Help I'm taking a k12 quiz 2.04: composite figures but, I can't figure these out can anyone help me?Question. 1 What is the area of this polygon? (Picture 1)Question. 2 What is the area of this polygon? 42.5 Units? 41.5 Units? 35.5 Units? 29.5 Units? (Picture 2)Question. 3 This figure is made up of a rectangle and parallelogram.What is the area of this figure? (Picture 3)Question. 4 What is the area of this composite shape? (Picture 4)Thank you angels who help me with these

All of these can be solved by separating complex figur into simpler one.
Needed formulas are:
Area of rectangle A=a*b
Area of triangle A=a*h/2
Area of paralelogram A= a*h

QUESTION 1.
When we connect points F and S we get a rectangle and a triangle.
Dimensions of sides of a rectangle are a=9 and b=2. 
A=9*2=18 [tex]units^{2} [/tex]
For a triangle we need to take one side and it's height. Let's take side FS because it's easiest to find the height.
a=9
h=6
A=9*6/2=27[tex]units^{2} [/tex]

Tot+al area of a figure: 
A=18+27=45[tex]units^{2} [/tex]

QUESTION 2.
We connect point (1,1) with (1,-2) and (-6,1). This way we get a rectangle and three triangles.
Rectangle:
a=3
b=4
A=3*4=12[tex]units^{2} [/tex]

Top right triangle:
a=4
h=3
A=4*3/2=6[tex]units^{2} [/tex]

Top left triangle:
a=7
h=2
A=7*2/2=7[tex]units^{2} [/tex]

Bottom left triangle:
a=7
h=3
A=7*3/2=10.5[tex]units^{2} [/tex]

Total area of a figure:
A=12+6+7+10.5=35.5[tex]units^{2} [/tex]

QUESTION 3.
We connect points (-4,2) and (-1,3). This way we get a paralelogram and a rectangle.

Paralelogram:
a=6
h=1
A=6*1=6[tex]units^{2} [/tex]

Rectangle:
a=distance between (-4,2) and (-1,3) = [tex] \sqrt{ (-1-(-4))^{2} + (3-2)^{2} } = \sqrt{10} [/tex]
b = distance between (-4,2) and (-2,-4) = [tex] \sqrt{ (-2-(-4))^{2} + (-2-2)^{2} } = \sqrt{20} [/tex]
A=a*b=[tex] \sqrt{10} * \sqrt{20} = \sqrt{200} =10 \sqrt{2} [/tex][tex]units^{2} [/tex]

Paralelogram:
a=6
h=1
A=6*1=6[tex]units^{2} [/tex]

Total area of a figure:
A=[tex]10 \sqrt{2} [/tex] + 6 [tex]units^{2} [/tex]

QUESTION 4.
We coonect missing side of a triangle. That side has a length of: 6in-3in=3in. Vertical side of triangle is: 13in-7in=6in
Area:
a=3
h=6
A=3*6/2=9[tex] in^{2} [/tex]

Rectangle:
a=7in
b=6in
A=7*6=42[tex] in^{2} [/tex]

Total area of a figure:
A=9+42=51[tex] in^{2} [/tex]