Scale, similarity and maps

Name both lengths and their order

A scale connects corresponding lengths. A room plan might pair the length of one drawn wall with that wall’s actual length. Before dividing, name which length is on top and which is below. In this lesson, the length factor from an original figure to a scaled figure is k = scaled length/original length, with the two lengths expressed in the same unit and the reference length positive.

If 6 cm on a drawing represents an actual length of 3 m, first convert 3 m to 300 cm. The drawing-to-actual ratio is (6 cm)/(300 cm) = 1/50. The units cancel, making this length scale dimensionless. Dividing the bare numbers 6 and 3 would lose the distinction between centimetres and metres. Reversing the ordered ratio gives actual length/drawing length = 50, which answers the reverse question.

Read 1:N as drawing length to actual length

The ordered drawing:actual scale 1:N means that one length unit on the drawing represents N of the same units in the actual object, with N > 0. Thus a 1:200 plan pairs 1 cm with 200 cm = 2 m. In the ideal model, actual length = N × drawing length, and drawing length = actual length/N. A scale 1:1 keeps lengths equal; N > 1 gives a reduced drawing, while 0 < N < 1 describes an enlarged drawing.

The first number need not be one. A drawing:actual ratio 2:5000 means two drawing units represent 5000 actual units, so dividing both entries by two gives 1:2500. Likewise, the statement 2 cm represents 5 m becomes 2 cm:500 cm = 1:250. Convert the units and simplify the whole ratio before applying it. A unit-bearing statement such as 1 cm represents 250 m is useful, but its dimensionless ratio is 1:25 000.

A positive factor has a reciprocal route back

For uniform scaling with k > 0, every corresponding length is multiplied by k. If k > 1 the figure is enlarged; if 0 < k < 1 it is reduced; if k = 1 its size is unchanged. A rectangle scaled by k = 2 has twice the original side lengths. Scaling the result by 1/k = 1/2 recovers those lengths, because k × (1/k) = 1. State the direction each factor follows.

A uniform scaling about a chosen centre also multiplies every point’s distance from that centre by k. A factor k = 0 would collapse the whole figure to the centre. The figure would be degenerate, its original shape could not be recovered, and 1/0 is undefined. Zero is therefore excluded from the invertible similarity scales used here. A drawing of a zero-length feature alone cannot establish a scale ratio either.

Similarity requires one shape-preserving correspondence

Similar figures have the same shape under a uniform change of size, possibly with a change of position or orientation. Match the parts before comparing them. For polygons, corresponding angles are equal and corresponding side lengths share one positive factor. A long side must be compared with its matching side, and each vertex must stay in the same correspondence throughout the calculation. Equal size is allowed: similar figures with factor one are congruent.

A 3 m by 2 m rectangle and a 6 m by 4 m rectangle are similar: the matching side ratios are 6/3 = 4/2 = 2, and their corresponding angles are right angles. A 6 m by 2 m rectangle instead has ratios 2 and 1. It has been stretched in one direction, so this correspondence is a distortion. Matching angles alone do not make those rectangles similar; one matching side ratio alone does not establish similarity of a general figure.

Lengths and perimeters receive the factor k

A perimeter is the sum of the boundary lengths. If every corresponding boundary length receives factor k, their sum receives factor k too: P′ = kP. The prime mark here identifies the scaled figure. The original 3 m by 2 m rectangle has perimeter 2 × (3 m + 2 m) = 10 m. At k = 2 the 6 m by 4 m rectangle has perimeter 20 m.

Use this linear factor for a corresponding side, a diagonal or a perimeter in genuinely similar figures. At k = 1/2 the same rectangle’s sides become 1.5 m and 1 m, and its perimeter becomes 5 m. Each result remains a length, with a length unit. A perimeter ratio is dimensionless when both perimeters use the same unit, just like the side-length ratio.

Area receives two length factors

For a rectangle of width w and height h, area A = wh. Uniform scaling gives width kw and height kh, so A′ = (kw)(kh) = k²A. For the 3 m by 2 m rectangle, A = 6 m². At k = 2 its area is 24 m², because k² = 4. At k = 1/2 its area is 1.5 m², because (1/2)² = 1/4.

The area factor k² applies to genuinely similar plane figures, not only rectangles. Their shape and corresponding lengths must support one uniform k. Stretching only the width of the 3 m by 2 m rectangle produces a 6 m by 2 m rectangle with area 12 m²; that change uses different factors in the two directions. An area ratio alone does not prove similarity. Establish the geometric relationship before selecting the factor.

Volume receives three length factors

For a rectangular prism with mutually perpendicular edge lengths w, h and d, volume V = whd. Scaling all three edge lengths by the same positive k gives V′ = (kw)(kh)(kd) = k³V. The stipulated 3 m by 2 m by 1 m prism has volume 6 m³. At k = 2 its dimensions are 6 m by 4 m by 2 m and its volume is 48 m³, eight times the original.

The factor k³ applies to genuinely similar solids whose corresponding lengths all share k. Doubling both front-face edge lengths while keeping the depth unchanged would not make a prism similar to the original. Corresponding surface areas of similar solids receive k², while their volumes receive k³. A flat drawing has an area; it does not acquire a volume merely because it depicts a solid. Use the solid’s stated dimensions and similarity relationship to calculate its volume.

Choose the factor for the quantity and retain its unit

Length, area and volume have dimensions L, L² and L³. The scale k is dimensionless, so the corresponding length, area and volume multipliers preserve each quantity’s dimension and unit. When a diagram is similar to an actual planar shape at drawing:actual scale 1:N, the actual area is N² times the diagram area after using compatible area units. For a genuinely similar three-dimensional model, the corresponding actual volume is N³ times the model volume.

Unit conversion is an additional step. At a stipulated 1:100 scale, a diagram area of 4 cm² corresponds to 4 × 100² cm² = 40 000 cm² = 4 m². The length conversion 1 m = 100 cm gives area conversion 1 m² = 10 000 cm². Dimensions check a quantity’s unit structure, but cannot distinguish the dimensionless factors k and k²; geometry supplies the correct area factor. The context also matters: the drawing must represent the stated planar shape faithfully, and a volume comparison needs an actual solid model or sufficient three-dimensional information.

Identify the distance that the map measurement represents

In an ideal flat model, a map ratio 1:25 000 pairs a drawn 3.6 cm separation with 90 000 cm = 900 m of corresponding ground distance. An actual measured map adds conditions: its printed size must match the numeric scale, the features and measurement must be identifiable, and the scale must be suitable for the location and direction being measured. A projection represents a curved surface on a plane, so a map’s stated scale does not guarantee the same exact ratio everywhere.

A straight connector between two mapped places describes their represented separation. A route follows its bends; measuring the straight connector does not measure that route. A traced route on a topographic map ordinarily gives a horizontal or plan distance under the map’s local assumptions. Terrain distance also depends on elevation changes, which the length ratio alone does not supply. Read the map’s projection, scale and relevant metadata, and describe the distance you have estimated instead of assigning it an unsupported precision.

Resize the scale indicator with the map

A numeric drawing:ground ratio belongs to a particular output size. If a map originally at 1:N is uniformly resized by a positive linear factor r, each drawing length becomes r times as large while its represented ground distance stays the same. The new ratio is r:N = 1:(N/r). A copy of a 1:25 000 map enlarged to 200% of its former width and height has r = 2 and ratio 1:12 500. Enlarging to 200% means factor two; increasing a length by 200% means factor three.

A graphic scale bar retains its ground-distance labels when it is uniformly resized together with the map. A bar originally 2 cm long and labelled 500 m on that 1:25 000 map becomes 4 cm long after enlargement by two and still represents 500 m. The old printed text 1:25 000 would need updating. If the bar is resized separately, or the map is stretched differently in its two directions, that simple correspondence fails. On a screen, a centimetre label in a schematic is not a physical ruler guarantee; use the intended scale representation and viewing context.

Compare uniform scaling with a separate map representation

Begin with the explorer’s stipulated rectangle 3 m by 2 m and its default positive factor k = 2. The scaled rectangle is 6 m by 4 m. Both diagrams use the same drawing scale per represented metre, so the size change is visible. Check the original and scaled perimeters, 10 m and 20 m, and areas, 6 m² and 24 m². The factor table distinguishes k = 2 from k² = 4.

Select k = 1/2, 1, 2 or 3. The associated rectangular prism starts with edge lengths 3 m, 2 m and 1 m; every one receives the selected k. At the default its edge lengths become 6 m, 4 m and 2 m, and its volume changes from 6 m³ to 48 m³. Use the displayed three-dimensional dimensions and volume table for this solid, rather than interpreting the rectangle’s shaded area as volume. All values are exact stipulations, and every factor is positive.

The separate map control begins at 1:10 000 and offers 1:10 000, 1:25 000 and 1:50 000 for one fixed stipulated ground distance of 1 km. Its drawing representations are 10 cm, 4 cm and 2 cm respectively. The bar is schematic and labels represented centimetres; do not measure your physical screen to recover these values. The two controls are independent, giving 4 × 3 = 12 combinations. Changing k changes the similar geometry, while changing the map ratio changes the representation of the same fixed ground distance. Neither module models terrain, projection distortion or measurement uncertainty.

Scale a stated reading uncertainty and preserve the map’s limits

An exact stipulated diagram and a ruler reading on an actual map carry different information. Suppose a map reading is reported as 3.60 cm with explicitly stated reading uncertainty 0.04 cm. If we stipulate the length ratio 1:25 000 as exact for this local calculation, multiplying both magnitudes by 25 000 gives 900 m with a reading-uncertainty contribution of 10 m. This contribution covers the stated reading uncertainty; it does not establish the total accuracy of the ground distance.

The map’s source data, feature placement, projection, printing and the chosen path can add limitations that this calculation has not quantified. Preserve their stated information; do not infer total uncertainty from the digits of a converted result. Magnifying a map does not improve its original data. If an existing measurement interval is simply enlarged with the drawing by factor two, both its value and its uncertainty magnitude double. A fresh ruler measurement is a new reading whose uncertainty must be assessed on its own.

Finish by naming the compared objects, corresponding parts, ratio direction, units and applicable model. Verify a positive k and its reciprocal, use the appropriate length, area or volume factor, and reverse an exact calculation where useful. For a map, verify the output size and scale indicator and specify whether you mean separation, route plan distance or terrain distance. Retain unrounded information until the final report and state any approximation. The next lesson, Length, area, volume and everyday shapes, develops the quantities and formulas used to describe familiar figures and solids.

Worked examples

Try it yourself

1. An exact drawing length of 4.5 cm represents 90 m. Find drawing:actual scale 1:N using common units. Separately simplify drawing:actual ratio 2:5000. At scale 1:200, how long is the drawing of an actual 8 m segment?

   Show the answer and reasoning

   4.5 cm:9000 cm = 1:2000. The ratio 2:5000 simplifies to 1:2500. At 1:200, an 8 m segment is drawn as 4 cm.

   Convert 90 m to 9000 cm before forming the dimensionless ratio. Simplify both entries of 2:5000. For the final drawing, 8 m = 800 cm and 800 cm/200 = 4 cm; multiplying back by 200 recovers 8 m.

2. Uniformly scale the explorer’s 3 m by 2 m rectangle by k = 1/2. Give its new side lengths and the reciprocal factor. Explain the cases k = 1 and k = 0.

   Show the answer and reasoning

   The new sides are 1.5 m and 1 m, and the reciprocal factor is 2. Factor one preserves size. Factor zero collapses the figure to a point and has no reciprocal.

   A positive factor below one reduces every corresponding length, and multiplication by its reciprocal recovers each length. Factor zero is degenerate and cannot recover the original shape, so it is excluded from the invertible similarity scales here.

3. Compare a 3 m by 2 m rectangle with a 6 m by 4 m rectangle and a 6 m by 2 m rectangle. Which is similar under the stated matching sides? Does having right angles establish similarity by itself?

   Show the answer and reasoning

   The 6 m by 4 m rectangle is similar with k = 2. The 6 m by 2 m rectangle has matching side ratios 2 and 1, so it is distorted. Right angles alone do not establish similarity.

   The correspondence must preserve both the matching angles and one common length factor. Doubling only one side direction changes the aspect ratio. A single matching ratio is also insufficient to establish similarity of a general polygon.

4. The exact explorer rectangle has perimeter 10 m and area 6 m², and its associated 3 m by 2 m by 1 m prism has volume 6 m³. Find all three quantities after uniform scaling by k = 3. Also find the area and volume at k = 1/2.

   Show the answer and reasoning

   At k = 3: perimeter 30 m, area 54 m², volume 162 m³. At k = 1/2: area 1.5 m² and volume 0.75 m³.

   Use k for perimeter, k² for area and k³ for the genuinely similar prism’s volume. The factors at k = 3 are 3, 9 and 27. At k = 1/2 the area and volume factors are 1/4 and 1/8. Keep length, square and cubic units distinct.

5. At an exact drawing:actual scale 1:100, a similar planar diagram has area 4 cm². What is the corresponding actual area in m²? Separately, a genuinely similar solid model at the same length scale has volume 1 cm³; find the actual volume in m³. Would a flat area drawing alone determine that volume?

   Show the answer and reasoning

   Actual area is 40 000 cm² = 4 m². Actual volume is 1 000 000 cm³ = 1 m³. A flat area drawing alone does not determine a solid’s volume.

   Reverse the drawing-to-actual length factor with 100, then square it for area and cube it for volume. Convert with 1 m² = 10 000 cm² and 1 m³ = 1 000 000 cm³. The volume result relies on the stated similar solid, not only on a planar picture.

6. Assume an exact local map ratio 1:50 000 for this calculation. A straight mapped separation is reported as 2.40 cm with stated reading uncertainty 0.05 cm. Find the represented horizontal distance and the reading-uncertainty contribution. Does this determine the route distance, terrain distance or total ground accuracy?

   Show the answer and reasoning

   The represented horizontal distance is 1200 m, and the reading-uncertainty contribution is 25 m. The result alone does not determine route distance, terrain distance or total ground accuracy.

   Multiply both stated magnitudes by 50 000 and convert centimetres to metres. A route’s bends and elevation changes require additional information; projection, map data and printing can also affect the estimate. The 25 m figure represents only the uncertainty contribution explicitly supplied by the reading in this stipulated local model.

7. Uniformly resize a 1:50 000 map by linear factor r = 1.25. Find its new numeric ratio. A bar originally 2 cm long represents 1000 m; what should its new length and ground-distance label be if resized with the map? What if the map width and height are stretched by different factors?

   Show the answer and reasoning

   The new ratio is 1:40 000. The bar becomes 2.5 cm long and retains its 1000 m label. Different width and height factors do not give one uniform length scale.

   Use N/r = 50 000/1.25 = 40 000. Every drawing length, including the bar, receives factor 1.25 while represented ground distance is unchanged. A separately resized bar or an unevenly stretched map does not preserve this simple correspondence.

What to carry forward

• Form an ordered ratio of corresponding lengths in the same unit; drawing:actual 1:N uses actual length N times drawing length.
• Uniform similarity preserves matching angles and one positive length factor k; its reciprocal reverses the scale, while k = 0 is degenerate.
• Corresponding lengths scale by k; for similar plane figures, perimeters scale by k and areas by k²; for similar solids, corresponding surface areas scale by k² and volumes by k³. Retain the appropriate units.
• A map scale needs a suitable location, direction and output size; a resized numeric ratio changes, while a bar resized with the map retains its ground-distance labels.
• Identify separation, route or terrain distance, scale explicitly stated reading uncertainty, and preserve other measurement and map limitations when reporting.