Question 485997
<pre><font face = "consolas" color = "indigo"  size=4><b>

We draw the pentagon:

{{{drawing(400,400,-8,12,-7,13,

graph(400,400,-8,12,-7,13),

line(-6,0,0,12), locate(10,1,"(10,0)"),

line(5,8,0,12), locate(5.5,8.2,"(5,8)"),

line(5,8,10,0), locate(.5,12.5,"(0,12)"),

line(10,0,0,-6), locate(-8,1,"(-6,0)"), locate(0,-6,"(0,-6)"),

line(-6,0,0,-6) ) )}}}

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:

{{{drawing(400,400,-8,12,-7,13,

green(line(5,8,5,0),line(5,8,0,8),line(0,12,0,-6),line(-6,0,10,0)),

locate(-2.5,4.5,"#1"),locate(2,5,"#3"), locate(6,3,"#4"),

locate(1,9.5,"#2"), locate(-2.5,-1.5,"#5"), locate(2.5,-1.5,"#6"),


graph(400,400,-8,12,-7,13),

line(-6,0,0,12), locate(10,1,"(10,0)"),

line(5,8,0,12), locate(5.5,8.2,"(5,8)"),

line(5,8,10,0), locate(.5,12.5,"(0,12)"), 

line(10,0,0,-6), locate(-8,1,"(-6,0)"), locate(0,-6,"(0,-6)"),

line(-6,0,0,-6) ) )}}}

The Area of a triangle is

A = {{{1/2}}}bh

We find area of right triangle #1:

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

A = {{{1/2}}}bh = {{{1/2}}}(6)(12) = 36

We find area of right triangle #2:

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

A = {{{1/2}}}bh = {{{1/2}}}(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 = {{{1/2}}}bh = {{{1/2}}}(5)(8) = 20

We find area of right triangle #5:

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

A = {{{1/2}}}bh = {{{1/2}}}(6)(6) = 18

We find area of right triangle #6:

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

A = {{{1/2}}}bh = {{{1/2}}}(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 {{{30/154}}}
which reduces to {{{15/77}}}

Edwin</pre>