PLEASE HELP Find the area and the perimeter of the shaded regions below. Give your answer as a completely simplified exact value in terms of π (no approximations). The figures below are based on semicircles or quarter circles and problems b), c), and d) are involving portions of a square.

perimeter is  4 sqrt(29) + 4pi  cm

area is 40 + 8pi cm^2

Step-by-step explanation:

We have a semicircle and a triangle

First the semicircle with diameter 8

A = 1/2 pi r^2 for a semicircle

r = d/2 = 8/2 =4

A = 1/2 pi ( 4)^2

  =1/2 pi *16

  = 8pi

Now the triangle with base 8 and height 10

A = 1/2 bh

  =1/2 8*10

  = 40

Add the areas together

A = 40 + 8pi cm^2

Now the perimeter

We have 1/2 of the circumference

1/2 C =1/2 pi *d

         = 1/2 pi 8

        = 4pi

Now we need to find the length of the hypotenuse of the right triangles

using the pythagorean theorem

a^2+b^2 = c^2

The base is 4 ( 1/2 of the diameter) and the height is 10

4^2 + 10 ^2 = c^2

16 + 100 = c^2

116 = c^2

sqrt(116) = c

2 sqrt(29) = c

Each hypotenuse is the same so we have

hypotenuse + hypotenuse + 1/2 circumference

2 sqrt(29) + 2 sqrt(29) + 4 pi

4 sqrt(29) + 4pi  cm

Step-by-step explanation:

First we need to deal with the half circle. The radius of this circle is 4, because the diameter is 8. The formula for the circumference of a circle is 2piR.

2pi4 so the perimeter for the half circle would be 8pi/2.

The area of that half circle would be piR^2 so 16pi/2.

Now moving on the triangle part, we need to find the hypotenuse side of AC. We will use the pythagoram theorem. 4^2+10^2=C^2

16+100=C^2

116=C^2

C=sqrt(116)

making the perimeter of this triangle 2×sqrt(116)

The area of this triangle is 8×10=80, than divided by 2 which is equal to 40.

We than just need to add up the perimeters and areas for both the half circle and triangle.

The area would be equal to 8pi+40

The perimeter would be equal to 4pi+4(sqrt(29))

trapezoid

step-by-step explanation:

first i plotted all my points on a grid graph

i made sure all of them were in the correct quadrant

then i saw it was a trapezoid or a quadrilateral