Example 2:

An artist made an aluminum scale model of a sculpture.  The scale model is 2 feet tall, and the actual sculpture, also made out of aluminum, is 8 feet tall.

a.  The scale model weighs 20 lbs.  What is the weight of the actual sculpture?

b.  The area of a face of the cube in the scale model is 16 in².  What is the area of the corresponding face in the actual sculpture?

c.  What is the volume of the cube in the actual sculpture?

 

Solution:

The scaling factor is 4 since 4 ∙ 2 ft = 8 ft.

a.  The weight is proportional to the volume, and the volume is scaled by a factor of 4³ = 64, so the weight of the actual sculpture is 64 ∙ 20 lbs = 1280 lbs.

b.  Areas are scaled by a factor of 4² = 16, so the area of the corresponding face in the actual sculpture is 16 ∙ 16 in² = 256 in².

c.  Since the area of a face of the cube in the scale model is 16 in², each side is 4 in.  In the actual sculpture, each face of the corresponding cube would therefore be 4 ∙ 4 in = 16 in, so the volume of that cube would be (16 in)³ = 4096 in³.

toc | return to top | previous page