Answer: 13.5 square units
Explanation: since we know that the area of a triangle = [tex]\frac{1}{2} (x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))[/tex]
let us give the name to the triangles ABC having vertices (4,10), (5,7), and (7,8), triangle DEF having vertices (3,6), (3,3) and (6,3) and triangle PQR having vertices (7,6), (7,1) and (9,4).
Thus, Area of [tex]\bigtriangleup[/tex] ABC= [tex]\frac{1}{2}[/tex]×(4(7-8)+5(8-10)+7(10-7))= [tex]\frac{1}{2}[/tex](-4-10+21)=8/2=4 unit.
Area of [tex]\bigtriangleup[/tex] DEF= [tex]\frac{1}{2}[/tex]×(3(3-3)+3(3-6)+6(6-3))=[tex]\frac{1}{2}[/tex]×(0-9+18)=9/2=4.5 unit.
Area of [tex]\bigtriangleup[/tex] PQR= [tex]\frac{1}{2}[/tex]×(7(1-4)+7(4-6)+9(6-1))=[tex]\frac{1}{2}[/tex]×(-21-14+45)=5 unit.
So, the total area of the three triangles=13.5 square units