15 POINTS WILL MARK BRAINLIEST Find the area of the following shape. You must show all work to receive credit.

shape with vertices located at negative 4 comma 0, 0 comma 4, 2 comma 2, 0 comma 2, and 0, 0

36a

Step-by-step explanation:

First label all the lines:

Line 1- (0,0) and (0,2)

Line 2- (0,2) and (2,2)

Line 3- (2,2) and (0,4)

Line 4- (0,4) and (-4,0)

Line 5- (-4,0) and (0,0)

Next split the shape into two 

Now you have two triangles

Find the area of the first triangle

First triangle is Line 5, 4, and 1+ a little more

First find the distance of 5

Square root of (0-0) squared + (0+4) squared

Simplify:

0+16→ 16 squared is 4

Line 5 is 4

Now we find the distance of line 4:

Square root of (-4-0) squared +(0-4) squared

Simplify

16+16 → 32 squared is about 5.6

Line 4 is 5.6

Now lets find Line 1 but change it to (0,0) and (0,4)

Square root of (0-0) squared + (4-0) squared

Simplify

0+16→16 squared is 4

Let's label them, so Line 1 is the height and Line 5 is the base

The triangle area formula is 1/2bh

Let's plug everything in:

(1/2)4 times 4

Simplify 1/2 times 16 = 8

The area of triangle 1 is 8

 Now triangle two consists of lines 2, 3, and another one that has not been labeled. 

Not labeled line: (0,2) and (0,4)

Find Line 2:

Square root of (2-0) squared + (2-2) squared

Simplify

4+0 →4 squared is 2 

Line 2 is 2

Now let's find line 3:

Square root of (0-2) squared + (4-2) squared

Simplify

4+4→ 8 squared is 2.8

Line 3 is 2.8

Not labeled line:

Square root of (0-0) squared + (4-2) squared 

Simplify

0+4-> 4 squared is 2

Now label the line as base and height

Not labeled line is height and the base is line 2

Put it into the equation:

(1/2) 2 times 2

(1/2)4

Are of second triangle is 2

Now add both of them together and the area of the shape is 10 units

sec ф = 5/3 (quadrant 4)

let x = base

y = perp

z = hypotenuse

sec ф = hypotenuse/base = 5/3

⇒ z = 5 and x = 3

finding y:

z² = x² + y²

5² = 3² + y²

y² = 25-9

y² = 16

y = 4

now finding all other geometric functions of theta:

sin ф = y/z = 4/5

sin ф = -4/5 (sin is negative in the 4th quadrant)

cos ф = x/z = 3/5

cos ф = 3/5 (cos is positive in 4th quadrant)

tan ф = y/x = 4/3

tan ф = -4/3 (tan is negative in 4th quadrant)

cosec ф = z/y = 5/4

cosec ф = -5/4 (cosec is negative in the 4th quadrant)

cot ф = x/y = 3/4

tan ф = -3/4 (tan is negative in 4th quadrant)

i don't know, but that looks pretty bad. my school passed us unless we failed in all the subjects, but idk abt ur school's rules.

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