Geometry

Scale factor and perimeter, area and volume

A model is built exactly like the real thing, one tenth of the size in every direction. Does it need one tenth of the paint, and hold one tenth of the water?

You need

Scale factorMath 7 Volume of prisms and cylindersMath 7

Where you are 10 in Similarity

Geometry11

Scale factor and perimeter, area and volume

Why this exists

A scale factor tells you what happens to lengths: make a copy with every length three times as long, and every side, every diagonal and every height is three times as long. But most questions people ask about a shape are not about one length. How much paint covers it? How much land is inside the fence? How much water does the tank hold? How heavy is the statue?

Those are areas and volumes, and they do not follow the scale factor. A model one tenth of the size needs far less than one tenth of the paint, and holds far less than one tenth of the water. This lesson finds the exact rule. It is short, it holds for every pair of similar figures and every pair of similar solids, whatever their shape, and it explains effects that look surprising until you know it: a large pizza that is only a little wider than a small one but has much more pizza, and a glass filled to half its height that is nowhere near half full.

Definition

Two figures are similar when one is a scaled copy of the other: every length in one is the same number k times the matching length in the other. That number is the scale factor,

k = \frac{\text{a length in the copy}}{\text{the matching length in the original}}

and k > 0. A copy with k > 1 is an enlargement, a copy with k < 1 is a reduction, and k = 1 gives a congruent copy.

“Every length” means every one, not only the sides. Heights, diagonals, radii, the distance between two marked points, and the perimeter, which is a sum of lengths, all change by the same factor k. The same definition covers solids: two solids are similar when every length in one is k times the matching length in the other. Two cylinders are similar exactly when their radii and their heights are in the same ratio; two boxes are similar exactly when all three edges are in the same ratio.

When the lengths of a solid change by different factors, say the length is doubled and the height halved, the new solid is not similar to the old one. This lesson covers that case too, under the name changing the dimensions.

Intuition

Double every length of a triangle. The bigger triangle is not twice the original: the three segments joining the midpoints of its sides cut it into four triangles, each exactly the size of the original. Triple every length of a square and the copy is made of nine squares the size of the original.

Figure 1Doubling every length makes four copies of the original. Tripling makes nine.

The pattern is that an area is built out of two lengths, a width and a height, and both of them grow. Doubling both multiplies the area by 2 \times 2 = 4; tripling both multiplies it by 3 \times 3 = 9.

A volume is built out of three lengths, so all three grow. A cube with every edge doubled is eight cubes of the original size, two along each edge.

Figure 2Doubling every edge of a cube makes eight copies of it.

So the rule counts how many lengths a measurement is made of. A length uses one, so it grows by k. An area uses two, so it grows by k^2. A volume uses three, so it grows by k^3. You may have met this rule already, found by counting squares and cubes, in scale factor, area and volume; here it is for any pair of similar figures and solids, with a reason.

Results

Scaling theorem. If figure B is similar to figure A with scale factor k, then

\begin{aligned} \frac{\text{perimeter of } B}{\text{perimeter of } A} &= k \\ \frac{\text{area of } B}{\text{area of } A} &= k^2 \end{aligned}

If solid B is similar to solid A with scale factor k, then

\begin{aligned} \frac{\text{surface area of } B}{\text{surface area of } A} &= k^2 \\ \frac{\text{volume of } B}{\text{volume of } A} &= k^3 \end{aligned}

As ratios: if the lengths are in the ratio a : b, the areas are in the ratio a^2 : b^2 and the volumes in the ratio a^3 : b^3.

The theorem runs backwards as well. An area ratio gives the scale factor as its square root, and a volume ratio gives it as its cube root:

k = \sqrt{\frac{\text{area of } B}{\text{area of } A}} \qquad k = \sqrt[3]{\frac{\text{volume of } B}{\text{volume of } A}}

The positive square root is the one to take, because a scale factor is positive.

Changing the dimensions. If a box or a cylinder has its dimensions multiplied by different factors, its volume is multiplied by the product of the factors, one for each length in the volume formula. For a box, multiplying the length by a, the width by b and the height by c multiplies the volume by abc. For a cylinder, multiplying the radius by a and the height by c multiplies the volume by a^2 c, since the radius appears twice in \pi r^2 h.

Why it is true

Perimeter. If the original has sides s_1, s_2, \dots, s_n, the copy has sides ks_1, ks_2, \dots, ks_n, and by the distributive property

ks_1 + ks_2 + \dots + ks_n = k(s_1 + s_2 + \dots + s_n)

The same holds for a curved boundary: a circle of radius r has circumference 2\pi r, and a circle of radius kr has circumference 2\pi(kr) = k(2\pi r).

Area of a triangle. A triangle has area \tfrac{1}{2}bh, where b is a base and h the height to it. In the similar copy the matching base is kb. The height is a length too, the distance from the opposite vertex to the base, and the similarity carries it to the matching height in the copy, with its right angle kept, so the matching height is kh. The copy’s area is

\tfrac{1}{2}(kb)(kh) = k^2 \cdot \tfrac{1}{2}bh

Area of a polygon. Any polygon can be cut into triangles. The scaled copy is cut into the scaled triangles, each with k^2 times the area of its original, and adding them up gives k^2 times the whole area, by the distributive property again.

Area of a curved region. Area is measured by covering with squares. Cover the region with a fine grid of small squares of side s. In the copy each little square becomes a square of side ks and area k^2s^2, which is k^2 times its old area. So the covering of the copy has exactly k^2 times the area of the covering of the original. As the grid gets finer the coverings get as close as we like to the true areas, and the ratio is k^2 at every stage, so it is k^2 for the true areas too. This is an argument by approximation, not a complete proof: it takes for granted that the coverings really do close in on the area, which is what area means for a curved region, and making that step airtight is a job for calculus. For a circle the result can be checked directly: \pi(kr)^2 = k^2 \cdot \pi r^2.

Surface area. The surface of a solid is a collection of flat or curved pieces, each of them an area. A flat piece is multiplied by k^2 by the argument above, and a curved piece is too, by the same kind of approximation, so their total is multiplied by k^2.

Volume. A prism has volume Bh, base area times height. In a similar prism the base is a similar figure, so its area is k^2B, and the height is kh:

(k^2B)(kh) = k^3 \cdot Bh

Any solid can be filled, as closely as we like, with small cubes of edge s, and each becomes a cube of edge ks and volume k^3s^3 in the copy, so the approximation argument for curved regions works here with cubes in place of squares, and it is an approximation argument here too, not a complete proof. The formulas you know agree: a sphere of radius kr has volume \tfrac{4}{3}\pi(kr)^3 = k^3 \cdot \tfrac{4}{3}\pi r^3.

Changing the dimensions. A box of length \ell, width w and height h has volume \ell wh. The new box has volume (a\ell)(bw)(ch) = abc \cdot \ell wh, using the commutative and associative properties of multiplication to collect the factors. When a = b = c = k this is the k^3 of the theorem. The cylinder is the same calculation with \pi(ar)^2(ch) = a^2c \cdot \pi r^2h.

Worked examples

Example 1

Perimeter and area from one pair of sides

The two right triangles in Figure 3 are similar. The smaller has sides 3, 4 and 5. In the larger, the side matching the side of length 4 is 10. Find the perimeter and the area of the larger triangle.

Figure 3Two similar right triangles. Only one side of the larger is given.

Solution

The scale factor from the smaller to the larger is the ratio of one pair of matching sides:

k = \frac{10}{4} = \frac{5}{2}

The smaller triangle has perimeter 3 + 4 + 5 = 12, and since it is a right triangle its legs serve as base and height, so its area is \tfrac{1}{2}(3)(4) = 6.

The perimeter is multiplied by k:

12 \times \tfrac{5}{2} = 30

The area is multiplied by k^2 = \tfrac{25}{4}:

6 \times \tfrac{25}{4} = \tfrac{75}{2} = 37.5

Check. The larger triangle’s sides are 3k = 7.5, 4k = 10 and 5k = 12.5. They add to 30, and its area is \tfrac{1}{2}(7.5)(10) = 37.5. The area grew by more than six times while the perimeter grew by two and a half times.

Example 2

From an area ratio back to lengths

Two similar pentagons have areas 50\text{ cm}^2 and 162\text{ cm}^2. A side of the smaller one is 7 cm. Find the matching side of the larger one, and the ratio of their perimeters.

Solution

The areas are in the ratio k^2, so

k^2 = \frac{162}{50} = \frac{81}{25}

The scale factor is positive, so take the positive square root: k = \tfrac{9}{5}.

The matching side is 7 \times \tfrac{9}{5} = \tfrac{63}{5} = 12.6 cm. The perimeters are in the ratio k, so the larger perimeter is to the smaller as 9 : 5.

Notice the order. It would be wrong to multiply 7 by \tfrac{81}{25}, because a side is a length, not an area. Find k first, then use k, k^2 or k^3 according to what is being measured.

Example 3

A floor plan

A floor plan is drawn at a scale of 1 : 50, so 1 cm on the plan stands for 50 cm in the house. A rectangular room measures 8 cm by 6.4 cm on the plan. Find the area of the real room, in square meters.

Solution

The real room is similar to its drawing with k = 50. On the plan the area is

8 \times 6.4 = 51.2\text{ cm}^2

Areas are multiplied by k^2 = 2500, so the real area is

51.2 \times 2500 = 128\,000\text{ cm}^2

A square meter is a square 100 cm on each side, so it holds 100^2 = 10\,000 square centimeters, the same rule once more with k = 100. The real area is 128\,000 \div 10\,000 = 12.8\text{ m}^2.

Check. Work with lengths first. The real room is 8 \times 50 = 400 cm = 4 m by 6.4 \times 50 = 320 cm = 3.2 m, and 4 \times 3.2 = 12.8. Using a plan’s scale for lengths is reading a map or plan; the new step here is that an area on the plan must be multiplied by the square of the scale.

Example 4

Two similar cans

Two cans are similar cylinders. The small can is 10 cm tall and holds 400\text{ cm}^3. The large can is 15 cm tall. How much does the large can hold? Both are made of sheet metal of the same thickness; how many times as much metal does the large can use?

Solution

The scale factor from small to large is the ratio of the heights:

k = \frac{15}{10} = \frac{3}{2}

Volume is multiplied by k^3 = \tfrac{27}{8}:

400 \times \tfrac{27}{8} = 1350\text{ cm}^3

The metal covers the surface of the can, and with the same thickness everywhere the amount of metal is proportional to the surface area, which is multiplied by k^2 = \tfrac{9}{4} = 2.25. So the large can uses 2.25 times as much metal.

The large can holds \tfrac{27}{8} = 3.375 times as much as the small one, for 2.25 times the metal. A larger container holds more for each square centimeter of its surface, which is why bulk goods come in big containers. Notice that neither radius was needed: similarity alone settled it.

Example 5

Two figurines of the same metal

Two solid figurines are cast from the same bronze and are similar. The small one has a mass of 270 g and a height of 6 cm. The large one has a mass of 1250 g. Find the height of the large one. The small one is covered with 108\text{ cm}^2 of gold leaf; how much does the large one need?

Solution

Both are solid and made of the same metal, so each cubic centimeter has the same mass, and the masses are in the same ratio as the volumes:

k^3 = \frac{1250}{270} = \frac{125}{27}

The cube root gives k = \tfrac{5}{3}, since 5^3 = 125 and 3^3 = 27.

The height is a length, so the large figurine is 6 \times \tfrac{5}{3} = 10 cm tall.

The gold leaf covers the surface, an area, so it is multiplied by k^2 = \tfrac{25}{9}:

108 \times \tfrac{25}{9} = 300\text{ cm}^2

Check. 10 is \tfrac{5}{3} of 6, and \left(\tfrac{5}{3}\right)^3 \times 270 = \tfrac{125}{27} \times 270 = 1250. The mass is the measurement that is easy to take, and the theorem turns it into a length and an area.

Example 6

Changing some dimensions, not all

A box measures 5 cm by 4 cm by 3 cm. Its length is doubled, its width tripled and its height halved. Find the new volume. Then, separately, a cylinder of radius 3 cm and height 5 cm has its radius doubled, keeping its height. By what factor does its volume change?

Solution

The box starts with volume 5 \times 4 \times 3 = 60\text{ cm}^3. The factors are 2, 3 and \tfrac{1}{2}, one for each length in the volume, so the volume is multiplied by

2 \times 3 \times \tfrac{1}{2} = 3

and the new volume is 180\text{ cm}^3. Directly: the new box is 10 by 12 by 1.5, and 10 \times 12 \times 1.5 = 180. The new box is not similar to the old one, because its edges were multiplied by different factors, so no single k describes it.

The cylinder starts with volume \pi(3)^2(5) = 45\pi\text{ cm}^3. The radius appears twice in \pi r^2h, so doubling it multiplies the volume by 2^2 = 4: the new volume is \pi(6)^2(5) = 180\pi\text{ cm}^3. Doubling the height instead would only double the volume, to 90\pi\text{ cm}^3. Which length changes matters as much as by how much.

Example 7

Filling a cone-shaped glass

A glass is a cone with its point at the bottom, 12 cm tall, and it holds 240\text{ cm}^3 when full. It is filled to a depth of 6 cm, as in Figure 4. How much liquid is in it? To what depth must it be filled to hold half of its full amount?

Figure 4The liquid is a cone of the same shape as the glass, half as tall.

Solution

The liquid takes the shape of a cone with the same point as the glass. Its surface is a circle parallel to the rim, and its sides lie along the sides of the glass, so the liquid is a smaller cone of the same shape: a reduction of the full glass, with every length in proportion to the depth.

Filled to 6 cm, the scale factor is k = \tfrac{6}{12} = \tfrac{1}{2}, and the volume is multiplied by k^3 = \tfrac{1}{8}:

240 \times \tfrac{1}{8} = 30\text{ cm}^3

Half the height holds only one eighth of the glass.

For half the amount, the volume factor must be \tfrac{1}{2}, so

k^3 = \tfrac{1}{2} \qquad\Longrightarrow\qquad k = \frac{1}{\sqrt[3]{2}}

and the depth is

\frac{12}{\sqrt[3]{2}} = 6\sqrt[3]{4} \approx 9.52\text{ cm}

Check. \left(\tfrac{9.52}{12}\right)^3 \approx 0.4993, which is very nearly \tfrac{1}{2}. The glass is half full only when the liquid is about four fifths of the way up, because the wide top of a cone holds most of its volume.

Boundaries

The figures must be similar. The rule needs every length multiplied by the same k. Stretch a square to twice its width but not its height and the area doubles, not quadruples. Two shapes with the same perimeter can have different areas, as perimeter versus area shows, and only similarity ties one to the other. When the dimensions change by different factors, multiply the factors one at a time, as in Example 6.

The exponent counts dimensions, not sides. Perimeter, height, radius and circumference all grow by k, because each is a length, however many sides the figure has. Surface area grows by k^2 even though it belongs to a solid, because it is an area.

Mass follows volume only for the same material, solid all the way through. The figurines in Example 5 could be compared by mass because each cubic centimeter weighed the same. A hollow object whose wall keeps the same thickness when it is enlarged uses material in proportion to its surface area, as the cans in Example 4 did, and then mass grows like k^2, not k^3.

Units are lengths in disguise. Changing centimeters to meters is a scaling with k = 100, so square centimeters to square meters is a factor of 10\,000 and cubic centimeters to cubic meters a factor of 1\,000\,000. Converting an area with 100 uses a length factor for an area, the reverse of the slip Example 2 warned against.

Shapes scale; what they are made of does not. A model built at one tenth of the size, in the same material, has \tfrac{1}{1000} of the weight standing on legs with \tfrac{1}{100} of the cross-section. The geometry is exact; whether the model behaves like the real thing is a question about materials that the geometry does not answer.

Where this goes next

Example 7 is the start of a method. A plane slice through a cone or pyramid parallel to its base cuts off a smaller similar solid, so the slice at a fraction t of the way from the point has area t^2 times the base. Slicing a pyramid or cone into thin layers to see why the volume formulas hold uses exactly that fact, and it is behind the \tfrac{1}{3} in the volume of a cone or pyramid.

Areas growing like the square of a length also gives a new view of the Pythagorean theorem. Similar figures drawn on the three sides of a right triangle have areas in the ratio a^2 : b^2 : c^2, so the two on the legs always add up to the one on the hypotenuse, whatever the shape (Exercise 22). The proof with similar triangles is built on the same idea.

Outside geometry the theorem is the reason surface and volume part company as things grow. Surface area per unit of volume is multiplied by \tfrac{1}{k} (Exercise 24), so large objects cool more slowly than small ones, and a small creature loses heat far faster for its size than a large one. Mass is volume times density, which is how Example 5 could weigh its way to a length.

Exercises

Lengths and perimeters

Two quadrilaterals are similar. The smaller has sides 2, 3, 4 and 5. The shortest side of the larger is 7. Find the perimeter of the larger.
Two similar triangles have perimeters 24 and 36. A side of the smaller is 10. Find the matching side of the larger.
A photo 10 cm by 15 cm is enlarged so that its longer side is 45 cm. Find its shorter side and its perimeter.

Areas

A triangle is enlarged with scale factor 4. The original has area 7\text{ cm}^2. Find the area of the enlargement.
Two pentagons are similar, and a side of length 6 in the smaller matches a side of length 9 in the larger. The larger has area 162. Find the area of the smaller.
Two similar rectangles have areas 20 and 45. Find the ratio of their perimeters, larger to smaller.
A round pizza 16 inches across and one 12 inches across have the same thickness. How many times as much pizza is in the larger one?
A map has scale 1 : 25\,000. A lake covers 6\text{ cm}^2 on the map. Find the area of the real lake in square kilometers.
A shape is enlarged, and its area goes from 18\text{ cm}^2 to 50\text{ cm}^2. Find the scale factor.

Volumes and surface areas

Two boxes are similar with scale factor 2. The smaller has volume 15\text{ cm}^3 and surface area 46\text{ cm}^2. Find the volume and surface area of the larger.
Two similar cones have heights 4 cm and 6 cm. The larger has volume 54\pi\text{ cm}^3. Find the volume of the smaller.
Two similar solids have surface areas 64\text{ cm}^2 and 144\text{ cm}^2. The smaller has volume 32\text{ cm}^3. Find the volume of the larger.
Two solid balls of the same material have radii 3 cm and 6 cm. The smaller has mass 120 g. Find the mass of the larger.
Two similar solids have volumes in the ratio 8 : 27. Find the ratio of their surface areas.
Two similar bottles hold 2 liters and 0.25 liters. The larger is 30 cm tall. How tall is the smaller?

Changing the dimensions

A rectangle has its length tripled and its width halved. By what factor does its area change?
A box measures 6 cm by 4 cm by 2 cm. Its length and width are doubled and its height is unchanged. Find the new volume.
A cylinder has its radius halved and its height multiplied by 4. By what factor does its volume change?
A cone has its radius doubled and its height tripled. By what factor does its volume change?

Reasoning

Explain why doubling every edge of a box does not double its volume, and find what it does to the volume.
Give two rectangles whose perimeters are in the ratio 2 : 1 but whose areas are not in the ratio 4 : 1. Why does this not contradict the theorem?
Semicircles are drawn outward on the three sides of a right triangle with sides 3, 4 and 5. Show that the two semicircles on the legs have a total area equal to the semicircle on the hypotenuse. Then explain why the same is true for any three similar figures drawn on the sides of any right triangle, with the sides as matching lengths.
A model of a building is made at a scale of 1 : 100. Someone says the model needs \tfrac{1}{100} of the glass of the real building. What fraction does it need? Explain.
A cube of edge 1 and a cube of edge 2 are compared. Find surface area divided by volume for each. Then show that for any two similar solids, surface area divided by volume is multiplied by \tfrac{1}{k}.
A dilation multiplies the area of a triangle by 9. Must it have multiplied the perimeter by 3? Then: a unit square is turned into a rectangle with 9 times its area by stretching it in one direction only. By what factor is its perimeter multiplied? Explain the difference.

Answers

49. The scale factor is \tfrac{7}{2}, matching the shortest sides, and the smaller perimeter is 2 + 3 + 4 + 5 = 14, so the larger is 14 \times \tfrac{7}{2} = 49.
15. The perimeters give k = \tfrac{36}{24} = \tfrac{3}{2}, and 10 \times \tfrac{3}{2} = 15.
The shorter side is 30 cm and the perimeter is 150 cm. The scale factor is \tfrac{45}{15} = 3, so the shorter side is 10 \times 3, and the perimeter is 3 \times 50 = 150, or directly 2(30 + 45) = 150.
112\text{ cm}^2. Area is multiplied by 4^2 = 16, and 7 \times 16 = 112.
72. From larger to smaller, k = \tfrac{6}{9} = \tfrac{2}{3}, so the area is multiplied by \tfrac{4}{9}: 162 \times \tfrac{4}{9} = 72.
3 : 2. The areas give k^2 = \tfrac{45}{20} = \tfrac{9}{4}, so k = \tfrac{3}{2}, and perimeters are in the ratio k.
\tfrac{16}{9} times, about 1.78. The scale factor is \tfrac{16}{12} = \tfrac{4}{3}, and the area, which is what the same thickness makes the amount of pizza depend on, is multiplied by \left(\tfrac{4}{3}\right)^2 = \tfrac{16}{9}.
0.375\text{ km}^2. One centimeter on the map is 25\,000 cm = 250 m = 0.25 km, so 1\text{ cm}^2 on the map is 0.25^2 = 0.0625\text{ km}^2, and 6 \times 0.0625 = 0.375.
\tfrac{5}{3}. k^2 = \tfrac{50}{18} = \tfrac{25}{9}, and the positive square root is \tfrac{5}{3}.
Volume 120\text{ cm}^3, surface area 184\text{ cm}^2. Volume is multiplied by 2^3 = 8 and surface area by 2^2 = 4.
16\pi\text{ cm}^3. From larger to smaller, k = \tfrac{4}{6} = \tfrac{2}{3}, so the volume is 54\pi \times \tfrac{8}{27} = 16\pi.
108\text{ cm}^3. k^2 = \tfrac{144}{64} = \tfrac{9}{4}, so k = \tfrac{3}{2} and 32 \times \tfrac{27}{8} = 108.
960 g. Same material, solid, so mass follows volume. k = 2, so the mass is multiplied by 2^3 = 8: 120 \times 8 = 960.
4 : 9. k^3 = \tfrac{8}{27} gives k = \tfrac{2}{3}, and surface areas are in the ratio k^2 = \tfrac{4}{9}.
15 cm. The volumes are in the ratio \tfrac{2}{0.25} = 8, so \displaystyle k = \sqrt[3]{8} = 2 from smaller to larger, and 30 \div 2 = 15.
\tfrac{3}{2}. The area is length times width, so it is multiplied by 3 \times \tfrac{1}{2} = \tfrac{3}{2}.
192\text{ cm}^3. The new box is 12 by 8 by 2, and 12 \times 8 \times 2 = 192, which is the old volume 48 times 2 \times 2 \times 1 = 4.
It does not change: the factor is 1. The radius appears squared, so the factor is \left(\tfrac{1}{2}\right)^2 \times 4 = 1.
12. The cone’s volume \tfrac{1}{3}\pi r^2h has the radius twice and the height once, so the factor is 2^2 \times 3 = 12.
It multiplies the volume by 8. Volume is length times width times height, and all three are doubled, so the volume is multiplied by 2 \times 2 \times 2 = 2^3 = 8, not 2. Equivalently, the larger box can be cut into 8 copies of the original.
For example, a 1 by 1 square and a 1 by 3 rectangle. Their perimeters are 4 and 8, ratio 2 : 1, but their areas are 1 and 3, ratio 3 : 1. The theorem is only about similar figures, and these two are not similar: one length was multiplied by 3 and the other by 1.
A semicircle on a side of length s has radius \tfrac{s}{2} and area \tfrac{1}{2}\pi\left(\tfrac{s}{2}\right)^2 = \tfrac{\pi s^2}{8}. On the legs this gives \tfrac{9\pi}{8} + \tfrac{16\pi}{8} = \tfrac{25\pi}{8}, the area on the hypotenuse. In general, if a figure drawn on a side of length 1 has area C, the similar figure on a side of length s has scale factor s and area Cs^2. With legs a and b and hypotenuse c, the areas are Ca^2, Cb^2 and Cc^2, and Ca^2 + Cb^2 = C(a^2 + b^2) = Cc^2 by the Pythagorean theorem.
\tfrac{1}{10\,000}. Glass covers an area, and areas are multiplied by k^2 = \left(\tfrac{1}{100}\right)^2 = \tfrac{1}{10\,000}. The claim uses the length factor for an area.
The cube of edge 1 has surface area 6 and volume 1, giving 6. The cube of edge 2 has surface area 24 and volume 8, giving 3. In general the copy’s surface area is k^2S and its volume k^3V, so its ratio is \tfrac{k^2S}{k^3V} = \tfrac{1}{k} \cdot \tfrac{S}{V}.
Yes: a dilation makes a similar figure, so k^2 = 9, and since k > 0, k = 3, which is the factor for the perimeter. The stretched square is 9 by 1, with perimeter 20 against the square’s 4, so its perimeter is multiplied by 5. The rectangle is not similar to the square, so the area factor says nothing about the perimeter; only for similar figures does k^2 = 9 force the length factor to be 3.