Question

Please someone that is good at math please help me!

1. Find the area of the pentagon below with the given information
2. Find the area of the shapes below.
3. Using the general graph of the triangle PQR below, fill out the table.

Answer 1

I'll do the first three problems to get you started. If you still need help with the fourth question, then please let me know and I'll update my solution.

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Problem 1

Answer: 200 square inches

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Work Shown:

The pentagon can be divided into five triangles as shown in figure 1 (attached image below). You start at the center of the pentagon and extend to each vertex. Each of these triangles are congruent, or the same. Each copy is a rotated version of another.

Each triangle has a base of 10, and a height of 8

A = area of one triangle

A = (1/2)*base*height

A = 0.5*10*8

A = 5*8

A = 40 square inches

There are n = 5 triangles, so the total area is therefore n*A = 5*40 = 200

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Problem 2

Answer: 30 square cm

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Work Shown:

base = 6 is the horizontal component

height = 5 cm is the vertical component

rule: the base and height are always perpendicular to one another

the area of the parallelogram is simply the product of base and height

area of parallelogram = base times height

A = B*H

A = 6*5

A = 30 square cm

The 7 cm isn't used at all. It's likely put in there to distract.

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Problem 3

Answer: 22 square meters

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Work Shown:

Have a look at figure 2 shown below. I divide up the original figure (which looks somewhat like an odd shoe of sorts) into a rectangle and a triangle.

The rectangle is 9 by 2, so it has area 9*2 = 18

The triangle has a base of 2 and height of 4 (these values are explained in the diagram I provided). So the triangle's area is 0.5*B*H = 0.5*2*4 = 4

The total area is therefore

(area of rectangle) + (area of triangle) = 18+4 = 22

Answer 2

Explanation:

Part 1). In a pentagon when attach the center to all vertices, five triangles are formed with the given height h = 8 inches and base b = 10 inches.

Now area of the pentagon = 5(Area of one triangle formed by center and two vertices of the pentagon)

= 5(
`(1)/(2))(Base)(height)`

=
`(5)/(2)(b)(h)`

= 2.5×10×8

= 200 inch²

Part 2). Area of parallelogram =
`(1)/(2)(\text{Sum of two parallel sides})(\text{Distance between the parallel sides})`

Area =
`(1)/(2)(6+6)(5)`

=
`(60)/(2)=30` cm²

Composite area of the figure = Area of a triangle + Area of the rectangle

Area of the triangle =
`(1)/(2)(Base)(height)`

=
`(1)/(2)(4-2)(9-5)`

=
`(1)/(2)(2)(4)`

= 4 m²

Area of rectangle = Length×width = 2×9 = 18m²

Area of the composite figure = 4 + 18 = 22 m²

Part 3). a.Coordinates of point x will be
`[(1)/(2)(x_(1)+x_(2)), (1)/(2)(y_(1)+y_(2))]`

=
`[(1)/(2)(0+2b), (1)/(2)(0+2c)]`

= (b, c)

b. Coordinates of Y will be =
`[(1)/(2)(x_(1)+x_(2)), (1)/(2)(y_(1)+y_(2))]`

=
`[(1)/(2)(2a+2b), (1)/(2)(0+2c)]`

= [(a + b), c]

c. Slope of XY =
`(y-y')/(x-x')`

=
`(c-c)/(a+b-b)`

= 0

d. Slope of PR =
`(y-y')/(x-x')`

=
`(0-0)/(2a-0)=0`