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
Obtuse quadrilaateral. midpoint(-1,3) with slooe 1/2
distance 2
Answer:
All the values of k are -4, -3, -2, -1, 0 , 1
Step-by-step explanation:
Given;
- 4≤ k ≤ 1
The range of value of k is between -4 and 1.
These values are listed as follows;
k = -4, -3, -2, -1, 0 , 1
Therefore, all the values of k are -4, -3, -2, -1, 0 , 1
Answer:
A right circular cone
Step-by-step explanation:
