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

The height of the new cones will be 16.5 inches.

Step-by-step explanation:

To find the height of the new cones, we can use the formula for the volume of a cone:

V = (1/3)πr^2h

Let's denote the radius of the new cones as r2 and the height of the new cones as h2. We know that the volume of the new cones will be the same as the volume of the existing cones:

(1/3)π(r2)^2h2 = (1/3)π(r1)^2h1

Substituting the given values, r1 = 22 inches (radius of existing cones) and h1 = 66 inches (height of existing cones), we can solve for h2:

(1/3)π(44)^2h2 = (1/3)π(22)^2(66)

Dividing both sides by (1/3)π(44)^2, we get:

h2 = h1(r1/r2)^2

Plugging in the values, h1 = 66 inches and r1 = 22 inches, and r2 = 44 inches, we can calculate:

h2 = 66(22/44)^2

h2 = 66(1/2)^2

h2 = 66(1/4)

h2 = 66/4

h2 = 16.5 inches

Answer 2

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=(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=(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=(1)/(3)\pi (44)^2(H)`

According to the question,

`V_2=V_1`

`\implies (1)/(3)\pi (44)^2(H)=(1)/(3)\pi (22)^2(66)`

`\implies H=(22^2(66))/(44^2)=((22)/(44))^2* 66 = (66)/(4)=16.5`

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