math.wiki
Measurement › Scale and size › Scale factor multiplies every length
Open on the map
Math 7 · 7.G.A.1 · G.SRT.A.1B

Scale factor multiplies every length

A shape is copied twice as large. Its sides double, but what about its diagonals, its height, its perimeter, and the distance between any two points on it?

You need
Scale factorMath 7
Unlocks
Scale factor, area and volumeMath 7
Where you are 3 in Scale and size
Measurement12
Scale and size3
Measuring what you cannot reach
Scale factor multiplies every length
Scale factor, area and volume

Why this exists

A scale copy is made by multiplying the sides by one number, the scale factor. But a shape carries far more lengths than its sides: a kite has diagonals, a triangle has a height, a frame has a perimeter, a track has curved ends, and any two marked points have a distance between them.

If each of those needed its own rule, every copy would have to be measured again. It does not. One number settles every length in the copy, straight or curved, drawn or not, and you can find any of them without a ruler.

Definition

When a figure is copied at scale factor k, every length in the copy is k times the matching length in the original.

“Every length” means every distance you could measure on the figure: sides, diagonals, a height, the perimeter, a curved edge, the distance between two marked points. In symbols, if L is any length in the original and L' is the matching length in the copy,

L' = kL

How to find k from one pair of matching sides is in the scale factor lesson; here it is put to work.

Intuition

Figure 1 shows a pentagon ABCDE on a grid and its copy at scale factor 2.

A B C D E A′ B′ C′ D′ E′
Figure 1The copy at scale factor 2. Each dot cuts a slanted length into two copies of the original one.

The sides along grid lines can be counted. AB covers 5 squares and A'B' covers 10. Every grid step has become two grid steps, so every length made of grid steps has doubled.

The slanted side cannot be counted. CD goes 3 squares across and 4 up; C'D' goes 6 across and 8 up. The dot at its middle splits it into two pieces, each going 3 across and 4 up: each piece is a copy of CD slid into place. So C'D' is two lengths of CD end to end, twice as long, whatever CD measures. The dashed diagonal tells the same story: AC goes 5 across and 2 up, and the dot splits A'C' into two copies of it.

Any path you trace on the copy is made of k copies of the matching path on the original, so any length there is k times as long. Notice what did not change: C'D' leans exactly as CD does, because its across and its up were both doubled.

Results

Perimeter. The perimeter is a length, so P' = kP.

Lengths inside one figure keep their ratio. If a and b are two lengths in the original,

\frac{a'}{b'} = \frac{ka}{kb} = \frac{a}{b}

Back to the original. Dividing by k undoes the copy, so the original is a copy of the copy at scale factor \displaystyle \frac{1}{k}, and \displaystyle L = \frac{L'}{k}.

A copy of a copy. Copying at k_1 and then at k_2 multiplies every length by k_1 and then by k_2, which is one copy at scale factor k_1 k_2.

Angles do not change. Every angle in the copy measures the same as its match in the original.

Why it is true

Straight segments. Copying at scale factor k places every grid point k times as far across and k times as far up from one fixed corner. So a segment going r across and s up becomes one going kr across and ks up.

When k is a whole number, the new segment cuts into k pieces that each go r across and s up. Each piece is the original slid to a new place, and sliding does not change a length, so the new segment is k times the old one.

When k is a fraction \displaystyle \frac{m}{n}, cut the original segment of length L into n equal pieces instead. Each goes \displaystyle \frac{r}{n} across and \displaystyle \frac{s}{n} up and has length \displaystyle \frac{L}{n}. The copy goes \displaystyle \frac{mr}{n} across and \displaystyle \frac{ms}{n} up, which is m of those pieces end to end:

L' = m \cdot \frac{L}{n} = \frac{m}{n}\,L = kL

Nothing here needed the segment to be a side. A diagonal, a height or the gap between two dots is a straight segment too.

Perimeter. A polygon’s perimeter is the sum of its sides, and the The distributive property gathers the k out of the sum:

P' = ka + kb + kc + \dots = k(a + b + c + \dots) = kP

Curved lengths. A curve can be followed as closely as you like by short straight pieces. Each piece is multiplied by k, so their total is too, and that total gets as close as you like to the curve’s length. So a curved length is multiplied by k as well.

Angles. A side going r across and s up becomes one going kr across and ks up, exactly as steep as before. Every side keeps its slant, so the angle between any two sides is kept.

Worked examples

Example 1

Sides and perimeter on a grid

In Figure 1 the copy is made at scale factor 2. Side CD measures 5 units with a ruler. Find the length of every side of the copy, and the perimeter of each pentagon.

Solution

Counting squares, AB = 5, BC = 2, DE = 2 and EA = 6, and CD = 5 as measured:

P = 5 + 2 + 5 + 2 + 6 = 20

Each length in the copy is 2 times its match: A'B' = 10, B'C' = 4, C'D' = 10, D'E' = 4 and E'A' = 12. So

P' = 10 + 4 + 10 + 4 + 12 = 40

Check. The perimeter is a length, so it should be 2P = 2(20) = 40. It is, and counting squares in the copy agrees for the four sides on grid lines.

Example 2

The diagonals of a kite

The kite on the left of Figure 2 has sides of 13 cm and 20 cm and diagonals of 24 cm and 21 cm, which cross 5 cm from the top corner. In a scale copy the long diagonal measures 30 cm. Find the copy’s other diagonal, its sides, its perimeter, and the distance from its top corner to the crossing point.

13 20 24 21 30
Figure 2A kite and a scale copy of it. Lengths in centimeters.
Solution

The long diagonals match, so

k = \frac{30}{24} = \frac{5}{4}

Every length in the copy is \displaystyle \frac{5}{4} of its match, side or not. The other diagonal is \displaystyle 21 \times \frac{5}{4} = 26.25 cm. The sides are \displaystyle 13 \times \frac{5}{4} = 16.25 cm and \displaystyle 20 \times \frac{5}{4} = 25 cm. The original’s perimeter is 2(13 + 20) = 66 cm, so the copy’s is \displaystyle 66 \times \frac{5}{4} = 82.5 cm. The distance from the top corner to the crossing point is a length too: \displaystyle 5 \times \frac{5}{4} = 6.25 cm.

Check. 2(16.25 + 25) = 82.5. The original’s short diagonal splits as 5 + 16 = 21, and the copy’s as 6.25 + 20 = 26.25, as it should.

Example 3

A picture frame

Winifred’s rectangular picture frame is 30 cm by 24 cm on the outside, with a border 3 cm wide all the way round, so the opening is 24 cm by 18 cm. She makes a scale copy at scale factor \displaystyle \frac{5}{6}. Find the copy’s outside perimeter, border width and opening perimeter.

Solution

Each is a length, so each is \displaystyle \frac{5}{6} of its match. The outside perimeter is 2(30 + 24) = 108 cm, so the copy’s is

108 \times \frac{5}{6} = 90 \text{ cm}

The border is \displaystyle 3 \times \frac{5}{6} = 2.5 cm wide. The opening’s perimeter is 2(24 + 18) = 84 cm, so the copy’s is \displaystyle 84 \times \frac{5}{6} = 70 cm.

Check. The copy’s outside is 25 by 20, with perimeter 2(25 + 20) = 90. Taking a 2.5 cm border off each side leaves an opening 20 by 15, with perimeter 2(20 + 15) = 70. Both routes agree, because the border and the sides were multiplied by the same number.

Example 4

A triangle’s height from its perimeter

The triangle in Figure 3 has sides of 10 cm, 17 cm and 21 cm, and a height of 8 cm to the 21 cm base. A scale copy of it has a perimeter of 36 cm. Find the copy’s height and base.

21 10 17 8
Figure 3The height is not a side, but it is a length. Lengths in centimeters.
Solution

The only length known in the copy is its perimeter, and that matches the original’s perimeter:

P = 10 + 17 + 21 = 48 \text{ cm} \qquad k = \frac{36}{48} = \frac{3}{4}

The height is not a side, but it is a length: \displaystyle 8 \times \frac{3}{4} = 6 cm. The base is \displaystyle 21 \times \frac{3}{4} = 15.75 cm.

Check. The other sides are \displaystyle 10 \times \frac{3}{4} = 7.5 and \displaystyle 17 \times \frac{3}{4} = 12.75, and 7.5 + 12.75 + 15.75 = 36.

Example 5

A running track drawn at two sizes

Zinaida draws a running track in her notebook (Figure 4). Each straight is 8 cm, each curved end measures 9.4 cm along the curve (she laid a thread along it and measured the thread), and the track is 6 cm across. She draws it again on a poster at scale factor 4. Find one lap of the line on each drawing, and how far across the poster track is.

8 6 9.4
Figure 4The notebook drawing of the track. Lengths in centimeters.
Solution

One lap is two straights and two curved ends. In the notebook that is

2(8) + 2(9.4) = 16 + 18.8 = 34.8 \text{ cm}

The lap is a length, curved parts and all, so on the poster it is

34.8 \times 4 = 139.2 \text{ cm}

The distance across is 6 \times 4 = 24 cm.

Check. On the poster each straight is 32 cm and each curved end is 9.4 \times 4 = 37.6 cm, and 2(32) + 2(37.6) = 139.2. A thread laid along the poster’s curve is 4 times as long as one laid along the notebook’s.

Example 6

A copy of a copy, and back again

Yorick draws a five-pointed star. Each of its five straight lines, from one point to another, is 12 cm, and the angle at each point is 36^\circ. He copies it at scale factor 3, then copies that enlargement at scale factor \displaystyle \frac{3}{4}. Find each line and each point’s angle in the final star, and the single scale factor that takes the final star back to the original.

Solution

The first copy has lines of 12 \times 3 = 36 cm, and the second \displaystyle 36 \times \frac{3}{4} = 27 cm. By the associative property the two factors can be multiplied first:

\begin{aligned} (12 \times 3) \times \frac{3}{4} &= 12 \times \left(3 \times \frac{3}{4}\right) \\ &= 12 \times \frac{9}{4} = 27 \end{aligned}

So the final star is one copy of the original at scale factor \displaystyle \frac{9}{4}. Neither step changes an angle, so each point is still 36^\circ. To go back, every length is divided by \displaystyle \frac{9}{4}, which is multiplying by \displaystyle \frac{4}{9}: the scale factor back is \displaystyle \frac{4}{9}.

Check. \displaystyle 27 \times \frac{4}{9} = 12, the original line.

Boundaries

Every length must be multiplied by the same number. If a 4 by 2 rectangle is stretched to 8 by 2, the length has doubled and the width has not, so there is no scale factor, and nothing here says what happened to the diagonal.

One matching pair is not enough. Two figures with perimeters in the ratio k, or one pair of sides in the ratio k, need not be scale copies. Every length has to match.

Only lengths are multiplied. Angles stay the same. An area does not grow by k: the copy in Figure 1 covers four times as much grid as the original, not twice, which is the subject of Scale factor, area and volume.

Both lengths in the same unit. A scale factor has no unit, so compare centimeters with centimeters. A drawing where 1 cm stands for 1 m is a map or plan scale, Using a scale on maps, plans and models.

A scale factor is positive. It multiplies distances, and a distance is never negative or zero.

Where this goes next

Area and volume grow faster than lengths, because an area needs two lengths and a volume needs three. Map and plan scales use this lesson’s fact with a change of unit on top, and the triangles made by two objects and their shadows are scale copies, which is how the height of something out of reach is found.

Later, figures that are scale copies of each other are named similar figures, and making a copy by moving every point k times as far from one fixed point, as the grid argument does from its corner, is a dilation. Because every circle is a scale copy of every other, the distance around a circle is always the same multiple of the distance across, which is where the number pi comes from.

Exercises

Sides and perimeters on a grid

  1. A rectangle 3 squares by 5 squares is copied at scale factor 4. Find the copy’s sides and perimeter.
  2. The pentagon in Figure 1 is copied at scale factor 3 instead of 2. Using CD = 5, find the length of every side of this copy and its perimeter.
  3. A right triangle on a grid has two sides along grid lines, 6 units and 8 units, and a slanted side of 10 units. It is copied at scale factor \displaystyle \frac{1}{2}. Find the copy’s sides and perimeter.
  4. A square has sides of 7 cm, and its diagonal measures 9.9 cm. It is copied at scale factor 3. Find the copy’s side, diagonal and perimeter.

Diagonals, heights and marked points

  1. A kite has diagonals of 10 cm and 7 cm. Find the diagonals of its copy at scale factor 2.4.
  2. In a five-pointed star, each straight line from one point to another is 20 cm. Find that length in a copy at scale factor 0.6.
  3. A triangle has a base of 10 cm and a height of 6 cm. Find the base and height of its copy at scale factor 1.2.
  4. A trapezoid has parallel sides of 8 cm and 14 cm and a height of 5 cm. In a scale copy the longer parallel side is 24.5 cm. Find the scale factor, the shorter parallel side and the height of the copy.
  5. Two dots marked inside a shape are 4.5 cm apart. How far apart are the matching dots in a copy at scale factor \displaystyle \frac{4}{3}?

Perimeter

  1. A picture frame has an outside perimeter of 140 cm. Find the outside perimeter of a copy at scale factor 0.4.
  2. A hexagon has a perimeter of 27 cm, and a scale copy of it has a perimeter of 36 cm. Find the scale factor, and the length in the copy of a side that is 6 cm in the original.
  3. A copy made at scale factor \displaystyle \frac{7}{5} has a perimeter of 84 cm. Find the perimeter of the original.
  4. A regular octagon has sides of 2.5 cm. Find the perimeter of its copy at scale factor 4.

Curved lengths

  1. A drawing of a running track has a lap of 40 cm, with two straights of 10 cm each and two curved ends of equal length. It is copied at scale factor 3. Find the copy’s lap, each straight and each curved end.
  2. The curved top of an arch in a drawing measures 7.5 cm along the curve. Find that length in a copy at scale factor 0.8.

Copies in both directions

  1. A copy is made at scale factor \displaystyle \frac{6}{5}. A length in the copy is 30 cm. Find the matching length in the original.
  2. Winifred copies a shape at scale factor 5, then copies that copy at scale factor \displaystyle \frac{1}{4}. A segment is 8 cm in the original. Find its length in the final copy, and the single scale factor that does both steps at once.
  3. Zinaida shrinks a shape at scale factor \displaystyle \frac{1}{2}, then shrinks the result at scale factor \displaystyle \frac{1}{2} again. The original’s perimeter is 64 cm. Find the final perimeter and the single scale factor.
  4. A triangle has sides of 6 cm, 8 cm and 9 cm. In a scale copy, the longest side is 11.25 cm. Find the copy’s other two sides and its perimeter.
  5. A rectangle is 16 cm long and 10 cm wide. A scale copy is 20 cm long. Use the ratio of width to length to find the copy’s width.

Reasoning

  1. A triangle has angles of 30^\circ, 60^\circ and 90^\circ. It is copied at scale factor 3. Ursula says the copy’s angles are 90^\circ, 180^\circ and 270^\circ. Explain why that cannot be, and give the copy’s angles.
  2. Use the distributive property to explain why a copy at scale factor k of a polygon with any number of sides has a perimeter k times the original’s.
  3. Explain why, in every scale copy of a triangle, the height divided by the base gives the same number.
  4. Every circle is a scale copy of every other circle. Explain why the distance around a circle divided by the distance across it gives the same number for every circle.
  5. A square has sides of 5 cm. A rectangle is 6 cm by 14 cm. Show that the rectangle’s perimeter is 2 times the square’s, and explain why it is still not a scale copy of the square.
Answers
  1. 12 by 20, with perimeter 64: 3 \times 4 = 12, 5 \times 4 = 20, and the perimeter is 4 \times 16 = 64.
  2. Sides 15, 6, 15, 6 and 18; perimeter 60. Each of 5, 2, 5, 2, 6 is multiplied by 3, and 3 \times 20 = 60.
  3. 3, 4 and 5 units; perimeter 12, half of 6 + 8 + 10 = 24.
  4. Side 21 cm, diagonal 29.7 cm, perimeter 84 cm. The diagonal is a length: 9.9 \times 3 = 29.7. The perimeter is 4 \times 21 = 84, which is also 3 \times 28.
  5. 24 cm and 16.8 cm: 10 \times 2.4 = 24 and 7 \times 2.4 = 16.8.
  6. 12 cm: 20 \times 0.6 = 12.
  7. Base 12 cm, height 7.2 cm: 10 \times 1.2 = 12 and 6 \times 1.2 = 7.2. The height is a length, so it is multiplied too.
  8. \displaystyle k = \frac{7}{4}; shorter side 14 cm; height 8.75 cm. The longer sides match, so \displaystyle k = \frac{24.5}{14} = \frac{7}{4}. Then \displaystyle 8 \times \frac{7}{4} = 14 and \displaystyle 5 \times \frac{7}{4} = 8.75.
  9. 6 cm: \displaystyle 4.5 \times \frac{4}{3} = 6. The distance between two points is a length.
  10. 56 cm: 140 \times 0.4 = 56.
  11. \displaystyle k = \frac{4}{3}, and the side is 8 cm. The perimeters match, so \displaystyle k = \frac{36}{27} = \frac{4}{3}, and \displaystyle 6 \times \frac{4}{3} = 8.
  12. 60 cm: \displaystyle 84 \div \frac{7}{5} = 84 \times \frac{5}{7} = 60. Check: \displaystyle 60 \times \frac{7}{5} = 84.
  13. 80 cm. The original’s perimeter is 8 \times 2.5 = 20, and 20 \times 4 = 80.
  14. Lap 120 cm, each straight 30 cm, each curved end 30 cm. In the drawing the curved ends share 40 - 2(10) = 20, so each is 10 cm. Every length is multiplied by 3, and 2(30) + 2(30) = 120 = 3 \times 40.
  15. 6 cm: 7.5 \times 0.8 = 6. A curved length is multiplied like any other.
  16. 25 cm: \displaystyle 30 \div \frac{6}{5} = 30 \times \frac{5}{6} = 25.
  17. 10 cm, and the single scale factor is \displaystyle \frac{5}{4}. The segment becomes 8 \times 5 = 40, then \displaystyle 40 \times \frac{1}{4} = 10, and \displaystyle 5 \times \frac{1}{4} = \frac{5}{4}.
  18. 16 cm, and the single scale factor is \displaystyle \frac{1}{4}: \displaystyle \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}, and \displaystyle 64 \times \frac{1}{4} = 16.
  19. 7.5 cm and 10 cm; perimeter 28.75 cm. The longest sides match, so \displaystyle k = \frac{11.25}{9} = 1.25; then 6 \times 1.25 = 7.5 and 8 \times 1.25 = 10, and 7.5 + 10 + 11.25 = 28.75, which is 1.25 \times 23.
  20. 12.5 cm. Width and length keep their ratio, \displaystyle \frac{10}{16} = \frac{5}{8}, so the width is \displaystyle \frac{5}{8} \times 20 = 12.5. (The scale factor is \displaystyle \frac{20}{16} = 1.25, and 10 \times 1.25 = 12.5 agrees.)
  21. The copy’s angles are 30^\circ, 60^\circ and 90^\circ. A scale factor multiplies lengths, not angles. Ursula’s angles cannot belong to any triangle: their sum is 540^\circ, while a triangle’s angles add to 180^\circ, and a 180^\circ angle is a straight line, not a corner.
  22. If the sides are a, b, c, \dots, the perimeter is P = a + b + c + \dots. In the copy every side is multiplied by k, so its perimeter is ka + kb + kc + \dots. By the distributive property that equals k(a + b + c + \dots) = kP, however many terms the sum has.
  23. The height and the base are both lengths, so in a copy at scale factor k they are kh and kb. Their quotient is \displaystyle \frac{kh}{kb} = \frac{h}{b}, because the common factor k cancels, so it is the same for every copy.
  24. Take any two circles. One is a copy of the other at some scale factor k. The distance around is a curved length and the distance across is a straight one, so both are multiplied by k: if they are C and d in the first circle, they are kC and kd in the second. Then \displaystyle \frac{kC}{kd} = \frac{C}{d}, so every circle gives the same number.
  25. The square’s perimeter is 4 \times 5 = 20 cm, and the rectangle’s is 2(6 + 14) = 40 cm, which is 2 \times 20. But a copy of the square at scale factor 2 would have every side 5 \times 2 = 10 cm. The rectangle’s sides are \displaystyle \frac{6}{5} and \displaystyle \frac{14}{5} of 5, two different factors, so it is not a scale copy.
kg1128 · updated 2026-10-07 · CC-BY-SA-4.0