a. To solve the first part, we are going to use the formula for the surface area of a sphere:
where
is the surface area of the sphere
is the radius of the sphere
We know from our problem that
; so lets replace that value in our formula:
To solve the second part, we are going to use the formula for the volume of a sphere:
Where
is the volume of the sphere
is the radius
We know form our problem that
, so lets replace that in our formula:
We can conclude that the surface area of the sphere is 314.16 square feet and its volume is 523.6 cubic feet.
b. To solve the first part, we are going to use the formula for the surface area of a square pyramid:
where
is the surface area
is the measure of the base
is the height of the pyramid
We know form our problem that
and
, so lets replace those value sin our formula:
To solve the second part, we are going to use the formula for the volume of a square pyramid:
where
is the volume
is the measure of the base
is the height of the pyramid
We know form our problem that
and
, so lets replace those value sin our formula:
We can conclude that the surface area of our pyramid is 266.39 square feet and its volume is 256 cubic feet.
c. To solve the first part, we are going to use the formula for the surface area of a circular cone:
where
is the surface area
is the radius of the circular base
is the height of the cone
We know form our problem that
and
, so lets replace those values in our formula:
![A= \pi (5ft)[(5ft)+ \sqrt{(8ft)^2+(5ft)^2}]](https://tex.z-dn.net/?f=A%3D%20%5Cpi%20%285ft%29%5B%285ft%29%2B%20%5Csqrt%7B%288ft%29%5E2%2B%285ft%29%5E2%7D%5D)
To solve the second part, we are going to use the formula for the volume os a circular cone:
where
is the volume
is the radius of the circular base
is the height of the cone
We know form our problem that
and
, so lets replace those values in our formula:
We can conclude that the surface area of our cone is 226.73 square feet and its surface area is 209.44 cubic feet.
d. To solve the first part, we are going to use the formula for the surface area of a rectangular prism:
where
is the surface area
is the width
is the length
is the height
We know from our problem that
,
, and
, so lets replace those values in our formula:
![A=2[(6ft)(10ft)+(16ft)(10ft)+(16ft)(6ft)]](https://tex.z-dn.net/?f=A%3D2%5B%286ft%29%2810ft%29%2B%2816ft%29%2810ft%29%2B%2816ft%29%286ft%29%5D%20)
To solve the second part, we are going to use the formula for the volume of a rectangular prism:
where
is the volume
is the width
is the length
is the height
We know from our problem that
,
, and
, so lets replace those values in our formula:
We can conclude that the surface area of our solid is 632 square feet and its volume is 960 cubic feet.
e. Remember that a face of a polygon is a side of polygon.
- A sphere has
no faces.
- A square pyramid has
5 faces.
- A cone has
1 face.
- A rectangular prism has
6 faces.
Total faces: 5 + 1 + 6 =
12 faces
<span>We can conclude that there are 12 faces in on the four geometric shapes on the holes.
</span>
f. Remember that an edge is a line segment on the boundary of the polygon.
-
A sphere has
no edges.
- A cone has no
edges.
- A rectangular pyramid has
8 edges.
- A rectangular prism has
12 edges.
Total edges: 8 + 20 =
20 edges
Since we have 20 edges in total, we can conclude that your boss will need 20 brackets on the four shapes.
g. Remember that the vertices are the corner points of a polygon.
-
A sphere has
no vertices.
- A cone has no vertices.
- A rectangular pyramid has
5 vertices.
- A rectangular prism has
8 vertices.
Total vertices: 5 + 8 =
13 vertices
We can conclude that there are 0 vertices for the sphere and the cone; there are 5 vertices for the pyramid, and there are are 8 vertices for the solid (rectangular prism). We can also conclude that your boss will need 13 brackets for the vertices of the four figures.
