Question

How does the volume of a cone change when the radius is quadrupled and the height id reduced to 1/5 of the original size?

Answer 1

The volume of a cone changes to 16/15 times the original volume when the radius is quadrupled and the height is reduced to one-fifth.

Step-by-step explanation:

The question asks how the volume of a cone changes when the radius is quadrupled and the height is reduced to one-fifth. The formula for the volume of a cone is V = (1/3)πr²h, where r is the radius and h is the height of the cone.

If we let the original radius be r and the original height be h, the original volume is V = (1/3)πr²h. Quadrupling the radius changes it to 4r, and reducing the height to one-fifth changes it to (1/5)h. The new volume is V' = (1/3)π(4r)²(1/5)h = (1/3)π(16r²)(1/5)h = (16/15)(1/3)πr²h.

Therefore, the new volume is 16/15 times the original volume. The volume increases by this factor when the radius is quadrupled and the height is reduced to one-fifth.

Answer 2

The volume (V) of the cone is one-third of the product of its height (h) and area (A).
V = A x h = πr²h
Now, if the radius is quadrupled and the height is reduced to 1/5, the equation will be,
V2 = π(4r)²(1/5 h) = π(3.2)r²h
Thus, the second volume would be 3.2 time the first volume.