral ABCD = 1764$mm^{2}$.