Question

A conical container can hold 120π cubic centimeters of water. The diameter of the base of the container is 12 centimeters.

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.