Question

A french fry stand at the fair serves their fries in paper cones. The cones have a radius of 22 inches and a height of 66 inches. It is a challenge to fill the narrow cones with their long fries. They want to use new cones that have the same volume as their existing cones but a larger radius of 44 inches. What will the height of the new cones be?

Answer 1

Answer:

The height of the new cones will be 16.5 inches.

Step-by-step explanation:

We know that,

The volume of a cone is,

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

Where, r is the radius of the cone,

h is the height of the cone,

In the original cone,

r = 22 inches,

h = 66 inches,

Thus, the volume would be,

$V_1=\frac{1}{3}\pi (22)^2(66)$

Also, for the new cone,

r = 44 inches,

Let H be the height,

So, the volume of the new cone would be,

$V_2=\frac{1}{3}\pi (44)^2(H)$

According to the question,

$V_2=V_1$

$\implies \frac{1}{3}\pi (44)^2(H)=\frac{1}{3}\pi (22)^2(66)$

$\implies H=\frac{22^2(66)}{44^2}=(\frac{22}{44})^2\times 66 = \frac{66}{4}=16.5$

Hence, the height of the new cones will be 16.5 inches.