Two cylinders have the same volume. If the radius of cylinder I is 3 times the radius of cylinder II, then the height of cylinder II is how many times the height of cylinder I?
A. 12
B. 9
C. 6
D. 3
Please explain how, if possible.
Cylinder 1 has nine times the base area of cylinder 2.
Volume is (base area) x (height)
For the volumes to be equal, cylinder 2 must have 9 times the height of cylinder 2.
msh fahma 7aga
To solve this question, we can use the formula for the volume of a cylinder, which is given by:
V = π * r^2 * h
where V is the volume, r is the radius, and h is the height.
Let's assume that the radius of cylinder II is "x" units. Therefore, the radius of cylinder I is 3 times the radius of cylinder II, which is 3x units.
Since both cylinders have the same volume, we can equate their volumes and solve for the height of cylinder II.
For cylinder I:
V1 = π * (3x)^2 * h1
For cylinder II:
V2 = π * x^2 * h2
Since V1 = V2 (the volumes are equal):
π * (3x)^2 * h1 = π * x^2 * h2
Canceling out the π and x^2 on both sides of the equation, we are left with:
(3x)^2 * h1 = x^2 * h2
Simplifying the equation:
9x^2 * h1 = x^2 * h2
We can divide both sides by x^2 to get rid of x^2:
9h1 = h2
Therefore, the height of cylinder II (h2) is 9 times the height of cylinder I (h1).
Hence, the answer is B. 9
pi*r^2*h2 = pi*(3r)^2*h1.
pi*r^2*h2 = pi*9r^2*h1
Divide both sides by pi*r^2:
h2 = 9*h1.