Respuesta :
Answer:
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))