ABJelly

Nevermind, I got the answer, it's 84

Basically you find the central angle which is 360/15. Then you do (15-8)*24, which equals 7*24 = 168. Finally you divide by 2 which gives you 84.

:)

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 \(180 - 20 = \boxed{160}.\)

Sorry, I didn't know how to add one