We draw the pentagon:



There are two ways of doing this, depending on what you have
studied.  If you have studied finding the area of a convex
polygon by means of a determinant, then post again and I will
show you that method.  But I will assume you haven't had that
yet and need to break the area down into right triangles and
a rectangle.

We petition off the pentagon into 5 right triangles
and 1 rectangle:



The Area of a triangle is

A = bh

We find area of right triangle #1:

Right triangle #1 has base 6 and height 12. Substituting

A = bh = (6)(12) = 36

We find area of right triangle #2:

Right triangle #2 has base 5 and height 4. Substituting

A = bh = (5)(4) = 10

Rectangle #3 is 8 by 5, so its area is 8×5 or 40

We find area of right triangle #4:

Right triangle #4 has base 5 and height 8. Substituting

A = bh = (5)(8) = 20

We find area of right triangle #5:

Right triangle #5 has base 6 and height 6. Substituting

A = bh = (6)(6) = 18

We find area of right triangle #6:

Right triangle #6 has base 10 and height 6. Substituting

A = bh = (10)(6) = 30

The area of the entire pentagon is

#1 has area 36
#2 has area 10
#3 has area 40
#4 has area 20
#5 has area 18
#6 has area 30
--------------
total =    154


The area that lies in the 4th quadrant is right
triangle #6, which has area 30.

Therefore the fraction of the total area that lies
in the 4th quadrant is 
which reduces to 

Edwin