Question

A conical container can hold 120π cubic centimeters of water. The diameter of the base of the container is 12 centimeters.
The height of the container is centimeters. If its diameter and height were both doubled, the container's capacity would be times its original capacity.

Answer 1

Answer:

A. 10cm

B. 8 times

Step-by-step explanation:

The question is on volume of a conical container

Volume of a cone= $\pi r^{2} h/3$

where r is the radius of base and h is the height of the cone

Given diameter= 12 cm, thus radius r=12/2 =6 cm

$v=\pi r^2h/3 \\120\pi =\pi *6*6*h/3\\120\pi =12\pi h\\10=h$

h=10 cm

B.

If height and diameter were doubled

New height = 2×10 =20 cm

New diameter = 2×12 = 24, r=12 cm

volume = $v=\pi r^2h/3\\v=\pi *12*12*20/3\\v=960\pi$

To find the number of times we divide new volume with the old volume

$N= 960\pi /120\pi \\\\N= 8$

Answer 2

Answer: The height of the container is 10 centimeters. If its diameter and height were both doubled, the container's capacity would be 8 times its original capacity.

Step-by-step explanation:

The volume of a cone can be calculated with this formula:

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

Where "r" is the radius and "h" is the height.

We know that the radius is half the diameter. Then:

$r=\frac{12cm}{2}=6cm$

We know the volume and the radius of the conical container, then we can find "h":

$120\pi cm^3=\frac{\pi (6cm)^2h}{3}\\\\(3)(120\pi cm^3)=\pi (6cm)^2h\\\\h=\frac{3(120\pi cm^3)}{\pi (6cm)^2}\\\\h=10cm$

The diameter and height doubled are:

$d=12cm*2=24cm\\h=10cm*2=20cm$

Now the radius is:

$r=\frac{24cm}{2}=12cm$

And the container capacity is

$V=\frac{\pi (12cm)^2(20cm)}{3}=960\pi cm^3$

Then, to compare the capacities, we can divide this new capacity by the original:

 $\frac{960\pi cm^3}{120\pi cm^3}=8$

Therefore,  the container's capacity would be 8 times its original capacity.