A Tangential Trapezoid

This is all thanks to the information in the link from the problem.

Say the point of tangency between the circle and the side $DC$ , then $AW\cdot DY = BW\cdot CY$ , so we can write $DY=9l$ and $CY=5l$ for some $l$ .

Since the pairs of tangent lengths from each vertex are equal (again, see the link for definitions), we have

$5k+9l=125$

and

$9k+5l=169$

Solving these, we find $k=16$ and $l=5$ , so the tangent lengths are $(e,f,g,h)=(80,144,25,45)$ . The area of a tangential quadrilateral is given by

$\sqrt{(e+f+g+h)(fgh+efh+egh+efg)}=\boxed{17640}$

Let the tangent point on $BC$ be $X$ , the tangent point on $CD$ be $Y$ , and the tangent point on $AD$ be $Z$ . Let $e = AW$ , $f = BW$ , $g = CY$ , and $h = DY$ . Since tangent segments to a circle from the same external point are congruent, $AZ = AW = e$ , $BX = BW = f$ , $CX = CY = g$ , and $DZ = DY = h$ . Also drop perpendiculars from $D$ to $AB$ at $U$ and from $C$ to $AB$ at $V$ . Let the radius of the incircle be $r$ , so the height of the trapezoid is $WY = CV = DU = 2r$ .

Using Pythagorean's Theorem on $\triangle BCV$ , $(2r)^2 + (f - g)^2 = (f + g)^2$ , which simplifies to $fg = r^2$ . Likewise, using Pythagorean's Theorem on $\triangle ADU$ , $(2r)^2 + (e - h)^2 = (e + h)^2$ , which simplifies to $eh = r^2$ .

Since the legs are $AD = 125$ and $BC = 169$ , $e + h = 125$ and $f + g = 169$ . Since $\frac{AW}{BW} = \frac{5}{9}$ , $\frac{e}{f} = \frac{5}{9}$ .

The five equations with five variables ( $fg = r^2$ , $eh = r^2$ , $e + h = 125$ , $f + g = 169$ , and $\frac{e}{f} = \frac{5}{9}$ ) solve to $e = 80$ , $f = 144$ , $g = 25$ , $h = 45$ , and $r = 60$ .

Therefore, the bases of the trapezoid are $b_1 = e + f = 80 + 144 = 224$ , $b_2 = g + h = 25 + 45 = 70$ , the height is $h = 2r = 2 \cdot 60 = 120$ , and the area of the trapezoid is $A = \frac{1}{2}(b_1 + b_2)h = \frac{1}{2}(224 + 70)120 = \boxed{17640}$ .