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Scale Drawings


AA map has a scale of 1 inch = 25 miles and two cities on the map are 5 inches apart. Show how to find the actual distance between the two cities.


BIn a scale drawing (or model), the scale or scale factor is always the ratio of a distance in the drawing to the corresponding distance in the real world.


CAll models use a scale of 1 to 24. Find the missing dimensions.




Model		Model dimensionActual dimension

Car		Length: 7 inLength:

Tree		Height:Height:

Desk		Width: 2.5 inWidth:

Whale		Length: 2 ftLength:

House		Height:Height: 14.4 ft





DSolve each problem:
  1. A map has a scale of 0.5 in = 25 miles. What is the actual distance when the distance on the map is 2.2 inches?
  2. Jackson made a scale drawing of a nearby park. In real life, the park is 120 feet wide. In his drawing, it is 8 inches wide. What scale did Jackson use for his drawing?
  3. Betty's scale drawing of her house uses a scale of 1 in = 5 feet. The actual dimensions of her house are 55 feet by 80 feet. What are the dimensions of Betty's house on the drawing?

Verified Answer :

  • Map scale \(1 \text{ in} = 25 \text{ mi}\) leads to an actual distance of \(25s\) miles for a map distance of \(s\).
  • Using the \(1:24\) scale, actual dimensions are found by multiplying model dimensions by \(24\), resulting in a \(168 \text{ in}\) car, \(60 \text{ in}\) desk, \(48 \text{ ft}\) whale, and \(7.2 \text{ in}\) house height.
  • The map distance problem yields \(\boxed{110 \text{ miles}}\), the park scale is \(\boxed{1 \text{ inch} = 15 \text{ feet}}\), and the house dimensions are \(\boxed{11 \text{ in} \times 16 \text{ in}}\).

Explanation

  1. Map Scale Calculations
ATo find the actual distance on a map with a scale of \(1 \text{ inch} = 25 \text{ miles}\), we use a proportion. If the map distance is \(s\) inches, the actual distance \(D_{actual}\) in miles satisfies the ratio \(\frac{1 \text{ inch}}{25 \text{ miles}} = \frac{s \text{ inches}}{D_{actual}}\). Solving for \(D_{actual}\), we get \(D_{actual} = 25 \times s\) miles.
  1. Scaling Dimensions (1:24)

B & C. We are given a scale factor of 1:24, meaning 1 unit in the model corresponds to 24 units in the real world (Actual = 24 \(\times\) Model).

  • Car: Length is \(7 \text{ in} \times 24 = 168 \text{ inches}\) (or \(14 \text{ feet}\)).
  • Desk: Width is \(2.5 \text{ in} \times 24 = 60 \text{ inches}\) (or \(5 \text{ feet}\)).
  • Whale: Length is \(2 \text{ ft} \times 24 = 48 \text{ feet}\).
  • House: Height is \(14.4 \text{ ft} \div 24 = 0.6 \text{ feet}\) (or \(7.2 \text{ inches}\)).
  • Tree: Since no dimensions were provided for the tree, any set of dimensions satisfying the 1:24 ratio (e.g., \(0.5 \text{ ft}\) model for a \(12 \text{ ft}\) tree) would be correct.
  1. Problem Solving Scale Problems

D.1. Using the map scale of \(0.5 \text{ in} = 25 \text{ miles}\), the actual distance for \(2.2 \text{ inches}\) is found by the proportion \(\frac{0.5}{25} = \frac{2.2}{D}\). Thus, \(D = \frac{2.2 \times 25}{0.5} = 110 \text{ miles}\).
D.2. For the park, the scale is \(\frac{8 \text{ inches}}{120 \text{ feet}}\). Simplifying, \(\frac{8}{120} = \frac{1}{15}\), so the scale is \(1 \text{ inch} = 15 \text{ feet}\).
D.3. For the house with scale \(1 \text{ in} = 5 \text{ feet}\), dimensions of \(55 \text{ feet}\) by \(80 \text{ feet}\) become \(55 \div 5 = 11 \text{ inches}\) by \(80 \div 5 = 16 \text{ inches}\).

Examples

Scale drawings are essential for architecture and engineering. For instance, if an architect designs a house layout on paper with a scale of \(1 \text{ inch} = 4 \text{ feet}\), they can easily calculate exactly how much physical space a wall will take up in the actual building, ensuring that furniture fits perfectly and the structure is safe. This skill allows us to visualize massive real-world projects on a small, manageable scale.

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