nvm I got the answer,
Since \(BQ = \frac{1}{3} AB\), \(BR = \frac{1}{3} BC\), and \(\angle QBR = \angle ABC\), triangles QBR and ABC are SAS-similar. Furthermore, since Q and R are trisection points, the side lengths are in a \(1:3\) ratio, so the areas are in a \(1:9\) ratio. This gives \([QBR] = \frac{1}{9} [ABC].\)
Likewise, triangles UDV and CDA are similar, and \([UDV] = \frac{1}{9} [CDA].\)
Therefore, \([QBR] + [UDV] = \frac{1}{9} ([ABC] + [CDA]) = \frac{1}{9} [ABCD] = 20.\)
The remainder of quadrilateral ABCD is hexagon \(AQRCUV\), so it has area \(180 - 20 = \boxed{160}.\)
.
+280 