Label each vertex: O(0,0), A(2,6), B(11,8) and C(9,2).
Draw a horizontal (y=8) passing thru B that crosses y-axis (x=0) at point D(0.8).
Draw a vertcal (x=11) passing thru B that crosses x-axis (y=0) at point E(11,0).
Connect A toD, and C to E.
In ΔOAD and ΔBAD, let the foot of altitude from A to OD be G(0,6), and the one from A to BD be H(2,8).
In ΔOAD and ΔBCE, OD∥BE and OA∥BC, so ∠AOD=∠CBE. OD=BE and OA=BC
Thus, ΔOAD≡ΔBCE (SAS) In the same manner, ΔBAD≡ΔOCE (SAS)
Therfore, area of parallelogram OABC=area of rectangle ODBE - 2(ΔOAD + ΔBAD)
Here, area of rectangle ODBE=ODxDB=8x11=88 (unit²),
ΔOAD=½xODxAG=½x8x2=8, and ΔBAD=½xBDxAH=½x11x2=11
Therefore, area of parallelogram OABC=88 - 2(8+11)=88 - 38=50 (unit²)
The answer is: area of parallelogram OABC is 50 (unit²)