Similar figures
One shape looks like a bigger, turned-round copy of another. What has to be true of their angles and sides for that to be exactly so?
Where you are 10 in Similarity
Why this exists
Similarity transformations say what “the same shape” means: two figures are similar when rigid motions and a dilation carry one exactly onto the other. That is the right meaning, but two quadrilaterals printed with their sides and angles labeled do not come with a list of moves. To decide whether they are the same shape, and how much bigger one is, we need a test that works from the measurements themselves.
For polygons there is one: two polygons are similar when their angles match and their sides all grow by the same factor. A missing side of a similar figure then comes from a Proportion, with no moves at all.
Definition
Two figures are similar when one can be carried exactly onto the other by rigid motions (slides, turns and flips) and a dilation. We write F \sim G, read “F is similar to G”.
For polygons this means the same as the following, which is the form used in practice. Two polygons are similar when their vertices can be matched up, one to one and in order around each polygon, so that
- every pair of matching angles is equal, and
- every pair of matching sides is in the same ratio.
That common ratio, a side of the second figure divided by the matching side of the first, is the scale factor k from the first figure to the second. It is always positive: k > 1 for an enlargement, k < 1 for a reduction, and k = 1 when the figures are congruent.
A similarity statement names the matching by its order: first letter with first letter, second with second, and so on. ABCD \sim PQRS claims \angle A = \angle P, \angle B = \angle Q, \angle C = \angle R and \angle D = \angle S, and that \tfrac{PQ}{AB}, \tfrac{QR}{BC}, \tfrac{RS}{CD} and \tfrac{SP}{DA} all equal k.
The two triangles in Figure 1 are similar, even though the second is larger, flipped over and turned. Example 1 reads off which vertex matches which.
Intuition
Shape is made of two things: how sharply the figure turns at each corner, and how long each side is compared with the others. Similar figures agree on both. Size is the one thing they may disagree on, and it changes the same way everywhere: a dilation cannot make one side longer without making every side longer by the same factor.
Comparing across the figures, side with matching side, every ratio is the scale factor k. Comparing within one figure, side with side, gives a ratio that the other figure repeats exactly: if a rectangle’s long side is twice its short side, so is every rectangle similar to it.
The two conditions do separate jobs: equal angles with sides out of proportion give a stretched shape, and sides in proportion with different angles a squashed one.
Results
Similar polygons. Polygons F and G are similar exactly when their vertices can be matched in order so that matching angles are equal and matching sides are all in one ratio k.
The scale factor goes both ways. If F \sim G with scale factor k, then G \sim F with scale factor \tfrac1k. A missing side of G is k times its match in F, and a missing side of F is its match in G divided by k.
Ratios within a figure are kept. If ABCD \sim PQRS, then
and the same holds for any other two sides and their matches.
Matching keeps the order of sizes. Multiplying every side by the same positive k keeps the longest side longest and the shortest shortest, so in similar polygons the longest side matches the longest side, and so on down.
Some families are always similar. Any two regular polygons with the same number of sides: their angles agree, and each has one side length, so all matching sides are in one ratio.
Triangles need less. For triangles one condition brings the other with it: two pairs of equal angles are enough (AA similarity), and so are three pairs of sides in one ratio (SSS similarity).
Why it is true
Moves give the two conditions. Suppose rigid motions and a dilation with scale factor k carry polygon F onto polygon G. Each vertex of F lands on a vertex of G, which gives the matching. Rigid motions change no length and no angle; a dilation keeps every angle and multiplies every length by k. So matching angles are equal and every side of G is k times its match in F.
The two conditions give moves. Suppose a matching makes every angle of G equal to its match in F and every side of G equal to k times its match. Dilate F with scale factor k about any center. The image F' has the angles of F, which are the angles of G, and sides k times those of F, which are the sides of G. Two polygons whose matching sides and matching angles are all equal are congruent, a fact about congruence used here without proof, so rigid motions carry F' onto G. A dilation followed by rigid motions is what the definition asks for, so F \sim G.
Ratios within a figure. If PQ = k \cdot AB and QR = k \cdot BC, then
because the common factor k cancels. Going back from G to F multiplies every side by \tfrac1k, the dilation that undoes the first.
Worked examples
Reading a similarity statement
The triangles in Figure 1 are similar. Write a similarity statement that begins \triangle JKL, say which side of \triangle XYZ matches \overline{JK}, and write the equal ratios of matching sides.
SolutionMatching angles are equal, so pair the vertices by their angles. \triangle JKL has 75^\circ at J, 40^\circ at K and 65^\circ at L. \triangle XYZ has 65^\circ at X, 75^\circ at Y and 40^\circ at Z. So J matches Y, K matches Z and L matches X. Writing the second triangle’s letters in that order,
The side \overline{JK} joins the first two vertices, so its match joins the first two vertices of the second name: \overline{YZ}. In the same way \overline{KL} matches \overline{ZX}, and \overline{LJ} matches \overline{XY}. The ratios of matching sides are all the scale factor:
\triangle JKL \sim \triangle XYZ would be false: it pairs the 75^\circ angle at J with the 65^\circ one at X. Any order that keeps the same pairs is correct, such as \triangle KLJ \sim \triangle ZXY.
Check. Both triangles have angles adding to 75^\circ + 40^\circ + 65^\circ = 180^\circ, as the angles of a triangle must.
Which trapezoid is the same shape
Each trapezoid in Figure 2 has angles of 60^\circ at both ends of its longer base and 120^\circ at both ends of its shorter base. Is ABCD similar to EFGH? Is it similar to JKLM?
The angles. The 60^\circ angles are at A and B, at E and F (the trapezoid is upside down), and at J and K. So matching the letters in order pairs every angle with an equal one, and everything rests on the sides.
EFGH. Divide each side by its match in ABCD:
All four ratios are equal and all four angles match, so ABCD \sim EFGH with scale factor \tfrac32.
JKLM. The first two ratios already disagree:
One similarity has one scale factor, so this matching fails. The only other matching that pairs 60^\circ angles with 60^\circ angles is the mirror one (A with K, B with J), and because the legs of each trapezoid are equal it gives the same ratios. So ABCD and JKLM are not similar, although every angle of one appears in the other: JKLM is relatively longer and flatter.
Each condition alone falls short
Use Figure 3 to decide (a) whether the square with sides 6 and the 6 by 15 rectangle are similar, and (b) whether the square with sides 10 and the rhombus with sides 8 and a 75^\circ angle are similar.
(a) Every angle of both shapes is a right angle, so the angles agree under any matching. The sides do not. The square’s sides are all 6, and the rectangle’s are 6, 15, 6, 15. Under any matching, some side of 6 meets a side of 6 and some side of 15 meets a side of 6:
Two different ratios, so the shapes are not similar. Equal angles are not enough.
(b) Every side of the square is 10 and every side of the rhombus is 8, so every ratio of matching sides is \tfrac{10}{8} = \tfrac54, whatever the matching. The angles do not agree. A rhombus is a parallelogram, so its neighboring angles add to 180^\circ and its angles are 75^\circ, 105^\circ, 75^\circ and 105^\circ, while every angle of the square is 90^\circ. The square has no 75^\circ angle to match, so the shapes are not similar. Sides in one ratio are not enough. Triangles are the exception, which is why they have shortcuts of their own.
A scale factor and the missing sides
In Figure 4, ABCD \sim PQRS. ABCD has AB = 10, BC = 5, CD = 7, DA = 4 and right angles at A and D, and RS = 8.4. Find the scale factor from ABCD to PQRS, the other three sides of PQRS, and the angles at P and S.
Read the matching from the statement: A with P, B with Q, C with R, D with S. So \overline{RS}, which joins the third and fourth vertices, matches \overline{CD}. That is the only pair with both lengths known, and it gives the scale factor:
Every side of PQRS is 1.2 times its match:
Matching angles are equal, so \angle P = \angle A = 90^\circ and \angle S = \angle D = 90^\circ.
Check. Within ABCD, \tfrac{AB}{CD} = \tfrac{10}{7}. Within PQRS, \tfrac{PQ}{RS} = \tfrac{12}{8.4} = \tfrac{10}{7}, as the ratios within a figure must agree.
The matching came from the statement, not from the flipped picture. From PQRS back to ABCD the scale factor is \tfrac{1}{1.2} = \tfrac56.
A phone screen and a tablet screen
Zuzana’s phone screen is a rectangle 6 cm wide and 13 cm tall. Her tablet’s screen has exactly the same shape and is 9.6 cm wide. (a) How tall is the tablet screen? (b) A square icon is 0.9 cm wide on the phone. When the same picture fills the tablet screen, how wide is the icon? (c) Ambrose’s tablet screen is 15 cm wide and 20 cm tall. Is it the same shape as Zuzana’s phone screen?
Solution“Exactly the same shape” means similar. Both screens are rectangles, so the angles agree, and width matches width.
(a) The scale factor from phone to tablet is
so the tablet screen is 1.6 \cdot 13 = 20.8 cm tall. The ratio within the figure agrees: 9.6 \cdot \tfrac{13}{6} = 20.8.
(b) The whole picture is enlarged by the same factor, icon included: 1.6 \cdot 0.9 = 1.44 cm.
(c) Compare width with width and height with height:
The ratios differ, so the screens are not similar: a picture filling the phone would have to be stretched, or leave empty strips, to fill this tablet.
Boundaries
Both conditions, under one matching. For polygons with four or more sides, equal angles alone and proportional sides alone each fall short (Example 3). Both must hold for the same matching of vertices: angles that agree under one matching and sides that agree under another prove nothing.
The order in the statement is part of the claim. ABCD \sim PQRS and ABCD \sim QRSP are different statements, and one can be true while the other is false.
Adding is not scaling. Similar figures have every side multiplied by the same number. Adding the same length to every side changes the ratios within the figure, so the result is in general not similar (Exercise 16).
The test is for polygons. Matching angles and proportional sides need corners and straight sides. For circles and other curved figures, similarity is decided by the definition with moves, as in similarity transformations.
A drawing is not a measurement. Similarity is decided from the stated lengths and angles, never by eye.
Where this goes next
Triangles get shortcuts. Two pairs of equal angles already make two triangles similar, which is AA similarity, and three proportional pairs of sides, or two with the angle between them equal, are SSS and SAS similarity. Once two triangles are known to be similar, the ratios of their sides find lengths that cannot be measured directly, such as the height of a tree from its shadow, which is proportions in similar triangles.
Perimeters scale by k but areas by k^2, which is scale factor and perimeter, area and volume and scale factor and area. Plans, maps and models are similar figures made on purpose (scale factor, scale drawings). And because right triangles with the same acute angle are similar, the ratio of two of their sides depends on the angle alone, which is where sine, cosine and tangent begin.
Exercises
Reading similarity statements
- Pentagon ABCDE \sim pentagon VWXYZ. List the five pairs of equal angles, and write the equal ratios of matching sides.
- \triangle KLM \sim \triangle TSR. Which side of \triangle TSR matches \overline{LM}? Which angle matches \angle K? Rewrite the statement so that the second triangle’s name begins with R.
- \triangle ABC has \angle A = 50^\circ, \angle B = 60^\circ and \angle C = 70^\circ. \triangle STU is similar to it, with \angle S = 70^\circ, \angle T = 50^\circ and \angle U = 60^\circ. Write a similarity statement that begins \triangle ABC.
Similar or not
- Is a 6 by 14 rectangle similar to a 15 by 35 rectangle? If so, give the scale factor from the first to the second.
- Is an 8 by 14 rectangle similar to a 12 by 20 rectangle? Explain.
- Parallelogram (a) has sides 4 and 6 and a 50^\circ angle. Parallelogram (b) has sides 6 and 9 and a 50^\circ angle. Parallelogram (c) has sides 6 and 8 and a 50^\circ angle. Which of (b) and (c) is similar to (a)? Give the scale factor where there is one.
Scale factors and missing sides
- ABCD \sim EFGH, with AB = 6, BC = 9, CD = 12, DA = 15 and EF = 4. Find the scale factor from ABCD to EFGH, and FG, GH and HE.
- PQRST \sim VWXYZ, with PQ = 5, ST = 7.5, \angle R = 110^\circ and VW = 8. Find YZ and \angle X.
- Two isosceles trapezoids are similar. The larger has parallel sides 20 and 12 and legs 10. The longer parallel side of the smaller is 15. Find the scale factor from the larger to the smaller, and the other sides of the smaller.
- A hexagon has sides 3, 4, 5, 3, 4, 5 in order. A similar hexagon has longest side 12. Find its sides in the same order.
Unknowns in similar figures
- ABCD \sim WXYZ, with AB = 8, BC = 12, WX = 10 and XY = 2x + 3. Find x.
- A 5 by x rectangle is similar to an 8 by 12 rectangle. Find every possible value of x.
- A rectangular sheet of paper is cut in half across its long side, so each half keeps the full short side and has half the long side. Each half is similar to the whole sheet. If the short side is 1 and the long side is r, find r. Is an A4 sheet, 21.0 cm by 29.7 cm, close to this shape?
Reasoning
- Explain why two rectangles are similar exactly when the ratio of short side to long side is the same in both.
- Dilnoza says that all right triangles are similar, since every one of them has a right angle. Is she right? Explain with an example.
- Bethan adds a border 1 unit wide all the way round a 4 by 6 rectangle, which makes a 6 by 8 rectangle. Is the new rectangle similar to the old one? Explain.
Answers
- \angle A = \angle V, \angle B = \angle W, \angle C = \angle X, \angle D = \angle Y and \angle E = \angle Z, and the ratios \tfrac{VW}{AB}, \tfrac{WX}{BC}, \tfrac{XY}{CD}, \tfrac{YZ}{DE} and \tfrac{ZV}{EA} are all equal. Letters in the same position match, and so do the sides joining them.
- \overline{SR} matches \overline{LM}, and \angle T matches \angle K. The matching is K with T, L with S and M with R; starting from M and keeping the pairs gives \triangle MKL \sim \triangle RTS.
- \triangle ABC \sim \triangle TUS. Pair equal angles: A (50^\circ) with T, B (60^\circ) with U, and C (70^\circ) with S.
- Yes, with scale factor \tfrac52. All the angles are right angles, short side to short side is \tfrac{15}{6} = \tfrac52, and long side to long side is \tfrac{35}{14} = \tfrac52.
- No. Short side to short side is \tfrac{12}{8} = \tfrac32 but long side to long side is \tfrac{20}{14} = \tfrac{10}{7}, and one similarity has one scale factor.
- (b) is similar to (a), with scale factor \tfrac32; (c) is not. All three have angles 50^\circ, 130^\circ, 50^\circ and 130^\circ, and matching keeps the order of sizes, so short side matches short side. For (b), \tfrac64 = \tfrac96 = \tfrac32. For (c), \tfrac64 = \tfrac32 but \tfrac86 = \tfrac43.
- k = \tfrac46 = \tfrac23, so FG = \tfrac23 \cdot 9 = 6, GH = \tfrac23 \cdot 12 = 8 and HE = \tfrac23 \cdot 15 = 10.
- YZ = 12 and \angle X = 110^\circ. The scale factor is \tfrac{VW}{PQ} = \tfrac85, and \overline{YZ} matches \overline{ST}, so YZ = \tfrac85 \cdot 7.5 = 12. \angle X matches \angle R.
- k = \tfrac{15}{20} = \tfrac34. The shorter parallel side is \tfrac34 \cdot 12 = 9 and each leg is \tfrac34 \cdot 10 = 7.5. The longer parallel side matches the longer one, since matching keeps the order of sizes.
- 7.2, 9.6, 12, 7.2, 9.6, 12. The longest side matches the longest side, 5, so k = \tfrac{12}{5} = 2.4, and 2.4 \cdot 3 = 7.2 and 2.4 \cdot 4 = 9.6.
- x = 6. The scale factor is \tfrac{10}{8} = \tfrac54, so XY = \tfrac54 \cdot 12 = 15, and 2x + 3 = 15 gives x = 6.
- x = 7.5 or x = \tfrac{10}{3}. If 5 is the short side, it matches 8 and x matches 12: \tfrac{x}{5} = \tfrac{12}{8} gives x = 7.5, which is indeed longer than 5. If 5 is the long side, x matches 8 and 5 matches 12: \tfrac{x}{5} = \tfrac{8}{12} gives x = \tfrac{10}{3}, which is indeed shorter than 5.
- \displaystyle r = \sqrt2 \approx 1.414, and A4 is very close: \tfrac{29.7}{21.0} \approx 1.414. A half has sides 1 and \tfrac r2. Its long side must be 1: if it were \tfrac r2, the half’s long over short would be \tfrac r2, less than the whole sheet’s r. Long over short in both gives r = \tfrac{1}{r/2} = \tfrac2r, so r^2 = 2, and a length is positive. The A paper sizes are built on this ratio, so each size halves into the next.
- All angles of a rectangle are right angles, so only the sides matter. Short sides s_1, s_2 match and long sides l_1, l_2 match (matching keeps the order of sizes), so similarity means \tfrac{s_2}{s_1} = \tfrac{l_2}{l_1}. Multiplying by \tfrac{s_1}{l_2} turns this into \tfrac{s_2}{l_2} = \tfrac{s_1}{l_1}, and multiplying by \tfrac{l_2}{s_1} turns it back.
- No. Take a right triangle with legs 3 and 4 and one with legs 1 and 1. Their right angles match, so the legs must match each other, and the ratios of matching legs are \tfrac31 = 3 and \tfrac41 = 4 whichever way round they are paired. Not one ratio, so not similar. One pair of equal angles is not enough; AA similarity needs two.
- No. The angles are all right angles, but short side to short side is \tfrac64 = \tfrac32 and long side to long side is \tfrac86 = \tfrac43. Adding 2 is a bigger share of the short side than of the long one, so the new rectangle is squarer: \tfrac46 = \tfrac23 but \tfrac68 = \tfrac34.