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Similar triangles have one common scale factor between corresponding lengths. That does not mean you can divide any side in one triangle by any side in the other. The sides must match the same positions in the triangles’ angle structure. A correct proportion starts with a correct correspondence.
If triangle ABC is similar to triangle PQR in that stated order, A matches P, B matches Q and C matches R. Therefore AB matches PQ, BC matches QR and AC matches PR. Turning either drawing on the page does not change those relationships. “Top side” and “left side” are unreliable substitutes for vertex names.
This guide shows how to establish the matching before calculating, then use it in missing-length, perimeter and area questions. The 2027 G3 Mathematics syllabus provides official curriculum context. Follow your school’s sequence for the particular similarity criteria and extensions you need now.
Contents
Sections 1–5
Sections 6–10
Sections 11–15
Sections 16–20
Sections 21–25
Sections 26–30
First establish that the triangles are similar using the information given and a valid criterion. Then match corresponding vertices through equal angles or the stated similarity order. Only after that write a consistent ratio, such as PQ/AB = QR/BC = PR/AC, where every fraction compares the second triangle with the first.
If AB is six centimetres and its partner PQ is nine centimetres, the scale factor from ABC to PQR is 9/6 = 3/2. A side BC of ten centimetres corresponds to QR = 10 × 3/2 = 15 centimetres. Pairing BC with PR instead would answer a different geometric relationship.
Check direction and size. A length in the enlarged triangle should be larger when the scale factor exceeds one. If a calculation gives a smaller partner, inspect whether you reversed one ratio or matched the wrong sides. The ratio method is dependable when the geometric correspondence and comparison direction are both preserved.
For triangles, similarity means corresponding angles agree and corresponding side lengths share a common positive ratio. The triangles may differ in size. A rotation, reflection or translation can change their appearance on the page while preserving the shape relationship. Their orientation is not part of the side-length scale factor.
The word corresponding identifies which parts play the same role. In a triangle with three different angles, the side opposite the smallest angle matches the side opposite the smallest angle in the similar triangle. The same applies to the other two angle positions. The correspondence belongs to the geometry, not to the page layout.
Begin by thinking about two complete triangles rather than searching for four nearby numbers to cross-multiply. A proportion is a consequence of similarity. It does not create similarity merely because the arithmetic produces a tidy answer. Geometry supplies the relationship that makes the numerical equation valid.
Two triangles that look alike in a sketch are not automatically similar. You need stated similarity or sufficient angle or side information. For example, two pairs of corresponding equal angles establish triangle similarity by the angle-angle criterion, because the remaining angle pair must also agree.
Other familiar routes include all three corresponding side pairs in one ratio, or two side pairs in one ratio with their included angles equal. Use the criterion appropriate to your course and the information actually provided. A pair of equal angles alone, meaning only one angle in each triangle, is not sufficient.
If a question already states triangle ABC is similar to triangle PQR, you may use that fact without proving it again unless a proof is requested. If it only supplies a drawing, inspect the marks and accompanying text. Clear reasoning begins by distinguishing a given relationship from one you still need to establish.
The notation triangle ABC similar to triangle PQR pairs the first named vertices, then the second, then the third. This is why vertex order matters. It communicates a mapping rather than merely listing the same two sets of three letters. Changing the order on one side can change the stated correspondence.
For that mapping, side AB joins the first and second vertices, so it matches PQ. Side BC joins the second and third, so it matches QR. Side AC joins the first and third, so it matches PR. The matching can be reconstructed without deciding which side looks horizontal.
When writing your own similarity statement, choose an order that reflects the angle matches you have established. Do not copy the vertices from the drawing in whichever order is easiest to read. A correct geometric proof followed by an incorrectly ordered statement can lead to wrong ratios in the next part.
Suppose A matches R, B matches P and C matches Q. Then AB matches RP, BC matches PQ and AC matches RQ. The similarity statement can be written ABC similar to RPQ. The triangles’ labels are intentionally out of alphabetical order to show why alphabetical matching is not a general rule.
A side is determined by the pair of endpoints it joins. Map both endpoints to identify its partner. AB maps to the side joining A’s partner R and B’s partner P, namely RP. Mapping only one endpoint does not identify a unique side because each vertex belongs to two sides.
This endpoint method is especially useful when a figure is rotated or when triangles overlap. You can write the vertex mapping in a short row beside the diagram, then derive every side match from it. It gives you a stable procedure when visual intuition is distracted by the drawing.
Triangle ABC is similar to triangle PQR in that order. The corresponding lengths AB and PQ are six and nine centimetres. Another side BC is ten centimetres, and AC is eight centimetres. Find QR and PR. All lengths are positive, and the stated correspondence is part of the problem.
The scale factor from ABC to PQR is PQ/AB = 9/6 = 3/2. Therefore QR = BC × 3/2 = 15 centimetres and PR = AC × 3/2 = 12 centimetres. Check the ratios: 9/6, 15/10 and 12/8 all simplify to three halves.
The original lengths six, eight and ten form a valid triangle, as do nine, twelve and fifteen. The final values fit both the scale relationship and the expectation that every length grows by fifty per cent. A result such as QR = 20/3 would suggest that the inverse scale factor was applied in the wrong direction.
The table follows the stated order ABC similar to PQR. It shows how each side’s endpoint pair determines its partner. This is useful working when several lengths are printed in one diagram: the correspondence can be checked before any arithmetic begins, reducing the chance of a neat calculation with the wrong pair.
| Part in triangle ABC | Corresponding part in PQR | Reason from vertex mapping |
|---|---|---|
| Angle A | Angle P | First vertices correspond |
| AB | PQ | A maps to P and B maps to Q |
| BC | QR | B maps to Q and C maps to R |
| AC | PR | A maps to P and C maps to R |
You do not need a full table for every short exercise. A labelled mapping or correctly ordered similarity statement may be enough. Use the amount of working that keeps the relationship reliable for you. The table is a tool for making correspondence visible, not a separate requirement to memorise.
From the worked example, second-triangle length divided by first-triangle length is 3/2. Therefore PQ/AB = QR/BC. If you instead write AB/PQ = QR/BC, one fraction compares first to second while the other compares second to first. They are generally reciprocals, not equal values.
Using numbers, 6/9 is two thirds while 15/10 is three halves. Their inequality exposes the direction mismatch. You can use the reverse direction consistently: AB/PQ = BC/QR = AC/PR = 2/3. Both conventions work when every ratio follows the same orientation.
Write a small label such as enlarged/original above a ratio while learning. It tells you what each numerator and denominator represents. Cross-multiplication will correctly solve even an incorrectly formed proportion, so this structural check must happen before the algebra. Calculator accuracy cannot repair a reversed relationship.
You can compare two sides inside each triangle rather than comparing across triangles. With ABC similar to PQR, AB/BC = PQ/QR. The first fraction compares one pair of roles in ABC, and the second compares their matching roles in PQR. Both shapes have the same internal proportions.
In the worked example, 6/10 = 9/15 = 3/5. This is a different ratio from the scale factor 3/2. It describes shape rather than size change. Confusing these two valid types of proportion can lead a student to expect every ratio in a similarity question to equal the same number.
Read what each fraction compares. Across-triangle corresponding ratios give the scale factor in a chosen direction. Matching within-triangle ratios give equal side relationships in the two shapes. Either can find a missing length, provided the roles align and the equation preserves the same comparison on both sides.
Imagine rotating triangle PQR so that its longest side, previously near the bottom of a sketch, now appears on the left. PR remains the side joining P and R. If A matches P and C matches R, AC still corresponds to PR. The rotation has not changed the names or angle relationships.
A student who pairs bottom with bottom may switch to a different side after rotation, even though the mathematical object is unchanged. This is why location language should be replaced by endpoint names or angle roles. A diagram is a representation you can turn; correspondence must survive that turn.
As practice, redraw a simple labelled triangle in a new orientation without changing the labels attached to its vertices. Then identify the partner of one side using the mapping alone. The exercise checks whether you are tracking the triangle’s structure or relying on its familiar position on the page.
A reflected triangle may appear to run clockwise where its partner runs anticlockwise. Similarity can still hold. The reflection preserves angles and changes every corresponding length by the same scale factor if an enlargement or reduction is included. The visible handedness of the drawing is not a reason to reject the relationship.
Match the marked equal angles or use the stated vertex order. Do not insist that the letters must travel around both drawings in the same clockwise direction on the paper. A reflection can reverse that direction while leaving the intended vertex mapping perfectly valid.
If a question asks about a specific transformation rather than similarity alone, further information may be needed to distinguish reflection, rotation and enlargement. The side ratios by themselves describe size relationships, not every detail of how one drawn position can be moved to the other. Answer the relationship actually requested.
In a triangle, each side lies opposite one vertex angle. If angle A corresponds to angle P, then the opposite sides BC and QR correspond. This offers another route to side matching, especially when the diagram labels make endpoints awkward to follow.
Suppose angle B corresponds to angle R and angle C corresponds to angle Q. Then the side opposite B, namely AC, matches the side opposite R, namely PQ. The full mapping is ABC similar to PRQ. The side opposite each matched angle retains its role despite the altered letter order.
Use either endpoint mapping or opposite-angle mapping as a check on the other. They should agree. If they produce different partners, inspect the original angle match or the vertex order you wrote. This small consistency check can locate a correspondence error before it spreads into several missing-length calculations.
In a larger triangle ABC, let D lie on AB and E lie on AC. The angle at A in triangle ADE equals the angle at A in triangle ABC because AD lies along AB and AE lies along AC. The two triangles use the same pair of rays at that vertex.
It is not enough that both triangles contain a letter A somewhere. Their actual sides forming the angle must line up appropriately. In more complicated intersecting diagrams, two angles at the same point may involve different rays and have different measures. Read the angle’s arms as well as its vertex.
Naming a three-letter angle can help: angle DAE equals angle BAC under the stated point locations. The middle letter identifies the vertex, while the outer letters identify its rays. Clear angle names make a proof inspectable and prevent the phrase “common angle” from being used where the geometry does not support it.
Continue the construction with D on AB, E on AC and DE parallel to BC. Corresponding angles give angle ADE equal to angle ABC, and the shared angle at A agrees. Thus triangle ADE is similar to triangle ABC by angle-angle similarity, with A matching A, D matching B and E matching C.
The justified side ratios are AD/AB = AE/AC = DE/BC. Note that AD corresponds to the full side AB, not to the remaining segment DB. DB may appear nearby and carry a numerical label, but it plays a different role in the larger triangle.
This is a common place where the geometry must guide the arithmetic. If AB is not given directly but AD and DB are, first obtain AB = AD + DB. Then use the full corresponding side. The nearest printed length is not necessarily the length required by the similarity relationship.
In the parallel-line construction, let AD = 4 centimetres, DB = 6 centimetres, AE = 5 centimetres and BC = 15 centimetres. Find AC, EC and DE. Since DE is parallel to BC, the triangles ADE and ABC are similar in the order already established.
The full side AB is 4 + 6 = 10 centimetres. The small-to-large scale ratio is AD/AB = 4/10 = 2/5. Therefore AE/AC = 2/5, giving AC = 5 × 5/2 = 12.5 centimetres. The remaining segment EC is 12.5 − 5 = 7.5 centimetres.
For the parallel side, DE/BC = 2/5, so DE = 15 × 2/5 = 6 centimetres. Check every reported quantity’s role: AC is a full large-triangle side, EC is a leftover segment, and DE is a small-triangle side. Their values cannot be substituted for one another merely because they lie along related lines.
Question A: ABC is similar to XYZ in that order. Which sides correspond to AB, BC and AC? Question B: the mapping is A to Z, B to X and C to Y. Write a correctly ordered similarity statement and identify the side matching AC.
Question C: ABC is similar to PQR, AB = 8, PQ = 12 and BC = 14 centimetres. Find QR. Question D: in the same correspondence, AC = 10 centimetres. Find PR. Question E: explain why AB/PQ = QR/BC is generally an inconsistent way to write the across-triangle ratio.
Question F: D lies on AB, E lies on AC and DE is parallel to BC. AD = 3, DB = 9 and AC = 20 centimetres. Find AE and EC. Before calculating, write the two triangle names in corresponding order and identify the full large-triangle side matching AD.
For A, AB matches XY, BC matches YZ and AC matches XZ. For B, write ABC similar to ZXY. Side AC matches ZY because A maps to Z and C maps to Y. The side’s name can be reversed to YZ without changing the segment it denotes.
For C, the scale factor from ABC to PQR is 12/8 = 3/2. Thus QR = 14 × 3/2 = 21 centimetres. For D, PR = 10 × 3/2 = 15 centimetres. For E, the first fraction compares the first triangle with the second, while the second reverses that direction.
For F, ADE is similar to ABC. AB is 3 + 9 = 12, so the small-to-large ratio is 3/12 = 1/4. Therefore AE = 20/4 = 5 centimetres and EC = 20 − 5 = 15 centimetres. Using AD/DB would compare a triangle side with a leftover segment and give the wrong scale.
Suppose two triangles each contain a side of length six centimetres. That agreement alone does not establish the same shape. The other sides and angles may vary widely. Even a claimed scale factor based on one chosen pair does not show that the remaining pairs obey it.
For example, triangles with side lengths four, five, six and six, eight, nine are not similar. If the shortest sides are paired, the ratio is 6/4 = 3/2. The middle ratio is 8/5, which differs. Both triples form valid triangles, but no common sorted-side scale factor exists.
Use enough information for a valid criterion. A neat ratio from one pair can be a necessary part of a similarity calculation after similarity is known, yet insufficient evidence before it is known. The logical order matters: establish the relationship, then use its consequences to calculate unknowns.
If two side pairs have the same ratio and the angle between those sides agrees, triangle similarity can be established through the proportional-side included-angle criterion. The angle’s location matters. An equal angle elsewhere in each triangle does not automatically provide the same sufficient information.
For instance, sides four and six around a fifty-degree angle match sides ten and fifteen around a fifty-degree angle. The side ratios are both 5/2, and the included angles agree, so the triangles are similar under this criterion. The corresponding third side would share the same factor.
Do not reduce the criterion to “two sides and an angle” without the word included. Geometry conditions are precise because different arrangements of apparently similar information can permit different shapes. When writing a reason, identify which sides surround the stated angle so that the sufficient evidence remains visible.
Triangles with side lengths five, seven and eight, and ten, fourteen and sixteen, have all sorted side lengths in ratio two. Their corresponding sides are proportional, so the triangles are similar. The larger triangle has twice every linear measurement, while its angles match the smaller triangle’s corresponding angles.
Sorting sides by length can help identify a candidate correspondence when no angle labels are supplied. A positive scale factor preserves the order of lengths: shortest matches shortest and longest matches longest. Still check all three ratios. One convenient pair cannot stand in for the complete criterion.
When equal sides occur, there may be more than one equivalent vertex mapping because of symmetry. That does not invalidate the triangle’s similarity. Choose a mapping that respects all the given conditions and maintain it consistently. Avoid treating a legitimate symmetric alternative as a mistake merely because it differs from one printed order.
If every corresponding side in a triangle is multiplied by k, its perimeter is multiplied by k. For sides six, eight and ten, the perimeter is twenty-four. Scaling by three halves gives nine, twelve and fifteen, whose perimeter is thirty-six. The ratio 36/24 is again three halves.
This follows by adding the scaled sides: k times the first side plus k times the second plus k times the third equals k times their total. The property is not a separate triangle formula to memorise. It comes from the same common linear multiplier applied to every boundary segment.
If a problem supplies perimeters of known similar triangles, their ratio can therefore recover the linear scale factor. Use that factor for a corresponding side. Do not square the perimeter ratio when finding a length; perimeter is a sum of lengths and retains the linear scale relationship.
For similar triangles, the corresponding base and perpendicular height both scale by k. The area one half times base times height therefore scales by k². If the linear factor is three halves, the area factor is nine quarters. A triangle enlarged by fifty per cent in length has more than a fifty per cent increase in area.
For a concrete example, a right triangle with perpendicular sides six and eight has area twenty-four square centimetres. Its scaled partner with perpendicular sides nine and twelve has area fifty-four. The ratio 54/24 is 9/4, matching the square of 3/2.
Correspondence still matters: the base in one triangle must pair with its corresponding base, and the corresponding height must be perpendicular in the matching way. You cannot take an arbitrary sloping side as a height. The series’ base-and-height lesson supports this condition when a rotated diagram makes the perpendicular direction harder to see.
Suppose two triangles are known to be similar and their areas are in the ratio nine to twenty-five. The linear ratio in the same direction is three to five, because the positive square root of 9/25 is 3/5. Lengths are positive here, so a negative scale ratio is not used for these side magnitudes.
If a side of the smaller triangle is twelve centimetres, its corresponding larger side is 12 × 5/3 = 20 centimetres. Multiplying by twenty-five ninths would incorrectly apply the area factor directly to a length. The dimension of the quantity tells you which scale relationship is appropriate.
Area information alone does not establish that arbitrary triangles are similar. The similarity must be given or proved before taking the area ratio’s square root as a common side ratio. Equal areas can belong to different shapes. Keep the shape condition visible when using a powerful scaling consequence.
In overlapping triangle questions, the same drawn segment may belong to both triangles. Its numerical length is of course the same segment length in each use. However, it need not correspond to itself under the similarity mapping. Correspondence is established through vertex roles, not merely through physical overlap on the page.
This can happen when a smaller triangle sits across part of a larger configuration or when a rotated copy shares one side. Write the vertex map before assigning a common ratio. An equation that treats the shared segment as matching itself would force scale factor one, which may not be justified by the actual mapping.
If the geometry does establish that a positive side corresponds to itself, then its across-triangle ratio is one, and the common scale factor is one. The triangles are then congruent as well as similar. The conclusion comes from a verified correspondence, not from seeing one common segment somewhere in the figure.
If the known scale factor is two and a corresponding side was originally seven centimetres, a proposed enlarged length of fourteen is consistent. A value of three and a half suggests the inverse factor. A value of nine suggests that two was added instead of used as a multiplier.
These checks identify likely errors, but they do not replace the correspondence proof. A wrong side paired with another wrong side can occasionally produce a plausible-looking factor. Return to the vertex mapping if an answer seems reasonable numerically but conflicts with the given angles or the ordered similarity statement.
Do not measure a not-to-scale sketch to settle the issue. Printed lengths and geometric conditions control the calculation. A diagram can help organise those conditions while intentionally distorting size for readability. Your solution should remain valid if the drawing is stretched on the screen or redrawn in a new orientation.
If the learner writes a correct scale factor but applies it to the wrong side, the arithmetic skill is working and the correspondence needs repair. Ask them to map the two endpoints of the side in question. This often resolves the error more directly than another exercise in solving proportions.
If the side matches are correct but the result shrinks an enlarged triangle, inspect direction. If they use a remaining segment instead of a full side, inspect the triangle boundary. If they square a length factor for a perimeter, inspect the dimension of the requested quantity. Each mistake belongs to a different relationship.
A useful correction note names that relationship: “AD matches AB, so I must add DB first,” or “both fractions must compare large to small.” Such notes create a specific next check. They also help a student recognise progress when one part of a multi-stage geometry question is already secure.
The broader similarity as a proof and scaling tool guide develops the proof structure. The congruence and similarity mastery guide connects equal shape with equal size. Read the route matching the condition you need to clarify.
Return to the Mathematics learning hub for scaling, diagram interpretation and geometric proof. These topics support the same habit: use the stated geometric evidence to justify a relationship before calculating. The current guide focuses on correspondence, so it can be revisited whenever the ratios seem familiar but the side pairing feels uncertain.
Families exploring support can read the Secondary 1 Mathematics Clementi tutorial page for the teaching approach. The learner’s current subject scope decides when these G3 similarity extensions are appropriate. A useful immediate checkpoint is whether they can name corresponding endpoints accurately on a rotated example.
Question G: similar triangles have corresponding side lengths twelve and eighteen centimetres. Their smaller perimeter is forty centimetres. Find the larger perimeter. Question H: their smaller area is sixty square centimetres. Find the larger area. Question I: explain why the factors in G and H differ.
Question J: similar triangles have areas sixteen and eighty-one square centimetres. A smaller side is eight centimetres. Find its larger partner. Question K: triangle ABC is similar to RQP. Identify the side corresponding to BC and state the vertex matching that justifies it.
For G, the linear factor is 18/12 = 3/2, giving perimeter sixty centimetres. For H, multiply sixty by 9/4 to get 135 square centimetres. For I, perimeter is linear while area uses two scaled dimensions. For J, the linear factor is 9/4, giving eighteen centimetres. For K, B maps to Q and C maps to P, so BC matches QP.
Try explaining a solution in three connected sentences. First say why the triangles are similar. Next state the vertex correspondence and one reliable corresponding side pair. Finally state the scale factor in a named direction and apply it to the requested partner. Each sentence supplies a reason for the next.
For the nested construction, that could be: “The parallel line gives an equal angle pair and the angle at A is shared. Thus ADE matches ABC with D corresponding to B and E to C. The small-to-large ratio is AD/AB, so I use it for AE/AC and DE/BC.”
The explanation does not need to repeat every theorem in the chapter. It should make the decisive connection visible. If you cannot say which large side matches AD, pause before calculating. The missing information is geometric correspondence, and a faster proportion calculation will not supply it.
After solving one example, rotate the drawing or relabel the second triangle in a different order. Keep the actual angle mapping unchanged and ask for the same corresponding side. This checks whether your method follows the geometry rather than a remembered arrangement of letters.
Next reverse the requested direction, finding a small length from a known large one. The mapping stays the same, while the scale factor becomes its reciprocal. Finally ask for perimeter and area from the same pair of shapes. That separates the stable correspondence from the quantity-dependent scaling rule.
A short set with these deliberate changes can teach more than many examples sharing one orientation. Record errors by their cause: wrong vertex mapping, reversed direction, partial side used, or wrong dimension factor. The aim is to make the next inspection step clear enough that you can perform it independently.
Suppose a problem gives two triangles, one equal angle pair and one side pair, then asks for a different corresponding length. Unless additional conditions establish similarity or otherwise determine the shapes, the requested length may not be uniquely fixed. A sketch that looks similar does not fill the gap.
State what extra information would make the calculation justified. A second corresponding equal angle could establish angle-angle similarity. Alternatively, suitable proportional side information with the correct included angle could support another criterion. Naming the missing condition is more useful than simply saying the question is impossible.
This is an honest mathematical response to incomplete data. Once similarity is established, ratios become powerful and efficient. Before that, they remain unverified assumptions. Learning to recognise the boundary between those situations protects you from applying an excellent method where its defining relationship has not actually been supplied.
If you are unsure whether enough evidence exists, list what is given without adding visual assumptions. A stated parallel pair can supply angle information, while a single printed side length cannot determine a whole triangle’s shape. Separating those facts on the page helps reveal the missing link. You can then say exactly which additional angle or side relationship would justify similarity, and which corresponding ratio would follow once that condition is available.
Use these answers when a ratio looks plausible but you are uncertain about its geometric meaning. Then return to the vertex names and map the endpoints of one side. That small action often resolves the problem before any new arithmetic is needed.
Can I match the longest sides?
For known similar triangles with positive side lengths, longest matches longest and shortest matches shortest. This can help check a mapping. If similarity is not yet known, compare all required side ratios or use another valid criterion. One matching pair alone does not establish the whole shape relationship.
Do reflected triangles count as similar?
Yes, if their corresponding angles and side ratios satisfy similarity. Reflection can reverse the apparent orientation on the page while preserving shape. Use the given angle matches or ordered statement to identify correspondence. A left-facing drawing does not need to match another left-facing drawing to be similar.
Does a scale factor of one mean congruence?
For similar triangles, a linear factor of one means every corresponding side has the same length, so the triangles are congruent. They can still appear in different positions or orientations. Equal size and shape do not require the two drawings to occupy the same place.
What is the most useful first check?
Establish or read the similarity, then write the vertex correspondence. Map both endpoints of each side before forming a ratio. Keep the comparison direction consistent, and use a linear or squared factor according to the quantity requested. These steps make the arithmetic answer a consequence of the geometry.