[tex]A=| \frac{-3(3-(-4))-3(-4-(-2))-2(-2-3)}{2}| = 2.5[/tex]
[tex]B=| \frac{4(3-(-4))-3(-4-(-4))-2(-4-3)}{2}|=21 [/tex]
[tex]C=| \frac{4(3-4)-3(4-(-4))-2(-4-3)}{2}|=7[/tex]
[tex]D=| \frac{4(4-4)+4(4-(-4))-2(-4-4)}{2}|=24 [/tex]
[tex]E=| \frac{4(4-3)+4(3-(-4))+5(-4-4)}{2}|=4 [/tex]
[tex]F=| \frac{4(-3-3)+5(3-(-4))+5(-4(-3))}{2}|=3 [/tex]
[tex]G = | \frac{-1(6-4)-1(4-4)+1(4-6)}{2}|=2 [/tex]
[tex]H=| \frac{1(6-4)-1(4-6)+1(6-6)}{2}|=2 [/tex]
[tex]A+B+C+...+H = \boxed{65.5}square units[/tex]
Area of "eyes and mouth":
[tex]\bold1=| \frac{-1(2-2)-2(2-3)-1(3-2)}{2}|=0.5 [/tex]
[tex]\bold2=| \frac{3(2-2)+4(2-3)+2(3-2)}{2}|=1 [/tex]
[tex]\bold3=| \frac{-1(2-(-2))+3(-2-(-1))-1(-1-(-2))}{2}|=2 [/tex]
[tex]\bold4=| \frac{-1(-2-(-1))+3(-1-(-1))+(-1-(-2))}{2}|=2 [/tex]
[tex]\bold1+\bold2+\bold3+\bold4=5.5square units[/tex]
IF we subtract the "holes" from the "head":
65.5 - 5.5 = 60 square units
