Question

Cones A and B both have volume cubic 48 π units, but have different dimensions. Cone A has radius 6 units and height 4 units. Find one possible radius and height for Cone B. Explain how you know Cone B has the same volume as Cone A.

Answer 1

Answer:

A possible solution is that radius of cone B is 2 units and height is 36 units

Step-by-step explanation:

The volume of a cone is given by

$V=\frac{1}{3}\pi r^2 h$

where

r is the radius

h is the height

Here we are told that both cones A and B have the same volume, which is:

$V_A=\frac{1}{3}\pi r_A^2 h_A = 48 \pi$

And

$V_B=\frac{1}{3}\pi r_B^2 h_B = 48 \pi$ (2)

We also know that cone A has radius 6 units:

$r_A=6$

and height 4 units:

$h_A=4$

For cone B, from eq.(2), we get

$\frac{1}{3}r_B^2 h_B = 48\\r_B^2 h_B = 48\cdot 3 =144$

One possible solution for this equation is

$r_B = 2\\h_B = 36$

In fact in this case, we get:

$r_B^2 h_B = (2)^2\cdot 36 = 4\cdot 36 = 144$

Therefore a possible solution is that radius of cone B is 2 units and height is 36 units, and we know that in this case Cone B has the same volume as cone A because it is told by the problem.

Answer 2

The possible dimensions of cone B will be; the radius of 3 units and height of 16 units.

Given information:

Cones A and B both have volume 48 π cubic units, but have different dimensions.

Cone A has a radius of 6 units and a height of 4 units.

It is required to find the possible values of radius and height of cone B.

Let the height of cone B be 16 units.

So, the radius of the cone B will be calculated as,

$V_B=\dfrac{1}{3}\pi r^2h\\48\pi=\dfrac{1}{3}\pi r^2\times 16\\9=r^2\\r=3$

Therefore, the possible dimensions of cone B will be; the radius of 3 units and height of 16 units.

For more details, refer to the link:

https://brainly.com/question/1984638