Find the perimeter of the following shapefind the perimeter of the following shape: shape abcd is shown. point a is at 7, 5. point b is at 6, 3. point c is at 3, 2. point d is at 4, 4.

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

You must find the surface of the block therefore 98187 dived by 5628 = 583m squared