Exercise 29 Page 731

Practice makes perfect

a Let's begin by drawing the original cone. Its radius is 3 ft, its height is 10 ft, and its volume is 30π ft^3.

Next, we will double the radius. Therefore, our new cone has a radius of 6 ft.

To find the volume of the latter cone, let's substitute r=6 and h=10 into the formula.

V = 1/3π r^2h

r= 6, h= 10

V = 1/3π* 6^2* 10

Calculate power

V = 1/3π* 36* 10

V = 120π

The volume of the new cone is 120π ft^3. Therefore, the volume of the original cone was quadrupled.

b In this part, we start with the first cone we drew in Part A and we will double its height. That is, the height of this cone is h=20 ft.

To find the volume of this cone, we substitute r=3 and h=20 into the formula.

V = 1/3π r^2h

r= 3, h= 20

V = 1/3π * 3^2* 20

Calculate power

V = 1/3π * 9* 20

V = 60π

The volume of this cone is 60π ft^3. That is, the volume of the original cone was doubled.

c Finally, we will start with the original cone and will double its radius and height. The new radius is 6 ft and the new height is 20 ft.

As we've done before, we will substitute r=6 and h=20 into the formula to find the volume of the latter cone.

V = 1/3π r^2h

r= 6, h= 20

V = 1/3π* 6^2 * 20

Calculate power

V = 1/3π* 36 * 20

V = 240π

The volume of this final cone is 240 π ft^3 which is 8 times the volume of the original one.

Lesson Check p.729

Practice and Problem-Solving Exercises p.729-732

Standardized Test Prep p.732

Mixed Review p.732

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