Let's find the perimeter and the area of the figure one at a time.

Perimeter

To determine the perimeter of the polygon, we must find the sum of its side lengths. This polygon has four vertices, so it is a $quadrilateral .$ Let's denote the vertexes of this polygon.

Before we can find the sum of the side lengths, we must find the length of each side. We can use the Distance Formula to do this. Let's start with

\overline{A D} .

A D = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

SubstitutePoints

Substitute $(- 1, - 1)$ & $(- 3, 1)$

A D = (- 1 - (- 3))^{2} + (- 1 - 1)^{2}

SubNeg

$a - (- b) = a + b$

A D = (- 1 + 3)^{2} + (- 1 - 1)^{2}

AddSubTerms

Add and subtract terms

A D = 2^{2} + (- 2)^{2}

CalcPow

Calculate power

A D = 4 + 4

AddTerms

Add terms

A D = 8

▼

Calculate root

SplitIntoFactors

Split into factors

A D = 4 \cdot 2

RootProd

$a \cdot b = a \cdot b$

A D = 4 \cdot 2

CalcRoot

Calculate root

A D = 22

We continue by calculating the length of the other three sides

\overline{A B},

\overline{B C},

and

\overline{C D} .

Side	Coordinates	$(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$	Length
$\overline{A B}$	$(- 3, 1)$ $(3, 1)$	$(3 - (- 3))^{2} + (1 - 1)^{2}$	$6$
$\overline{B C}$	$(3, 1)$ $(3, - 1)$	$(3 - 3)^{2} + (- 1 - 1)^{2}$	$2$
$\overline{C D}$	$(3, - 1)$ $(- 1, - 1)$	$(- 1 - 3)^{2} + (- 1 - (- 1))^{2}$	$4$

Now, let's calculate the quadrilateral's perimeter. We do so by adding the four sides.

P = A D + A B + B C + C D

SubstituteValues

Substitute values

P = 22 + 6 + 2 + 4

AddTerms

Add terms

P = 22 + 12

CommutativePropAdd

Commutative Property of Addition

P = 12 + 22

The quadrilateral's perimeter is equal to

12 + 2

units.

Area

Now, we can find the area of our quadrilateral. To do it, let's divide it into a triangle and a rectangle.

The area of our

quadrilateral

is the sum of areas of the

triangle

and the

rectangle .

A_{A B C D} = A_{A E D} + A_{E B C D}

Let's begin by calculating the area of a triangle. To do so, we will use the formula for area of a triangle.

A = \frac{1}{2} b h

Here,

A E

is the base

b

of our triangle and

E D

is the height

h .

We can use the Distance Formula again to find these lengths.

Side	Coordinates	$(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$	Length
$\overline{A E}$	$(- 3, 1)$ $(- 1, 1)$	$(- 1 - (- 3))^{2} + (1 - 1)^{2}$	$2$
$\overline{E D}$	$(- 1, 1)$ $(- 1, - 1)$	$(- 1 - (- 1))^{2} + (- 1 - 1)^{2}$	$2$

By substituting these lengths into our formula, we can calculate the area.

A_{A E D} = \frac{1}{2} b h

SubstituteII

$b = 2$ , $h = 2$

A_{A E D} = \frac{1}{2} (2) (2)

▼

Multiply

A_{A E D} = \frac{1}{2} (4)

MoveRightFacToNumOne

$\frac{1}{b} \cdot a = \frac{a}{b}$

A_{A E D} = \frac{4}{2}

CalcQuot

Calculate quotient

A_{A E D} = 2

The area of the triangle is

2 square units .

Next, we will find the area of the rectangle. We can do it by multiplying its base by its height.

A = b h

Here,

C D

is the base

b

and

E D

is the height

h .

We already know these lengths, so we can substitute them into the formula to find the area.

A_{E B C D} = b h

SubstituteII

$b = 4$ , $h = 2$

A_{E B C D} = (4) (2)

Multiply

A_{E B C D} = 8

The area of the triangle is

8 square units .

Now, we can add the areas of the

triangle

and the

rectangle,

to find the area of the

quadrilateral .

A_{A B C D} = A_{A E D} + A_{E B C D}

SubstituteII

$A_{A E D} = 2$ , $A_{E B C D} = 8$

A_{A B C D} = 2 + 8

AddTerms

Add terms

A_{A B C D} = 10

The quadrilateral's area is

10 square units .

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