Curriculum Alignment
- Strand
- 2.0 Measurements and Geometry
- Sub-Strand
- 2.1 Similarity and Enlargement
- Specific Learning Outcomes
- Apply similarity and enlargement to real-life situations
Subsection 2.1.4 Similarity and Enlargement in Real-Life
Teacher Resource 2.1.64.
To assist your teaching, we have prepared lesson resources, aligned with this textbook and the CBC. The Lesson Plan links to syllabus learning outcomes and provides suggest time allocations. The Step-by-Step Guide provides more detailed guidance on how to teach the content, including suggested questions to ask learners, and possible answers.
Learner Experience 2.1.11.
A local community is planning to design a new park. The architects have created a small blueprint of the park that measures \(10\) feet by \(15\) feet. They need to enlarge the blueprint to an actual size that fits a vacant lot that is \(200\) feet by \(300\) feet. The architects will use a scale factor to determine how much they need to enlarge their blueprint to fit the available space while maintaining the same proportions and layout.
Work in pairs:
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Write down the dimensions of the blueprint and the dimensions of the actual park.
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Calculate the linear scale factor (LSF) by dividing the actual length by the blueprint length. Verify by dividing the actual width by the blueprint width. Are they the same?
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Calculate the area of the blueprint and the area of the actual park.
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Find the area scale factor (ASF) by dividing the area of the actual park by the area of the blueprint.
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Compare the ASF with the square of the LSF. What do you notice?
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The blueprint shows a rectangular swimming pool measuring \(1\) foot by \(2\) feet by \(0.5\) feet deep. Using the LSF, calculate the actual dimensions of the swimming pool.
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Calculate the volume of the blueprint pool and the volume of the actual pool.
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Find the volume scale factor (VSF) by dividing the actual pool volume by the blueprint pool volume.
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Compare the VSF with the cube of the LSF. What do you notice?
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Discuss: How do these scale factors affect the amount of materials, water, and cost needed for the real park? Share your findings with the class.
Similarity and enlargement are key concepts in geometry, particularly when it comes to shapes, figures, and their properties. Here are some practical applications of these concepts:
(i)\(\textbf{Architecture and design:}\)
In architecture, architects often use enlargement and similarity to create scale models of buildings, bridges, or other structures. By maintaining the same proportions but adjusting the size, architects can test designs or present ideas in a more manageable form.
Similarity is used to create blueprints for large structures. These blueprints are scaled-down versions of the final building, preserving the ratio of dimensions in the original structure.
Suppose an architect is designing a scale model of a building. The actual building is \(150\, m\) tall, and the architect creates a scale model where the height of the model is \(0.75\, m\text{.}\)To find the scale factor, use the formula for similarity:
scale factor \(=\frac{\text{Height of model}}{\text{Height of actual building}}\)
Scale factor \(=\frac{0.75}{150}=\frac{1}{200}\)
So, the scale factor is \(1:200\text{,}\) meaning the model is \(\frac{1}{200}\) the size of the actual building. This scale factor ensures that the proportions are similar.
(ii) \(\textbf{Map scale:}\)
Let’s say a map is drawn with a scale of \(1:50,000\text{,}\) meaning \(1\) unit on the map represents \(50,000\) units in real life. If a road is \(4\, cm\) long on the map, how long is the actual road?
We can calculate the actual length using the scale factor:
Actual Length = Length on map \(\times\) scale factor
Actual length \(=4\, cm \times 50,000= 200,000\, cm\)
Key Takeaway 2.1.65.
When architects enlarge a blueprint to create the actual structure, every dimension is multiplied by the same linear scale factor. This ensures that the enlarged design is SIMILAR to the original blueprint all proportions and the layout are preserved. The area and volume, however, do not scale by the same factor as the length.
| Dimension | Scale Factor | Park Design Example (LSF = 20) |
|---|---|---|
| Length (1D) | Linear Scale Factor (LSF) = k k = Actual length \(\div\) Blueprint length |
\(k = 200 \div 10 = 20\) A 1 ft path on the blueprint becomes 20 ft |
| Area (2D) |
\(\text{Area Scale Factor (ASF)} = k^2\) ASF \(=\) Actual area \(\div\) Blueprint area |
ASF = \(20^2 = 400\) Blueprint area \(150\) sq ft Actual area \(60,000\) sq ft. Affects: land coverage, paving, grass, garden beds. |
| Volume (3D) |
\(\text{Volume Scale Factor (VSF)} = k^3\) VSF \(=\) Actual volume \(\div\) Blueprint volume |
VSF = \(20^3 = 8000\) Blueprint pool \(1\) cu ft Actual pool \(8000\) cu ft. Affects: water features, soil, concrete, excavation. |
Checkpoint 2.1.66.
Checkpoint 2.1.67.
Exercises Exercises
1.
An architect is creating a scale model of a building. The actual height of the building is \(120\) meters, and the height of the model is \(0.6\) meters.
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What is the scale factor of the model?
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If the width of the actual building is \(50\) meters, what is the width of the model?
2.
A map scale is given as \(1:25,000\text{,}\) meaning \(1\) cm on the map represents \(25,000\) cm in real life.
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A river on the map measures \(8\) cm in length. What is the actual length of the river in kilometers?
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If a road on the map measures \(12.5\) cm, how long is the actual road in meters?
3.
A photograph has a size of \(5\) cm by \(7\) cm. It needs to be enlarged so that the width becomes \(20\) cm. The height will also increase proportionally. What is the new height of the photograph after the enlargement?
4.
A pole of height \(2.4\) meters casts a shadow of length \(1.6\) meters. A tree casts a shadow of length \(12\) meters.
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Using the concept of similar triangles, find the height of the tree.
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If the tree’s shadow increases to \(15\) meters, what would be the new height of the tree, assuming the proportion remains the same?
Answer.
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\begin{gather*} \frac{\text{pole height}}{\text{pole shadow}} = \frac{\text{tree height}}{\text{tree shadow}}\\ \frac{2.4}{1.6} = \frac{h}{12}\\ h = \frac{2.4}{1.6} \times 12 = 1.5 \times 12 = 18\; \text{m} \end{gather*}
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\(\displaystyle h = \frac{2.4}{1.6} \times 15 = 1.5 \times 15 = 22.5\; \text{m}\)
