Question

After painting his porch, Jamil has \dfrac14 4 1 ​ start fraction, 1, divided by, 4, end fractionof a can of paint remaining. The can has a radius of 888 cm and a height of 202020 cm. He wants to pour the remaining paint into a smaller can for storage. The smaller can has a radius of 555 cm. What does the height of the smaller can need to be to hold all of the paint?

Answer 1

The volume of the remaining paint is:
V1 = (1/4) * (pi) * (r ^ 2) * (h)
Where,
r: radio
h: height
Substituting values we have:
V1 = (1/4) * (3.14) * (8 ^ 2) * (20)
V1 = 1004.8 cm ^ 3
The smallest can volume is:
V2 = pi * r ^ 2 * h
Substituting values:
V2 = (3.14) * 5 ^ 2 * h
Matching the volumes:
V2 = V1
(3.14) * 5 ^ 2 * h = 1004.8
Clearing h:
h = (1004.8) / ((3.14) * 5 ^ 2)
h = 12.8 cm
Answer:
The height of the smallest part of the paint is:
h = 12.8 cm