Question

A rolling pin has two identical cylindrical handles attached to a larger handle that is 12 inches long. The large cylinder measures 5 inches in diameter and the handles measure 1.5 inches in diameter. What is the volume of the rolling pin in cubic inches? Round to the nearest tenth

Answer 1

Answer:

Volume of the rolling pin =  246.09 $in^{3}$  

Step-by-step explanation:

Length of larger cylinder L = 12 in

Diameter of larger cylinder D = 5 in

Diameter of two handles = 1.5 in

Length of two handles = 3 in

Volume of the rolling pin = Volume of two handles + Volume of larger cylinder

Volume of two handles = 2 ($\pi r^{2}L$)

Volume of two handles =  2 (3.14 × $0.75^{2}$ × 3 )

Volume of two handles = 10.59 $in^{3}$  

Volume of larger cylinder = ($\pi r^{2}L$)

Volume of larger cylinder = (3.14 × $2.5^{2}$ × 12 )

Volume of larger cylinder = 235.5 $in^{3}$  

Volume of the rolling pin =  10.59 + 235.5

Volume of the rolling pin =  246.09 $in^{3}$