Rectangular Parallelepiped Problem, 8

Category: Solid Geometry, Physics

"Published in Newark, California, USA"

A tank, open at the top, is made of sheet iron 1 in. thick. The internal dimensions of the tank are 4 ft. 8 in. long; 3 ft. 6 in. wide; 4 ft. 4 in. deep. Find the weight of the tank when empty, and find the weight when full of salt water. (Salt water weighs 64 lb. per cu. ft., and iron is 7.2 times as heavy as salt water).

Solution:

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

Photo by Math Principles in Everyday Life

There are two rectangular parallelepiped in the figure, the outside dimensions and the inside dimensions. The volume of a tank that is made of sheet iron is equal to the difference of the two rectangular parallelepiped.

For the outside dimensions, the volume of a rectangular parallelepiped is

For the inside dimensions, the volume of a rectangular parallelepiped which is also the volume of a salt water is

Hence, the volume of a tank is

Therefore, the weight of the empty tank is

The weight of the salt water is

Therefore, the weight of a tank filled with salt water is

 

Rectangular Parallelepiped Problem, 7

Category: Solid Geometry

"Published in Newark, California, USA"

How many cubic yards of material are needed for the foundation of a barn 40 ft. by 80 ft., if the foundation is 2 ft. thick and 12 ft. high?

Solution: 

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

Photo by Math Principles in Everyday Life

There are two rectangular parallelepiped in the figure, the outside dimensions and the inside dimensions. Their height is the same. The volume or the amount of material needed for the foundation of a barn is equal to the difference of the two rectangular parallelepiped. 

For the outside dimensions, the volume of a rectangular parallelepiped is

For the inside dimensions, the volume of a rectangular parallelepiped is

Hence, the volume or the amount of material needed for the foundation of a barn is

Therefore, the volume in cubic yards is equal to

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