Answer:
Volume refers to the amount of space inside a 3 - dimensional figure. This space is computed by multiplying the dimensions of the given figure. For instance, if the given figure is a sphere, the volume is four - third of the product of pi and the cube of its radius. For the cylinder, the formula is V = πr²h. For the pyramid, the volume is one - third of the product of the area of the base and height.
Answers:
307.72 m³
33 493.33 dm³
1 025.73 m³
7 in³
Solutions:
1. Asked: The volume of the cylinder.
Given: r = 3.5 m, h = 8 m, π = 3.14
Operation: multiplication
Number Sentence: V = πr²h
Solutions: V = πr²h
V = (3.14)(3.5 m)²(8 m)
V = (3.14)(12.25 m²)(8 m)
V = (38.465 m²)(8 m)
Answer: V = 307.72 m³
2. Asked: The volume of the sphere.
Given: r = 20 dm, π = 3.14
Operation: multiplication
Number Sentence: V = \frac{4}{3}
3
4
πr³
Solutions: V = \frac{4}{3}
3
4
πr³
V = \frac{4}{3}
3
4
(3.14)(20 dm)³
V = \frac{4}{3}
3
4
(3.14)(8000 dm³)
V = \frac{4}{3}
3
4
(25 120 dm³)
Answer: V = 33 493.33 dm³
3. Asked: The volume of the cone.
Given: h = 20 m, r = 7 m, π = 3.14
Operation: multiplication
Number Sentence: V = \frac{1}{3}
3
1
πr²h
Solutions: V = \frac{1}{3}
3
1
(3.14)(7 m)²(20 m)
V = \frac{1}{3}
3
1
(3.14)(49 m²)(20 m)
V = \frac{1}{3}
3
1
(3.14)(980 m³)
V = \frac{1}{3}
3
1
(3 077.2 m³)
Answer: V = 1 025.73 m³
4. Asked: The volume of the rectangular pyramid.
Given: L = 3.5 in, W = 1.5 in, H = 4 in
Operation: multiplication
Number Sentence: V = \frac{1}{3}
3
1
Bh
Solutions: V = \frac{1}{3}
3
1
(3.5 in)(1.5 in)(4 in)
V = \frac{1}{3}
3
1
(5.25 in²)(4 in)
V = \frac{1}{3}
3
1
(21 in³)
Answer: V = 7 in³
Things to Remember:
There are special relations that exists between volumes of 3 - dimensional figures.
