One of the ways to find the area of ABCD is by enclosing it in a rectangle. Then from the area of the rectangle, we subtract areas outside the shape, but inside the rectangle, as follows:
Area of rectangle AFCE=3*4=12 sq. units.
Area of AFCB (one square and two triangles)
= 1*1+(2*1)/2+(3*1)/2
= 1+1+1.5
=3.5
Similarly, area of ADCE
= 1*1+(2*1)/2+(3*1)/2
= 1+1+1.5
=3.5
Therefore area of ABCD
= 12-3.5-3.5
= 5 sq. units.
Alternatively, we can solve by the coordinates of points A,B,C & D.
We will prepare a table according to the x and y coordinates of each point A,B,C,D and repeating the first point A.
Pt X Y
A 7 5
B 6 3
C 3 2
D 4 4
A 7 5
Twice the area is found by summing products, according to a pattern
Ax*By-Bx*Ay = 7*3-6*5 = 21-30 = -9
Bx*Cy-Cx*By = 6*2-3*3 = 12-9 = 3
Cx*Dy-Dx*Cy = 3*4-4*2 = 12-8 = 4
Dx*Ay-Ax*Dy = 4*5-7*4 = 20-28 = -8
Sum of the products
=-9+3+4-8
=-10
= 10 (by ignoring the sign for area)
which is twice the area of the polygon.
Therefore area of ABCD = 10/2 = 5 sq. units
