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