Question 166152
First plot the points and see what you've got.
{{{drawing( 300, 300, -6, 6, -6, 6,grid( 1 ),
circle( 1, -4, .2 ),
circle( 4, -1, .2 ),
circle( 4, 5, .2 ),
circle( -1, 4, .2 ),
circle( -2, -1, .2 ),
line(1,-4,4,-1),
line(4,-1,4,5),
line(4,5,-1,4),
line(-1,4,-2,-1),
line(-2,-1,1,-4),
locate(1.1,-4.2,A),
locate(4.1,-1.2,B),
locate(4.2,5.2,C),
locate(-1.5,4.2,D),
locate(-2.5,-1.2,E)
)}}}
Define a couple of extra points that will help simplify the solution. 
X at (4,4)
Y at (-1,-1)
{{{drawing( 300, 300, -6, 6, -6, 6,grid( 1 ),
circle( 1, -4, .2 ),
circle( 4, -1, .2 ),
circle( 4, 5, .2 ),
circle( -1, 4, .2 ),
circle( -2, -1, .2 ),
circle(4,4,.2),
circle(-1,-1,.2),
green(line(4,4,-1,4)),
green(line(-2,-1,4,-1)),
green(line(-1,-1,-1,4)),
green(line(4,-1,4,4)),
locate(-1.2,-1.2,Y),
locate(4.2,4,X),
line(1,-4,4,-1),
line(4,4,4,5),
line(4,5,-1,4),
line(-1,4,-2,-1),
line(-2,-1,1,-4),
locate(1.1,-4.2,A),
locate(4.1,-1.2,B),
locate(4.2,5.2,C),
locate(-1.5,4.2,D),
locate(-2.5,-1.2,E)
)}}}
Now we can find the areas of the individual triangles and square and add them together.
Area of a square is s*s where s is the side length.
Area of a triagle is half the product of base and height.
.
.
.
Square BXDY : {{{A=s^2=5^2=25}}}
Triangle XCD :{{{A=(1/2)bh=(1/2)(5)(1)=5/2}}}
Triangle DEY :{{{A=(1/2)bh=(1/2)(1)(5)=5/2}}}
Triangle EAB :{{{A=(1/2)bh=(1/2)(6)(3)=9}}}

Now add all of the areas,
{{{A[p]=25+5/2+5/2+9}}}
{{{A[p]=39}}}