Rectangular Parallelepiped Problem, 6

Category: Solid Geometry

"Published in Newark, California, USA"

A packaging box 2.2 ft. by 4.9 ft. by 5.5 ft. is to be completely covered with tin. How many square feet of the metal are needed? (Neglect waste for seams, etc.)

Solution:

To illustrate the problem, it is better to draw the figure as follows

Photo by Math Principles in Everyday Life

If a packaging box is completely covered with tin, then the amount of metal used is equal to the total area of a packaging box. The total area of a packaging box is equal to

Rectangular Parallelepiped Problem, 5

Category: Solid Geometry

"Published in Newark, California, USA"

Building bricks are closely stacked in a pile 7 ft. high, 36 ft. long, and 12 ft. wide. If the bricks are 2 in. by 4 in. by 9 in., how many bricks are in the pile?

Solution:

To illustrate the problem, it is better to draw the figure as follows

Photo by Math Principles in Everyday Life

Let n = be the number of bricks in a pile
     V₁ = be the volume of a brick
     V₂ = be the volume of a pile of bricks

The volume of a brick is equal to

   
The volume of a pile of bricks is equal to
 

In cubic inches, the volume of a pile of bricks is

Therefore, the number of bricks in a pile is

Rectangular Parallelepiped Problem, 4

Category: Solid Geometry

"Published in Vacaville, California, USA"

Compute the cost of the lumber necessary to resurface a foot-bridge 16 ft. wide, 150 ft. long with 2-in. planks, if lumber is $40 per 1000 board feet. Neglect waste. (One board foot = 1 ft. by 1 ft. by 1 in.)

Solution:

To illustrate the problem, it is better to draw the figure as follows

Photo by Math Principles in Everyday Life

The volume of a rectangular parallelepiped is given by the formula

Substitute the values of length, width, and height of a lumber to resurface a foot-bridge, we have 

Therefore, the cost of the lumber necessary to resurface a foot-bridge is