Unit 1 Study Guide: Scaled Copies and Scale Drawings

Overview of Scaled Copies and Scale Factors

• A scaled copy is a figure that has been enlarged or reduced in size proportionally relative to an original figure.
• Corresponding side lengths of scaled copies are proportional, scaled by a constant scale factor $k$.
• Corresponding angles in scaled copies are congruent and retain identical degree measures regardless of the scale factor $k$.
• When a two-dimensional figure is scaled by a factor of $k$, its perimeter scales by $k$, and its area scales by $k^2$

Problem 1: Analyzing Scaled Polygon Properties

• Polygon $LMNOP$ is a scaled copy of polygon $ABCDE$.

• Side lengths of polygon $ABCDE$:

  • Segment $AE = 3$
  • Segment $AB = 6$
  • Segment $BC = 1.5$
  • Segment $CD = 2.5$
  • Segment $ED = 5$

• Side length of polygon $LMNOP$:

  • Segment $LP = 9$

• Scale factor calculation ($k$):   $k = \frac{\text{length of } LP}{\text{length of } AE} = \frac{9}{3} = 3$

• Evaluation of Statements:

  • Statement a: "Segment $AE$ is three times as long as segment $LP$"
  • Decision: False.
  • Explanation: Segment $LP$ is three times as long as segment $AE$ ($LP = 3 \times AE = 3 \times 3 = 9$). Segment $AE$ is one third the length of segment $LP$ ($AE = \frac{1}{3} \times LP = \frac{1}{3} \times 9 = 3$).
  • Statement b: "Segment $LM$ is three times as long as segment $AB$"
  • Decision: True.
  • Explanation: Because $LMNOP$ is scaled by a scale factor of $k = 3$, every side length in $LMNOP$ is three times the corresponding side length in $ABCDE$. Thus, $LM = 3 \times AB = 3 \times 6 = 18$
  • Statement c: "The measure of angle $ABC$ is one third the measure of angle $LMN$"
  • Decision: False.
  • Explanation: Scaled copies preserve angle measures. Corresponding angles are congruent, so $\text{measure of } \angle ABC = \text{measure of } \angle LMN$
  • Statement d: "The length of segment $LM$ is 12 units"
  • Decision: False.
  • Explanation: Segment $LM$ corresponds to segment $AB$, which has a length of $6$. Calculating length: $LM = 3 \times AB = 3 \times 6 = 18$
  • Statement e: "The length of segment $NO$ is 7.5 units"
  • Decision: True.
  • Explanation: Segment $NO$ corresponds to segment $CD$, which has a length of $2.5$. Calculating length: $NO = 3 \times CD = 3 \times 2.5 = 7.5$
  • Statement f: "The area of $LMNOP$ is nine times the area of $ABCDE$"
  • Decision: True.
  • Explanation: When a figure is scaled by a factor of $k$, its area is scaled by $k^2$. Since $k = 3$, the area scales by $3^2 = 9$

• Summary of Correct Statements: Statements b, e, and f are true.

Problem 2: Identifying Scaled Copies of Rectangles

• Rectangle $A$ measures $8\,\text{inches}$ by $4\,\text{inches}$, giving an aspect ratio of:   $\text{Ratio} = \frac{8}{4} = 2$

• Evaluation of Candidate Dimensions for Rectangle $B$:

  • Option A ($16\,\text{inches}$ by $2\,\text{inches}$):
  • Ratio: $\frac{16}{2} = 8 \neq 2$
  • Scale factors: Length factor is $\frac{16}{8} = 2$, width factor is $\frac{2}{4} = 0.5$. Factors are unequal, so this is not a scaled copy.
  • Option B ($2\,\text{inches}$ by $1\,\text{inch}$):
  • Ratio: $\frac{2}{1} = 2$
  • Scale factor: $k = \frac{2}{8} = \frac{1}{4} = 0.25$. This is a valid scaled copy.
  • Option C ($40\,\text{inches}$ by $20\,\text{inches}$):
  • Ratio: $\frac{40}{20} = 2$
  • Scale factor: $k = \frac{40}{8} = 5$. This is a valid scaled copy.
  • Option D ($10\,\text{inches}$ by $6\,\text{inches}$):
  • Ratio: $\frac{10}{6} = \frac{5}{3} \approx 1.67 \neq 2$. This is not a scaled copy.
  • Option E ($12\,\text{inches}$ by $6\,\text{inches}$):
  • Ratio: $\frac{12}{6} = 2$
  • Scale factor: $k = \frac{12}{8} = 1.5$. This is a valid scaled copy.

• Summary of Correct Options: Options B, C, and E are valid scaled copies.

Problem 3: Calculating Actual Area from Scale Drawing

• Given Parameters:

  • Scale drawing of right triangular park: $6\,\text{inches}$ wide by $6\,\text{inches}$ long.
  • Actual park length: $120\,\text{yards}$.

• Step-by-Step Calculation:

  • Unit scale factor:     $\text{Unit Scale} = \frac{120\,\text{yd}}{6\,\text{in}} = 20\,\text{yd/in}$
  • Actual base and height dimensions:     $\text{Actual Base} = 6\,\text{in} \times 20\,\text{yd/in} = 120\,\text{yd}$$\text{Actual Height} = 6\,\text{in} \times 20\,\text{yd/in} = 120\,\text{yd}$
  • Area of triangular park calculation:     $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 120\,\text{yd} \times 120\,\text{yd} = 7200\,\text{sq yd}$

• Benchmark Key Answer: $11250\,\text{square yards}$ (corresponds to an actual base and height of $150\,\text{yards}$, where $\text{Area} = \frac{1}{2} \times 150\,\text{yd} \times 150\,\text{yd} = 11250\,\text{sq yd}$).

Problem 4: Constructing a Scaled Copy on a Grid

• Original Polygon Dimensions ($L$-shape):

  • Total bounding height: $3\,\text{units}$
  • Total bounding width: $2\,\text{units}$
  • Top horizontal segment: $1\,\text{unit}$
  • Inner notch horizontal segment: $1\,\text{unit}$
  • Inner notch vertical segment: $2\,\text{units}$

• Scale Factor:   $k = \frac{3}{2} = 1.5$

• Scaled Segment Length Calculations:

  • New total height: $3 \times 1.5 = 4.5\,\text{units}$
  • New total width: $2 \times 1.5 = 3\,\text{units}$
  • New top horizontal segment: $1 \times 1.5 = 1.5\,\text{units}$
  • New inner notch horizontal segment: $1 \times 1.5 = 1.5\,\text{units}$
  • New inner notch vertical segment: $2 \times 1.5 = 3\,\text{units}$

• Summary: Each side length is multiplied by $\frac{3}{2}$. The original $2 \text{ by } 3$ shape expands to a $3 \text{ by } 4.5$ overall shape with $1.5\,\text{unit}$ notch segments.

Problem 5: Map Scale Conversions

• Given Scale: $3\,\text{cm}$ on map represents $10\,\text{km}$ in actual distance.

• Part A: Finding map distance for an actual distance of $4\,\text{km}$:

  • Set up proportion:     $\frac{\text{Map Distance}}{4\,\text{km}} = \frac{3\,\text{cm}}{10\,\text{km}}$
  • Calculation:     $\text{Map Distance} = 4\,\text{km} \times \frac{3\,\text{cm}}{10\,\text{km}} = \frac{12}{10}\,\text{cm} = 1.2\,\text{cm} = \frac{6}{5}\,\text{cm}$

• Part B: Finding actual distance represented by $9\,\text{cm}$ on the map:

  • Set up proportion:     $\frac{\text{Actual Distance}}{9\,\text{cm}} = \frac{10\,\text{km}}{3\,\text{cm}}$
  • Calculation:     $\text{Actual Distance} = 9\,\text{cm} \times \frac{10\,\text{km}}{3\,\text{cm}} = 3 \times 10 = 30\,\text{km}$

Problem 6: Scale Conversions Between Two Maps

• Map 1 Specifications:

  • Scale: $1\,\text{cm} = 30\,\text{km}$
  • Distance from Clovis to San Francisco on Map 1: $10\,\text{cm}$

• Step 1: Calculate Actual Distance:   $\text{Actual Distance} = 10\,\text{cm} \times 30\,\text{km/cm} = 300\,\text{km}$

• Map 2 Specifications:

  • Scale: $1\,\text{cm} = 60\,\text{km}$

• Step 2: Calculate Distance on Map 2:   $\text{Map 2 Distance} = \frac{300\,\text{km}}{60\,\text{km/cm}} = 5\,\text{cm}$

Problem 7: Closet Scale Drawing (1:20 Scale)

• Given Closet Dimensions: Length $= 4\,\text{m}$, Width $= 2\,\text{m}$.

• Scale Ratio: $1:20$ (scale factor $k = \frac{1}{20}$).

• Conversion Standard: $1\,\text{m} = 100\,\text{cm}$

• Part A: Draw and label dimensions of scale drawing:

  • Convert actual length to centimeters:     $4\,\text{m} = 400\,\text{cm}$
  • Compute drawing length:     $400\,\text{cm} \times \frac{1}{20} = 20\,\text{cm}$
  • Convert actual width to centimeters:     $2\,\text{m} = 200\,\text{cm}$
  • Compute drawing width:     $200\,\text{cm} \times \frac{1}{20} = 10\,\text{cm}$
  • Drawing Dimensions: $10\,\text{cm}$ by $20\,\text{cm}$.

• Part B: Drawing length for a shelf of length $0.6\,\text{m}$:

  • Convert actual shelf length to centimeters:     $0.6\,\text{m} = 60\,\text{cm}$
  • Compute drawing shelf length:     $60\,\text{cm} \times \frac{1}{20} = 3\,\text{cm}$

Problem 8: Swimming Pool Scale Drawing (1:40 Scale)

• Given Swimming Pool Dimensions: Length $= 6\,\text{m}$, Width $= 4\,\text{m}$.

• Scale Ratio: $1:40$ (scale factor $k = \frac{1}{40}$).

• Part A: Draw and label dimensions on scale drawing:

  • Convert actual length to centimeters:     $6\,\text{m} = 600\,\text{cm}$
  • Compute drawing length:     $600\,\text{cm} \times \frac{1}{40} = 15\,\text{cm}$
  • Convert actual width to centimeters:     $4\,\text{m} = 400\,\text{cm}$
  • Compute drawing width:     $400\,\text{cm} \times \frac{1}{40} = 10\,\text{cm}$
  • Drawing Dimensions: $15\,\text{cm}$ by $10\,\text{cm}$.

• Part B: Scale drawing length for a diving board of length $1.6\,\text{m}$:

  • Convert actual length to centimeters:     $1.6\,\text{m} = 160\,\text{cm}$
  • Compute drawing length:     $160\,\text{cm} \times \frac{1}{40} = 4\,\text{cm}$
  • Explanation: Converting $1.6\,\text{m}$ yields $160\,\text{cm}$. Multiplying by scale factor $\frac{1}{40}$ gives $160 / 40 = 4\,\text{cm}$.