Question

Help I'm taking a k12 quiz 2.04: composite figures but, I can't figure these out can anyone help me?

Answer 1

Answer:

1. 39 units²

2. 35.5 units²

3. i didn't have that question

Step-by-step explanation:

Answer 2

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 $units^{2}$
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$units^{2}$

Tot+al area of a figure: 
A=18+27=45$units^{2}$

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$units^{2}$

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

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

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

Total area of a figure:
A=12+6+7+10.5=35.5$units^{2}$

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$units^{2}$

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

Paralelogram:
a=6
h=1
A=6*1=6$units^{2}$

Total area of a figure:
A=$10 \sqrt{2}$ + 6 $units^{2}$

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$in^{2}$

Rectangle:
a=7in
b=6in
A=7*6=42$in^{2}$

Total area of a figure:
A=9+42=51$in^{2}$