The hexagon ABCDEF with the vertices #A(-2, 4)#, #B(0,4)#, #C(2, 1)#, #D(5, 1)#, #E(5, 2)# & #F(-2, 2)# can be divided into four triangles #\Delta ABC. \Delta ACD, \Delta ADE# & #\Delta AEF#
The area #\Delta_1# of #\Delta ABC# with vertices #A(-2, 4)\equiv(x_1, y_1), B(0, 4)\equiv(x_2, y_2)# & #C(2, 1)\equiv(x_3, y_3)# is given by following formula
#\Delta_1=1/2|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|#
#=1/2|-2(4-1)+0(1-4)+2(4-4)|#
#=3#
Similarly, the area #\Delta_2# of #\Delta ACD# with vertices #A(-2, 4), C(2, 1)# & #D(5, 1)# is given as follows
#\Delta_2=1/2|-2(1-1)+2(1-4)+5(4-1)|#
#=4.5#
Similarly, the area #\Delta_3# of #\Delta ADE# with vertices #A(-2, 4), D(5, 1)# & #E(5, 2)# is given as follows
#\Delta_3=1/2|-2(1-2)+5(2-4)+5(4-1)|#
#=3.5#
Similarly, the area #\Delta_4# of #\Delta AEF# with vertices #A(-2, 4), E(5, 2)# & #F(-2, 2)# is given as follows
#\Delta_4=1/2|-2(2-2)+5(2-4)-2(4-2)|#
#=7#
hence, the area of hexagon ABCDEF is equal to the sum of areas of above triangles, given as
#\Delta_1+\Delta_2+\Delta_3+\Delta_4#
#=3+4.5+3.5+7#
#=18#
