MathsCBSE Class 10 · Triangles

Similar Triangles: Definition, Examples and Counter Examples

Updated 14 October 2026 · Fact-checked

Two triangles are similar when their corresponding angles are equal and their corresponding sides are in the same ratio, so they have the same shape but may be different sizes. The board asks you to test pairs of figures, spot counter examples, write correct correspondence, and use the scale factor to find missing sides and perimeters.

Matching angles carry the same number of arcs, and matching sides (same colour) are in one ratio, here 1 : 2, so ABC ~ PQR.

Illustration drawn for QuestionBuddy.

Understand Similar Triangles: Definition and Examples

Two figures are similar if they have the same shape, though not necessarily the same size. A photo and its enlargement, or a map and the ground it shows, are everyday examples. One figure is just a scaled copy of the other.

For polygons with the same number of sides, similarity needs two conditions together: (i) corresponding angles are equal, and (ii) corresponding sides are in the same ratio. This common ratio is the scale factor $k$ . If $k > 1$ the copy is bigger, if $k < 1$ it is smaller, and if $k = 1$ the figures are congruent. So every pair of congruent figures is similar, but similar figures need not be congruent.

Why both conditions? Counter examples show it. A square and a rhombus of the same side have sides in ratio $1 : 1$ but different angles, so they are not similar. A $6 \times 4$ rectangle and a $6 \times 6$ square have all angles $9 0^{\circ}$ but sides not in one ratio, so they are not similar either. Any one condition alone can fail for quadrilaterals.

Triangles are special. For two triangles, if all angles match then the sides are automatically proportional, and if all sides are proportional the angles automatically match. That is why the criteria later in the chapter (AA, SSS, SAS) let you check only a few facts. On this page you use the full definition: $△ A B C \sim △ P QR$ means $∠ A = ∠ P$ , $∠ B = ∠ Q$ , $∠ C = ∠ R$ and $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{C A}{R P}$ .

The order of letters is the correspondence. First matches first, second matches second. To find the side matching $B C$ , take the second and third letters of the other triangle: $QR$ . Because every side is multiplied by $k$ , the perimeters are also in ratio $k$ .

Some families are always similar: all circles, all squares, all equilateral triangles. One length fixes their shape. Families like rectangles, rhombuses and isosceles triangles are not always similar, because their shape can change.

Similar Triangles: Definition and Examples: formulas to remember

Definition of similar triangles: $△ A B C \sim △ P QR ⟺ ∠ A = ∠ P, ∠ B = ∠ Q, ∠ C = ∠ R, \frac{A B}{P Q} = \frac{B C}{QR} = \frac{C A}{R P}$; Letter order fixes the correspondence.
Scale factor: $k = \frac{A B}{P Q} = \frac{B C}{QR} = \frac{C A}{R P}$; k = 1 means congruent.
Ratio of perimeters: $\frac{perimeter of △ A B C}{perimeter of △ P QR} = k$; Every side is multiplied by k, so the sum of the sides is too. Do not use for areas here.
Angle sum: $∠ A + ∠ B + ∠ C = 18 0^{\circ}$; Finds the third angle when two are known.

How to solve Similar Triangles: Definition and Examples questions

Use this on any question asking whether figures are similar, or asking for a missing side from a similarity statement.

1Read the similarity statement and write the three angle pairs and three side ratios in letter order.
2If testing similarity, check angles: find any missing angle using the angle sum of $18 0^{\circ}$ .
3Check sides: sort each figure's sides by size and divide matching sides; all ratios must be equal.
4Both conditions must hold for polygons; for triangles one full condition is enough.
5Find the scale factor $k$ from a pair of sides that are both known.
6Multiply or divide by $k$ to get the missing sides, and use $k$ for perimeters.
7State units and check one ratio again with your answers.

Similar Triangles: Definition and Examples in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Similar triangles: scale factor

Suppose you are making a triangular kite. Your small paper model has sides 12 cm, 15 cm and 18 cm. The real kite must have the same shape, a border of 180 cm and one side of 48 cm.

Find the other two sides of the real kite and the cost of edging the border at Rs 2 per 10 cm. Your friend has already cut a frame with sides 50 cm, 60 cm and 70 cm. Can it be used for a kite of the same shape? Justify.

Try it first, then show the solutionSolution

Same shape, so the kite and the model are similar triangles.
Model perimeter $= 12 + 15 + 18 = 45$ cm.
Perimeters are in the ratio of sides: $k = \frac{180}{45} = 4$ .
$\frac{48}{4} = 12$ , so the 48 cm side matches the 12 cm side of the model.
Other sides: $15 \times 4 = 60$ cm and $18 \times 4 = 72$ cm; check $48 + 60 + 72 = 180$ cm.
Cost $= \frac{180}{10} \times 2 = 36$ , so Rs 36.
Friend's frame: $\frac{50}{12} \neq = \frac{60}{15} = 4$ , so its sides are not in one ratio with the model; not similar.

Answer: Sides 48 cm, 60 cm, 72 cm; edging cost Rs 36; the 50 cm, 60 cm, 70 cm frame is not the same shape, so it cannot be used.

Memory hook: Kite trick: the border ratio gives the scale factor, and the scale factor gives every side.

Counter examples of similarity

Imagine a shop has four kinds of floor tiles: A is 30 cm by 20 cm, B is 45 cm by 30 cm, C is 40 cm by 30 cm, and D is a square 30 cm by 30 cm. You want a border made of tiles that all have the same shape.

Which tiles are similar to each other? Give the scale factor for each similar pair. A designer wants a new tile similar to C with shorter side 45 cm. Find its longer side.

Try it first, then show the solutionSolution

All four are rectangles, so every angle is $9 0^{\circ}$ ; only the side ratios decide.
Long side to short side: A $\frac{30}{20} = \frac{3}{2}$ , B $\frac{45}{30} = \frac{3}{2}$ , C $\frac{40}{30} = \frac{4}{3}$ , D $\frac{30}{30} = 1$ .
Only A and B have equal ratios, so only A and B are similar.
Scale factor from A to B: $\frac{45}{30} = \frac{30}{20} = \frac{3}{2}$ .
C and D match neither A, B nor each other, even though all their angles are right angles.
New tile similar to C: $\frac{L}{45} = \frac{4}{3}$ , so $L = 60$ cm.

Answer: Only A and B are similar, with scale factor 3/2 from A to B; C and D are not similar to any other tile. The new tile is 60 cm by 45 cm.

Memory hook: Right angles alone do not make tiles similar: the long-to-short ratio must match too.

Tricks to remember Similar Triangles: Definition and Examples

Short memory aids. Say each one aloud once, then use it on a question the same day.

Same shape, any size
Similar = photocopy with zoom. Congruent = photocopy at 100%.
Use it when: Deciding similar versus congruent.
Two locks
Polygons need both keys: angles equal AND sides in one ratio.
Use it when: Spotting counter examples like square vs rhombus.
Letter position
Position 1-2 gives the first side, 2-3 the second, 3-1 the third.
Use it when: Writing correct ratios from a statement.
One-number shapes
If one number fixes the shape (circle radius, square side, equilateral side), the family is always similar.
Use it when: MCQs asking which figures are always similar.

Similar Triangles: Definition and Examples: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Always similar figures · 1 mark · MCQBoard level

Which pair is always similar?
(a) Two right triangles
(b) Two equilateral triangles
(c) Two rectangles
(d) Two isosceles triangles

Try it first, then show the solutionSolution

Equilateral triangles have all angles $6 0^{\circ}$ and sides in one ratio, so always similar.
Right, isosceles triangles and rectangles can change shape, so not always similar.

Answer: (b) Two equilateral triangles

Example 2: Showing two shapes are NOT similar · 2 marks · Very short answerBoard level

Are a rectangle 8 cm by 5 cm and a rectangle 12 cm by 9 cm similar? Justify.

Try it first, then show the solutionSolution

Every angle of each rectangle is $9 0^{\circ}$ , so the angle condition holds. (½)
Corresponding sides: longer $\frac{8}{12} = \frac{2}{3}$ , shorter $\frac{5}{9}$ . (1)
$\frac{2}{3} \neq = \frac{5}{9}$ , so the sides are not proportional and the rectangles are not similar, even though their angles are equal. (½)

Answer: Not similar: the angles match but the side ratios 2/3 and 5/9 differ.

Example 3: Deciding similarity and naming the correspondence · 2 marks · Very short answerBoard level

In $△ A B C$ , $A B = 4$ cm, $B C = 6$ cm and $C A = 8$ cm. In $△ P QR$ , $P Q = 12$ cm, $QR = 9$ cm and $R P = 6$ cm. Are the triangles similar? If yes, write the similarity statement with the correct correspondence.

Try it first, then show the solutionSolution

Sides in increasing order: $4, 6, 8$ and $6, 9, 12$ . (½)
$\frac{4}{6} = \frac{6}{9} = \frac{8}{12} = \frac{2}{3}$ , so the sides are proportional and the triangles are similar (SSS criterion). (½)
$A B = 4$ matches $R P = 6$ (common ratio $\frac{2}{3}$ ), $B C = 6$ matches $R Q = 9$ , $C A = 8$ matches $P Q = 12$ , so $A \leftrightarrow P$ , $B \leftrightarrow R$ , $C \leftrightarrow Q$ . (½)
Statement: $△ A B C \sim △ P R Q$ . (½)

Answer: Yes, they are similar: $△ A B C \sim △ P R Q$ , with scale factor $\frac{2}{3}$ .

Example 4: Using a similarity statement with perimeter · 3 marks · Short answerBoard level

$△ P QR \sim △ X Y Z$ . $P Q = 6$ cm, $QR = 8$ cm, $R P = 10$ cm and $X Y = 9$ cm. Find YZ and ZX. Then find the ratio of the perimeter of $△ P QR$ to that of $△ X Y Z$ , and the perimeter of $△ X Y Z$ .

Try it first, then show the solutionSolution

$\frac{P Q}{X Y} = \frac{QR}{Y Z} = \frac{R P}{Z X}$ by letter order. (½)
$k = \frac{X Y}{P Q} = \frac{9}{6} = \frac{3}{2}$ . (½)
$Y Z = 8 \times \frac{3}{2} = 12$ cm. (½)
$Z X = 10 \times \frac{3}{2} = 15$ cm. (½)
Perimeter ratio $P QR : X Y Z = 2 : 3$ . (½)
Perimeter of $△ X Y Z = 9 + 12 + 15 = 36$ cm. (½)

Answer: YZ = 12 cm, ZX = 15 cm, ratio 2 : 3, perimeter 36 cm.

Example 5: Angles and unknown from similarity · 3 marks · Short answerBoard level

$△ A B C \sim △ D E F$ with $∠ A = 5 0^{\circ}$ , $∠ E = 7 0^{\circ}$ . $A B = (x + 1)$ cm, $B C = 10$ cm, $D E = 9$ cm, $E F = 15$ cm. Find $∠ C$ , $∠ F$ , $x$ and AB.

Try it first, then show the solutionSolution

$∠ B = ∠ E = 7 0^{\circ}$ by correspondence. (½)
$∠ C = 18 0^{\circ} - 5 0^{\circ} - 7 0^{\circ} = 6 0^{\circ}$ . (½)
$∠ F = ∠ C = 6 0^{\circ}$ . (½)
$\frac{A B}{D E} = \frac{B C}{E F}$ , so $\frac{x + 1}{9} = \frac{10}{15}$ . (½)
$x + 1 = 6$ , so $x = 5$ . (½)
AB = 6 cm; check $\frac{6}{9} = \frac{10}{15} = \frac{2}{3}$ . (½)

Answer: $∠ C = ∠ F = 6 0^{\circ}$ , $x = 5$ , AB = 6 cm.

Example 6: Congruent as the special case k = 1 · 2 marks · Very short answerBoard level

$△ A B C \sim △ P QR$ with $A B = 5$ cm, $B C = 7$ cm and $C A = 8$ cm. The perimeter of $△ P QR$ is 20 cm. Find the sides of $△ P QR$ and show that the two triangles are congruent.

Try it first, then show the solutionSolution

Perimeter of $△ A B C = 5 + 7 + 8 = 20$ cm. (½)
Ratio of perimeters $= \frac{20}{20} = 1$ , so the scale factor $k = 1$ . (½)
$P Q = 5$ cm, $QR = 7$ cm, $R P = 8$ cm. (½)
All three pairs of corresponding sides are equal, so $△ A B C ≅ △ P QR$ (SSS congruence): similar triangles with $k = 1$ are congruent. (½)

Answer: PQ = 5 cm, QR = 7 cm, RP = 8 cm; k = 1, so the triangles are congruent.

Example 7: Algebraic unknown in the ratio · 3 marks · Short answerHarder board level

$△ A B C \sim △ D E F$ with $A B = (2 x - 1)$ cm, $B C = (x + 3)$ cm, $C A = (x + 5)$ cm, $D E = 10$ cm, $E F = 12$ cm and $F D = 16$ cm. Find $x$ , check that the third pair of sides is consistent, and find the perimeter of $△ A B C$ .

Try it first, then show the solutionSolution

$\frac{A B}{D E} = \frac{B C}{E F}$ by letter order, so $\frac{2 x - 1}{10} = \frac{x + 3}{12}$ . (½)
$12 (2 x - 1) = 10 (x + 3)$ , so $24 x - 12 = 10 x + 30$ . (½)
$14 x = 42$ , so $x = 3$ . (½)
$A B = 5$ cm, $B C = 6$ cm, $C A = 8$ cm. (½)
Check: $\frac{5}{10} = \frac{6}{12} = \frac{8}{16} = \frac{1}{2}$ , so the third side is consistent. (½)
Perimeter of $△ A B C = 5 + 6 + 8 = 19$ cm. (½)

Answer: $x = 3$ ; AB = 5 cm, BC = 6 cm, CA = 8 cm; perimeter 19 cm.

Example 8: Perimeters and a sum of sides · 3 marks · Short answerHarder board level

$△ A B C \sim △ P QR$ and their perimeters are 36 cm and 60 cm. If $A B + P Q = 16$ cm and $B C = 15$ cm, find all six sides.

Try it first, then show the solutionSolution

$\frac{A B}{P Q} = \frac{36}{60} = \frac{3}{5}$ , since perimeters are in the ratio of sides. (½)
Let $A B = 3 t$ and $P Q = 5 t$ ; then $8 t = 16$ , so $t = 2$ . (½)
$A B = 6$ cm and $P Q = 10$ cm. (½)
$QR = 15 \times \frac{5}{3} = 25$ cm. (½)
$C A = 36 - 6 - 15 = 15$ cm and $R P = 60 - 10 - 25 = 25$ cm. (½)
Check: $\frac{15}{25} = \frac{3}{5}$ . (½)

Answer: AB = 6 cm, BC = 15 cm, CA = 15 cm; PQ = 10 cm, QR = 25 cm, RP = 25 cm.

Example 9: Sides as expressions in both triangles · 3 marks · Short answerHarder board level

$△ A B C \sim △ D E F$ with $A B = (x + 1)$ cm, $B C = 2 x$ cm, $C A = 8$ cm, $D E = (x + 4)$ cm and $E F = 3 x$ cm. Find $x$ , all the sides, and the ratio of the perimeters.

Try it first, then show the solutionSolution

$\frac{A B}{D E} = \frac{B C}{E F}$ by letter order, so $\frac{x + 1}{x + 4} = \frac{2 x}{3 x}$ . (½)
Cross-multiplying: $3 x (x + 1) = 2 x (x + 4)$ , so $3 x^{2} + 3 x = 2 x^{2} + 8 x$ and $x^{2} - 5 x = 0$ . (1)
$x (x - 5) = 0$ ; $x = 0$ makes BC = 0, so reject it. Hence $x = 5$ . (½)
AB = 6, BC = 10, CA = 8; DE = 9, EF = 15; the ratio is $\frac{6}{9} = \frac{10}{15} = \frac{2}{3}$ , so $F D = 8 \times \frac{3}{2} = 12$ cm. (½)
Perimeters 24 cm and 36 cm, ratio $2 : 3$ . (½)

Answer: $x = 5$ ; AB = 6, BC = 10, CA = 8 cm and DE = 9, EF = 15, FD = 12 cm; perimeters in the ratio 2 : 3.

Example 10: Nested triangles with an unknown · 5 marks · Long answerHarder board level

In $△ A B C$ , D is a point on side AB and E is a point on side AC, and DE is joined (DE is not parallel to BC). $∠ A D E = ∠ A C B$ . $A B = 16$ cm, $A C = 12$ cm, $B C = 14$ cm, $A D = x$ cm and $A E = (x + 2)$ cm.
(i) Prove that $△ A D E \sim △ A C B$ and write the correspondence.
(ii) Find $x$ , DE, EC and DB.

Try it first, then show the solutionSolution

In $△ A D E$ and $△ A C B$ : $∠ D A E = ∠ C A B$ (common angle). (½)
$∠ A D E = ∠ A C B$ (given). (½)
So $△ A D E \sim △ A C B$ by AA, with $A \leftrightarrow A$ , $D \leftrightarrow C$ , $E \leftrightarrow B$ . (1)
Hence $\frac{A D}{A C} = \frac{A E}{A B} = \frac{D E}{C B}$ . (½)
$\frac{x}{12} = \frac{x + 2}{16}$ . (½)
$16 x = 12 x + 24$ , so $x = 6$ . (½)
$A D = 6$ cm, $A E = 8$ cm, and the ratio is $\frac{6}{12} = \frac{8}{16} = \frac{1}{2}$ . (½)
$D E = 14 \times \frac{1}{2} = 7$ cm. (½)
$E C = 12 - 8 = 4$ cm and $D B = 16 - 6 = 10$ cm. (½)

Answer: $△ A D E \sim △ A C B$ (AA); $x = 6$ , DE = 7 cm, EC = 4 cm, DB = 10 cm.

Example 11: Case study: model of a roof truss · 4 marks · Case-basedHOTS / competency

A carpenter builds a scale model of a triangular roof truss ABC with $A B = A C = 10$ cm and $B C = 12$ cm. The real truss PQR is similar to the model, has perimeter 16 m, and its side QR is 6 m long.
(i) Find the scale factor from model to real truss. (1)
(ii) Show that QR must correspond to BC, and find PQ and PR. (1)
(iii) Wood costs Rs 250 per metre. Find the cost of wood for the three sides of the real truss. A second design has sides 6 m, 6 m and 8 m. Is it similar to the model? Justify. (2)

Try it first, then show the solutionSolution

(i) Model perimeter $= 10 + 10 + 12 = 32$ cm; real perimeter $= 1600$ cm, so $k = \frac{1600}{32} = 50$ . (1)
(ii) $\frac{600}{50} = 12$ cm $= B C$ (a 10 cm side would give 5 m), so QR matches BC, with $Q \leftrightarrow B$ , $R \leftrightarrow C$ , $P \leftrightarrow A$ ; then $P Q = P R = 10 \times 50 = 500$ cm $= 5$ m. (1)
(iii) Wood $= 5 + 5 + 6 = 16$ m; cost $= 16 \times 250 = 4000$ , so Rs 4000. (1)
Second design: $\frac{600}{10} = 60$ but $\frac{800}{12} = \frac{200}{3} \neq = 60$ , so its sides are not proportional to the model's; not similar. (1)

Answer: (i) 50; (ii) QR corresponds to BC, PQ = PR = 5 m; (iii) Rs 4000; the 6 m, 6 m, 8 m design is not similar to the model.

Where students lose marks in Similar Triangles: Definition and Examples

The slips examiners see most often, and what to write instead.

Mistake: Wrong correspondence from the statement
Students match sides by size or position on paper, not by letter order.
Wrong working
$△ A B C \sim △ P R Q$ , so $\frac{A B}{P Q} = \frac{B C}{QR}$ .
Right working
Letter order gives $\frac{A B}{P R} = \frac{B C}{R Q} = \frac{C A}{QP}$ .
Do this: Underline letter pairs 1-2, 2-3, 3-1 in both names before writing ratios.
Usually costs: Usually the whole numerical answer.
Mistake: Calling figures similar from angles alone
Forgetting that polygons need both conditions.
Wrong working
All angles of both rectangles are $9 0^{\circ}$ , so they are similar.
Right working
Angles match, but $\frac{8}{12} \neq = \frac{5}{9}$ , so not similar.
Do this: For quadrilaterals always check both angles and side ratios.
Usually costs: 1 mark in MCQ or assertion-reason.
Mistake: Inverting the scale factor
Dividing the wrong way between small and big triangle.
Wrong working
$k = \frac{6}{9}$ , so $Y Z = 8 \times \frac{2}{3}$ .
Right working
Going from PQR to XYZ, $k = \frac{9}{6}$ , so $Y Z = 12$ cm.
Do this: Ask: should the new side be bigger or smaller? Check against your answer.
Usually costs: 1 to 2 marks.
Mistake: Saying similar figures must be congruent, or vice versa
Mixing the two definitions.
Wrong working
Similar triangles have equal sides.
Right working
Similar triangles have proportional sides; equal sides only when k = 1.
Do this: Remember: congruent is the special case k = 1.
Usually costs: 1 mark in reasoning questions.

Exam tips

Write the full similarity statement in correct order before computing any ratio.
In always-similar MCQs, ask whether one measurement fixes the shape.
Give a specific counter example with numbers when asked why figures are not similar.
Use the ratio of perimeters freely, but do not use area ratios; that result is not in the syllabus.
Show the scale factor as a separate line; examiners award a step mark for it.

Practice questions on Similar Triangles: Definition and Examples

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · Assertion-reasonBoard level
Assertion (A): A square of side 4 cm and a rhombus of side 4 cm with an angle of $7 0^{\circ}$ are similar.
Reason (R): For two polygons to be similar, corresponding sides must be proportional and corresponding angles equal.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Check the angles of both figures.
Answer
(d). Sides are in ratio 1 : 1 but angles $9 0^{\circ}$ and $7 0^{\circ}$ differ, so A is false; R is the correct definition.
Q2 · 2 marks · Very short answerBoard level
$△ L M N \sim △ R S T$ . The perimeters are 30 cm and 45 cm. If $S T = 18$ cm and $L N = 8$ cm, find MN and RT.
Hint
The ratio of perimeters equals the scale factor; MN matches ST, LN matches RT.
Answer
$k = \frac{30}{45} = \frac{2}{3}$ . $M N = 18 \times \frac{2}{3} = 12$ cm; $R T = 8 \times \frac{3}{2} = 12$ cm.
Q3 · 3 marks · Short answerHarder board level
$△ A B C \sim △ P QR$ with $A B = 2 x$ cm, $B C = 9$ cm, $P Q = 8$ cm, $QR = (x + 5)$ cm and $x > 0$ . Find $x$ and all four lengths. What can you say about the two triangles?
Hint
Set $\frac{A B}{P Q} = \frac{B C}{QR}$ and solve the quadratic.
Answer
$\frac{2 x}{8} = \frac{9}{x + 5}$ gives $2 x (x + 5) = 72$ , so $x^{2} + 5 x - 36 = 0$ , i.e. $(x + 9) (x - 4) = 0$ . Since $x > 0$ , $x = 4$ . AB = 8 cm, BC = 9 cm, PQ = 8 cm, QR = 9 cm. The ratio is $\frac{8}{8} = \frac{9}{9} = 1$ , so $k = 1$ and the triangles are congruent.
Q4 · 3 marks · Short answerBoard level
$△ D E F \sim △ K L M$ with $∠ D = 4 5^{\circ}$ and $∠ M = 7 5^{\circ}$ . $D E = 6$ cm, $E F = 9$ cm, $K L = 10$ cm. Find $∠ E$ , $∠ L$ , $L M$ and the ratio of the perimeters of $△ D E F$ and $△ K L M$ .
Hint
$∠ F = ∠ M$ ; use the angle sum, then $k = \frac{K L}{D E}$ .
Answer
$∠ F = 7 5^{\circ}$ , so $∠ E = ∠ L = 6 0^{\circ}$ . $k = \frac{10}{6} = \frac{5}{3}$ , $L M = 9 \times \frac{5}{3} = 15$ cm. Perimeter ratio $3 : 5$ .
Q5 · 3 marks · Short answerHOTS / competency
Aman says: "Two isosceles triangles each having a $4 0^{\circ}$ angle are always similar." Test his claim by taking one triangle with vertex angle $4 0^{\circ}$ and another with base angles $4 0^{\circ}$ . Then state a correct version of the claim.
Hint
Find all three angles of each triangle.
Answer
First: $4 0^{\circ}, 7 0^{\circ}, 7 0^{\circ}$ . Second: $4 0^{\circ}, 4 0^{\circ}, 10 0^{\circ}$ . Angles differ, so not similar; claim false. Correct: two isosceles triangles with equal vertex angles are similar (all angles then match).
Q6 · 3 marks · Short answerHarder board level
$△ A B C \sim △ P QR$ . Their perimeters are 24 cm and 32 cm. $A B = (x + 1)$ cm, $P Q = (2 x - 2)$ cm and $B C = 9$ cm. Find $x$ and all six sides.
Hint
The ratio of perimeters equals $\frac{A B}{P Q}$ .
Answer
$\frac{x + 1}{2 x - 2} = \frac{24}{32} = \frac{3}{4}$ , so $4 x + 4 = 6 x - 6$ and $x = 5$ . AB = 6 cm, PQ = 8 cm. $QR = 9 \times \frac{4}{3} = 12$ cm. $C A = 24 - 6 - 9 = 9$ cm and $R P = 32 - 8 - 12 = 12$ cm. Check: $\frac{9}{12} = \frac{3}{4}$ .
Q7 · 3 marks · Short answerHarder board level
A rectangle measures 6 cm by 4 cm. Another rectangle measures $(k + 3)$ cm by $(k - 1)$ cm, where $k > 1$ . (i) Find $k$ if the rectangles are similar, and give the scale factor. (ii) Show that for $k = 5$ they are not similar, even though all their angles are equal. (iii) Can the second rectangle ever be a square? What does that tell you about a square and the first rectangle?
Hint
Since $k + 3 > k - 1$ , the longer sides must correspond.
Answer
(i) $\frac{k + 3}{k - 1} = \frac{6}{4}$ gives $4 k + 12 = 6 k - 6$ , so $k = 9$ : the rectangle is 12 cm by 8 cm, scale factor 2. (ii) $k = 5$ gives 8 cm by 4 cm, ratio $\frac{8}{4} = 2 \neq = \frac{6}{4} = 1.5$ , so not similar. (iii) A square needs $k + 3 = k - 1$ , which is impossible, so it is never a square. In any case a square (ratio 1) could not be similar to the 6 by 4 rectangle (ratio 1.5).
Q8 · 5 marks · Long answerHarder board level
In $△ P QR$ , S is a point on side QR and PS is joined. $∠ QP S = ∠ QR P$ . $QS = 4$ cm, $S R = 5$ cm and $P R = 12$ cm.
(i) Prove that $△ QP S \sim △ QR P$ .
(ii) Show that $P Q^{2} = QS \times QR$ .
(iii) Find PQ and PS.
Hint
Angle Q is common to both triangles; write the correspondence Q-Q, P-R, S-P.
Answer
(i) $∠ P QS = ∠ R QP$ (common) and $∠ QP S = ∠ QR P$ (given), so $△ QP S \sim △ QR P$ by AA. (ii) Hence $\frac{QP}{QR} = \frac{QS}{QP} = \frac{P S}{R P}$ , so $P Q^{2} = QS \times QR$ . (iii) $QR = 4 + 5 = 9$ cm, $P Q^{2} = 4 \times 9 = 36$ , so $P Q = 6$ cm. $\frac{P S}{R P} = \frac{QP}{QR} = \frac{6}{9} = \frac{2}{3}$ , so $P S = 12 \times \frac{2}{3} = 8$ cm.
Q9 · 2 marks · Very short answerHOTS / competency
In $△ A B C$ , $A B = 4$ cm, $B C = 6$ cm and $C A = 5$ cm. In $△ D E F$ , $D E = 8$ cm, $E F = 10$ cm and $F D = 12$ cm. Riya writes: 'AB/DE is 1/2 but BC/EF is 3/5, so the triangles are not similar.' Find her error and write the correct similarity statement.
Hint
Pair the sides by size, not by the letters you were given.
Answer
She paired sides by letter position without a similarity statement. Sorted sides $4, 5, 6$ and $8, 10, 12$ give ratios all $\frac{1}{2}$ , so the triangles are similar (SSS). AB matches DE, CA matches EF and BC matches FD, so $A \leftrightarrow E$ , $B \leftrightarrow D$ , $C \leftrightarrow F$ and $△ A B C \sim △ E D F$ .
Q10 · 4 marks · Case-basedHOTS / competency
Suppose at 4 pm a girl 1.5 m tall stands near a vertical pole on level ground. Her shadow is 2 m long and the pole's shadow is 12 m long.
(i) Why are the triangle formed by the girl and her shadow and the triangle formed by the pole and its shadow similar? (1)
(ii) Find the height of the pole. (1)
(iii) An hour later the girl's shadow is 3 m long. Find the length of the pole's shadow then, and how much it has grown. (2)
Hint
At one moment the sun's rays make the same angle with the ground everywhere.
Answer
(i) Both triangles have a right angle at the ground and the same angle of elevation of the sun at that moment, so they are similar by AA. (ii) $\frac{h}{1.5} = \frac{12}{2}$ , so $h = 9$ m. (iii) Shadow to height is now $\frac{3}{1.5} = 2$ , so the pole's shadow is $9 \times 2 = 18$ m; it has grown by $18 - 12 = 6$ m.
Q11 · 3 marks · Short answerBoard level
A triangle has sides 6 cm, 8 cm and 10 cm. A similar triangle has perimeter 36 cm and its longest side is $(2 y + 1)$ cm. Find $y$ and the other two sides of the second triangle.
Hint
Use the ratio of perimeters to get the scale factor first.
Answer
First perimeter $= 24$ cm, so $k = \frac{36}{24} = \frac{3}{2}$ . Longest side $= 10 \times \frac{3}{2} = 15$ cm, so $2 y + 1 = 15$ and $y = 7$ . Other sides $6 \times \frac{3}{2} = 9$ cm and $8 \times \frac{3}{2} = 12$ cm; check $9 + 12 + 15 = 36$ cm.
Q12 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a town map shows three triangular plots, with the scale 1 cm = 20 m. Plot T1 has sides 3 cm, 4 cm and 5 cm; plot T2 has sides 4.5 cm, 6 cm and 7.5 cm; plot T3 has sides 6 cm, 8 cm and 9 cm.
(i) Which plots are similar to T1? Justify. (1 mark)
(ii) Find the real perimeter of plot T2. (1 mark)
(iii) A new plot T4 is to be similar to T1, with real longest side 200 m. Find its sides on the map and its real perimeter, and state the scale factor from T1 to T4. (2 marks)
Hint
Match shortest with shortest and longest with longest before comparing ratios.
Answer
(i) T2: $\frac{4.5}{3} = \frac{6}{4} = \frac{7.5}{5} = 1.5$ , so T2 ~ T1 (SSS). T3: $\frac{6}{3} = \frac{8}{4} = 2$ but $\frac{9}{5} = 1.8$ , so T3 is not similar. (1)
(ii) Map perimeter of T2 $= 18$ cm, so real perimeter $= 18 \times 20 = 360$ m. (1)
(iii) 200 m is 10 cm on the map, matching T1's longest side 5 cm, so the scale factor is 2. T4 on the map: 6 cm, 8 cm, 10 cm; real perimeter $= 24 \times 20 = 480$ m. (2)

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Two conditions for polygons to be similar?Show answer
Corresponding angles equal and corresponding sides in the same ratio.
2.Are all congruent figures similar?Show answer
Yes, with scale factor 1.
3.Name three families that are always similar.Show answer
Circles, squares, equilateral triangles.
4.If $△ A B C \sim △ P QR$ , which side matches CA?Show answer
RP.
5.Ratio of perimeters of similar triangles?Show answer
Equal to the ratio of corresponding sides.

Similar Triangles: Definition and Examples: frequently asked questions

What is the difference between similar and congruent triangles?

Similar triangles have the same shape with sides in a fixed ratio. Congruent triangles have the same shape and the same size, so the ratio is 1. Every congruent pair is similar, but not every similar pair is congruent.

Why is a square not similar to a rhombus of the same side?

Their sides are in the same ratio, but the angles of a rhombus are not all right angles. Similarity needs both equal angles and proportional sides, so they are not similar.

Do I need to prove the similarity criteria?

No. Only the Basic Proportionality Theorem is proved in this chapter. The converse and the AA, SSS and SAS criteria are stated and used without proof.

Is the theorem on areas of similar triangles in the syllabus?

No, the areas of similar triangles theorem is not in the 2026-27 syllabus.