Given:
The scale factor of the actual sports utility vehicle to the toy model is 50:1.
Actual length = 7 feet
Actual width = 5 feet
To find:
The length and width of the toy model.
Solution:
a. Let l be the length of the toy model, then
[tex]\dfrac{\text{Actual length}}{\text{Length of the toy model}}=\dfrac{50}{1}[/tex]
[tex]\dfrac{7}{l}=\dfrac{50}{1}[/tex]
[tex]7=50l[/tex]
[tex]\dfrac{7}{50}=l[/tex]
[tex]0.14=l[/tex]
Therefore, the length of the toy model is 0.14 feet.
b. Let w be the width of the toy model, then
[tex]\dfrac{\text{Actual width}}{\text{Width of the toy model}}=\dfrac{50}{1}[/tex]
[tex]\dfrac{5}{w}=\dfrac{50}{1}[/tex]
[tex]5=50w[/tex]
[tex]\dfrac{5}{50}=w[/tex]
[tex]0.10=w[/tex]
Therefore, the width of the toy model is 0.10 feet.
