MathsCBSE Class 10 · Triangles

SSS Similarity Criterion for Class 10 Triangles

Updated 14 October 2026 · Fact-checked

The SSS similarity criterion says that if the three sides of one triangle are proportional to the three sides of another, the triangles are similar and their corresponding angles are equal. The board asks you to check the three ratios, write the correct vertex correspondence, find unknown sides or perimeters, and name equal angles.

Understand SSS Similarity Criterion

Two triangles are similar when their corresponding angles are equal and their corresponding sides are in one ratio. Checking all six facts is slow. The SSS similarity criterion lets you check only the three sides: if $\frac{A B}{D E} = \frac{B C}{E F} = \frac{C A}{F D}$ , then $△ A B C \sim △ D E F$ , and so $∠ A = ∠ D$ , $∠ B = ∠ E$ , $∠ C = ∠ F$ . In your syllabus this is stated without proof.

Why does it work? Three side lengths fix a triangle completely: you cannot build two different-looking triangles from the same three sticks. If every side of one triangle is $k$ times the matching side of another, the bigger triangle is just an enlargement of the smaller one, like a photocopy zoomed by $k$ . Zooming changes size, never angles, so the angles stay equal.

This is special to triangles. A square and a rhombus that is not a square, both of side 5 cm, have sides in ratio 1 : 1 but are not similar, because a quadrilateral can flex. A triangle cannot flex, so for triangles proportional sides are enough.

The hard part is the correspondence. Sides are often listed in a jumbled order. Sort each triangle's sides from smallest to largest and divide smallest by smallest, middle by middle, largest by largest. If the three ratios are equal, the triangles are similar. Then match vertices: the angle opposite the smallest side matches the angle opposite the smallest side, and so on. Equivalently, a vertex is where two sides meet, so match vertices by the pair of sides that meet there.

SSS sits beside the other criteria: AA uses angles only, SAS uses one included angle and two sides, and SSS uses sides only. Use SSS when you know all six sides and no angles. Once similarity is proved, every corresponding length (perimeter, altitude, median) is in the same ratio $k$ .

Conditions: all three ratios must be equal. Two equal ratios with no angle information prove nothing. And if $k = 1$ , the triangles are congruent, which is SSS congruence as a special case.

SSS Similarity Criterion: formulas to remember

SSS similarity criterion: $\frac{A B}{D E} = \frac{B C}{E F} = \frac{C A}{F D} \Rightarrow △ A B C \sim △ D E F$; Stated without proof. All three ratios must be equal.
Equal angles from SSS: $△ A B C \sim △ D E F \Rightarrow ∠ A = ∠ D, ∠ B = ∠ E, ∠ C = ∠ F$; Read angles from the letter order of the statement.
Sorted-side test: $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}, a_{1} \leq b_{1} \leq c_{1}, a_{2} \leq b_{2} \leq c_{2}$; Match smallest with smallest, middle with middle, largest with largest.
Perimeter ratio: $\frac{perimeter of △ A B C}{perimeter of △ D E F} = \frac{A B}{D E} = k$; The scale factor $k$ applies to every pair of corresponding sides, so the perimeters are in the same ratio too.

How to solve SSS Similarity Criterion questions

This method works for any SSS question, whether it asks you to prove similarity, name an angle or find a missing side.

1List the three sides of each triangle and sort each list from smallest to largest.
2Divide smallest by smallest, middle by middle, largest by largest, and simplify each ratio.
3If all three ratios are equal, write: the sides are proportional, so the triangles are similar by SSS.
4Find the vertex opposite each side in both triangles, and match vertices opposite matching sides.
5Write the similarity statement in that order, for example $△ A B C \sim △ F D E$ .
6Read off equal angles or use the common ratio $k$ to find unknown sides, perimeters, altitudes or medians, with units.

SSS Similarity Criterion 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.

SSS similarity criterion

Imagine your school is building a triangular garden bed. The model on the architect's plan has edges 5 cm, 12 cm and 13 cm. The gardener has fencing pieces of 2.5 m, 6 m and 6.5 m ready for the real bed, and the angle in the model between the 5 cm and 12 cm edges is a right angle.

Show that the bed built from the fencing pieces has the same shape as the model, state which angle of the real bed is a right angle, and find the total fencing used and the length of the shortest altitude of the real bed.

Try it first, then show the solutionSolution

Sorted model sides: 5, 12, 13 cm. Real sides: 250, 600, 650 cm.
Ratios: $\frac{250}{5} = 50$ , $\frac{600}{12} = 50$ , $\frac{650}{13} = 50$ .
All ratios equal, so the triangles are similar by SSS with $k = 50$ .
The right angle lies between the two shorter sides, so in the real bed it is between the 2.5 m and 6 m pieces.
Fencing: $2.5 + 6 + 6.5 = 15$ m.
The shortest altitude is the one to the longest side, because area $= \frac{1}{2} \times side \times altitude$ is fixed, so the longest side has the shortest altitude. Area of the real bed $= \frac{1}{2} \times 2.5 \times 6 = 7.5 m^{2}$ (the right angle is between these sides).
Altitude to the 6.5 m side $= \frac{2 \times 7.5}{6.5} = \frac{15}{6.5} = \frac{30}{13}$ m, about 2.31 m.

Answer: Similar by SSS (ratio 1 : 50); the right angle is between the 2.5 m and 6 m sides; fencing 15 m; shortest altitude $\frac{30}{13}$ m (about 2.31 m).

Memory hook: Plan to garden: every edge times 50, so the shape and the corner angles stay the same.

Tricks to remember SSS Similarity Criterion

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

Sort, Divide, Match
S-D-M: Sort the sides, Divide in order, Match vertices opposite matching sides.
Use it when: Any SSS question with jumbled sides.
Triangles cannot flex
Three sticks make only one triangle shape, so proportional sticks make the same shape.
Use it when: Remembering why SSS works for triangles but not for quadrilaterals.
Opposite partners
Smallest side faces the smallest angle; partners across the triangle stay partners in the copy.
Use it when: Writing the vertex order correctly.

SSS Similarity Criterion: solved examples

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

Example 1: Choosing the correct similarity statement · 1 mark · MCQBoard level

In $△ P QR$ , $P Q = 4$ cm, $QR = 6$ cm, $R P = 5$ cm. In $△ X Y Z$ , $X Y = 10$ cm, $Y Z = 8$ cm, $Z X = 12$ cm. Which statement is correct?
(a) $△ P QR \sim △ X Y Z$
(b) $△ P QR \sim △ Y Z X$
(c) $△ P QR \sim △ Z Y X$
(d) $△ P QR \sim △ Y X Z$

Try it first, then show the solutionSolution

Sorted: PQ = 4, RP = 5, QR = 6; YZ = 8, XY = 10, ZX = 12. Each ratio is $\frac{1}{2}$ .
PQ ↔ YZ, RP ↔ XY, QR ↔ ZX.
P is common to PQ and RP, so it matches the vertex common to YZ and XY, which is Y.
Q is common to PQ and QR, matching the vertex common to YZ and ZX, which is Z. So R ↔ X.

Answer: (b) $△ P QR \sim △ Y Z X$

Example 2: Proving similarity and naming angles · 3 marks · Short answerBoard level

In $△ A B C$ , $A B = 9$ cm, $B C = 12$ cm, $C A = 15$ cm. In $△ L M N$ , $L M = 10$ cm, $M N = 6$ cm, $N L = 8$ cm. Show that the triangles are similar, write the correct statement, and if $∠ N = 9 0^{\circ}$ , find which angle of $△ A B C$ is $9 0^{\circ}$ .

Try it first, then show the solutionSolution

Sorted: 9, 12, 15 and 6, 8, 10. Ratios: $\frac{9}{6} = \frac{12}{8} = \frac{15}{10} = \frac{3}{2}$ . (1 mark)
So the triangles are similar by SSS, with AB ↔ MN, BC ↔ NL, CA ↔ LM. (½ mark)
B is between AB and BC, matching the vertex between MN and NL, which is N. A is between AB and CA, matching the vertex between MN and LM, which is M. So C ↔ L. (½ mark)
$△ A B C \sim △ M N L$ . (½ mark)
$∠ B = ∠ N = 9 0^{\circ}$ . (½ mark)

Answer: $△ A B C \sim △ M N L$ by SSS; $∠ B = 9 0^{\circ}$ .

Example 3: Finding an unknown so the triangles are similar · 3 marks · Short answerBoard level

The sides of $△ A B C$ are $A B = 2 x - 1$ , $B C = 2 x + 2$ , $C A = 3 x$ (in cm) and the sides of $△ P QR$ are $P Q = 3$ cm, $QR = 4$ cm, $R P = 5$ cm. Given $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{C A}{R P}$ , find $x$ , the sides of $△ A B C$ and the ratio of perimeters.

Try it first, then show the solutionSolution

From $\frac{2 x - 1}{3} = \frac{2 x + 2}{4}$ : $8 x - 4 = 6 x + 6$ , so $x = 5$ . (1 mark)
Sides: AB = 9 cm, BC = 12 cm, CA = 15 cm. (½ mark)
Check third ratio: $\frac{15}{5} = 3 = \frac{9}{3} = \frac{12}{4}$ . All equal, so similar by SSS. (1 mark)
Ratio of perimeters = 3 : 1, i.e. 36 : 12. (½ mark)

Answer: $x = 5$ ; sides 9 cm, 12 cm, 15 cm; perimeter ratio 3 : 1.

Example 4: SSS inside a figure, then a median · 5 marks · Long answerHarder board level

In $△ A B C$ , $∠ A = 9 0^{\circ}$ , $A B = 6$ cm, $A C = 8$ cm and $B C = 10$ cm. D, E, F are the midpoints of BC, CA and AB respectively, and $△ D E F$ is formed by joining them.
(i) Find the sides of $△ D E F$ (use the midpoint theorem).
(ii) Prove $△ D E F \sim △ A B C$ with the correct correspondence.
(iii) Find the perimeter of $△ D E F$ .
(iv) The median AD of $△ A B C$ is 5 cm. Find the corresponding median of $△ D E F$ , naming the vertex it is drawn from and the side it meets.
(v) Name the angle of $△ D E F$ equal to $∠ B$ , and find $∠ E D F$ .

Try it first, then show the solutionSolution

Midpoint theorem: $E F = \frac{1}{2} B C = 5$ cm, $F D = \frac{1}{2} C A = 4$ cm, $D E = \frac{1}{2} A B = 3$ cm. (1 mark)
$\frac{E F}{B C} = \frac{F D}{C A} = \frac{D E}{A B} = \frac{1}{2}$ . (1 mark)
EF ↔ BC, FD ↔ CA, DE ↔ AB. D is opposite EF and A is opposite BC, so D ↔ A; E is opposite FD and B is opposite CA, so E ↔ B; F ↔ C. So $△ D E F \sim △ A B C$ by SSS. (1 mark)
Perimeter of $△ D E F = 3 + 4 + 5 = 12$ cm (half of 24 cm). (½ mark)
Since D ↔ A, $∠ E D F = ∠ A = 9 0^{\circ}$ , so EF is the hypotenuse of right $△ D E F$ . The median from D goes to the midpoint of the hypotenuse EF, and the midpoint of the hypotenuse is equidistant from all three vertices, so this median is $\frac{1}{2} \times E F = 2.5$ cm, which is half of AD = 5 cm, as the scale factor predicts. (1 mark)
Since E ↔ B, $∠ E = ∠ B$ ; and $∠ E D F = ∠ A = 9 0^{\circ}$ as found above. (½ mark)

Answer: (i) EF = 5 cm, FD = 4 cm, DE = 3 cm; (ii) $△ D E F \sim △ A B C$ by SSS (D ↔ A, E ↔ B, F ↔ C, ratio 1 : 2); (iii) 12 cm; (iv) 2.5 cm, from D to the midpoint of EF; (v) $∠ E$ , and $∠ E D F = 9 0^{\circ}$ .

Example 5: Case study: triangular kite frames · 4 marks · Case-basedHOTS / competency

Suppose a craft club builds triangular kite frames. Frame 1 has sticks 24 cm, 32 cm and 40 cm. Frame 2 has sticks 36 cm, 48 cm and 60 cm. Frame 3 has sticks 30 cm, 40 cm and 45 cm.
(i) Which two frames have the same shape? (1 mark)
(ii) What is the ratio of perimeters of those two frames? (1 mark)
(iii) In Frame 1 the angle between the 24 cm and 32 cm sticks is $9 0^{\circ}$ . A student says Frame 3 also has a right angle because its two shorter sticks are in ratio 3 : 4 like Frame 1. Is she correct? Justify. (2 marks)

Try it first, then show the solutionSolution

Frame 1 to Frame 2: $\frac{36}{24} = \frac{48}{32} = \frac{60}{40} = \frac{3}{2}$ , so similar by SSS. (1 mark)
Perimeters 96 cm and 144 cm, ratio 2 : 3. (1 mark)
Frame 1 to Frame 3: $\frac{30}{24} = \frac{5}{4}$ , $\frac{40}{32} = \frac{5}{4}$ , $\frac{45}{40} = \frac{9}{8}$ . Not all equal, so not similar. (1 mark)
Check by Pythagoras (known result): $3 0^{2} + 4 0^{2} = 2500 \neq = 4 5^{2} = 2025$ , so no right angle there. Two proportional sides alone do not give similarity. She is wrong. (1 mark)

Answer: (i) Frames 1 and 2; (ii) 2 : 3; (iii) No, Frame 3 is not similar to Frame 1 (third ratio $\frac{9}{8}$ ), and $3 0^{2} + 4 0^{2} \neq = 4 5^{2}$ .

Where students lose marks in SSS Similarity Criterion

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

Mistake: Matching sides in the order they are listed
Sides are often jumbled on purpose.
Wrong working
AB = 6, BC = 9, CA = 12 and DE = 16, EF = 8, FD = 12: $\frac{6}{16} \neq = \frac{9}{8}$ , so not similar.
Right working
Sort: 6, 9, 12 and 8, 12, 16. Ratios all $\frac{3}{4}$ , so similar by SSS.
Do this: Always sort both lists before dividing.
Usually costs: The whole question, since the conclusion is wrong.
Mistake: Wrong vertex order in the statement
Students write $△ A B C \sim △ D E F$ by habit.
Wrong working
Sides proportional, so $△ A B C \sim △ D E F$ and $∠ A = ∠ D$ .
Right working
Match vertices opposite matching sides, e.g. $△ A B C \sim △ E F D$ , so $∠ A = ∠ E$ .
Do this: Match each vertex by the pair of sides meeting there.
Usually costs: ½ to 1 mark for the statement and any angle read from it.
Mistake: Checking only two ratios
Two equal ratios look convincing.
Wrong working
$\frac{2 x - 1}{3} = \frac{2 x + 2}{4}$ , so similar.
Right working
After finding $x$ , verify the third ratio too before concluding SSS.
Do this: Write all three ratios in the final line.
Usually costs: ½ to 1 mark for incomplete justification.
Mistake: Treating SSS similarity as SSS congruence
Students mix up the two SSS rules and look for equal sides.
Wrong working
Sides 3, 4, 5 cm and 6, 8, 10 cm are not equal, so the triangles are not similar.
Right working
Similarity needs proportional sides, not equal ones: $\frac{6}{3} = \frac{8}{4} = \frac{10}{5} = 2$ , so the triangles are similar by SSS.
Do this: For similarity, divide matching sides and compare the ratios; equal sides are only the special case $k = 1$ (congruence).
Usually costs: 1 mark, or the whole proof if the wrong conclusion is reached.

Exam tips

Write the three ratios in one line, then the words 'by SSS similarity criterion'.
Always write the full similarity statement in the correct order before using an angle or side.
In MCQs on correspondence, check each option's vertex order against the sides that meet at each vertex.
If an unknown appears, find it from two ratios and verify with the third.
Use SSS only when no angles are given; if one angle is given with two sides, think SAS.

Practice questions on SSS Similarity Criterion

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 · MCQBoard level
In $△ A B C$ , $A B = 6$ cm, $B C = 8$ cm and $C A = 10$ cm. In $△ P QR$ , $P Q = 12$ cm, $QR = 15$ cm and $R P = 9$ cm. Which statement is correct?
(a) $△ A B C \sim △ P QR$
(b) $△ A B C \sim △ R P Q$
(c) $△ A B C \sim △ QR P$
(d) $△ A B C \sim △ P R Q$
Hint
Sort both sets of sides, pair smallest with smallest, then match each vertex by the two sides that meet there.
Answer
(b). Sorted: 6, 8, 10 and 9, 12, 15, ratio $\frac{3}{2}$ each. So AB ↔ RP, BC ↔ PQ, CA ↔ QR. A (AB, CA) ↔ R (RP, QR); B (AB, BC) ↔ P (RP, PQ); C ↔ Q. Hence $△ A B C \sim △ R P Q$ .
Q2 · 1 mark · Assertion-reasonBoard level
Assertion (A): A triangle with sides 3 cm, 4 cm and 5 cm is similar to a triangle with sides 6 cm, 8 cm and 11 cm.
Reason (R): For SSS similarity, all three pairs of corresponding sides must be in the same ratio.
(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 third ratio, not just the first two.
Answer
(d) A is false but R is true. $\frac{6}{3} = \frac{8}{4} = 2$ , but $\frac{11}{5} = 2.2$ , so the third pair is not in the same ratio and the triangles are not similar. Two proportional pairs are not enough; R states the correct condition.
Q3 · 3 marks · Short answerHarder board level
The sides of $△ D E F$ are in the ratio 3 : 5 : 7 and its perimeter is 60 cm. $△ A B C \sim △ D E F$ , and the longest side of $△ A B C$ is 24 cm more than its shortest side. (i) Find the sides of $△ D E F$ . (ii) Find the sides and the perimeter of $△ A B C$ . (iii) Find the scale factor from $△ D E F$ to $△ A B C$ and check that the perimeters are in the same ratio.
Hint
Similar triangles keep the side ratio 3 : 5 : 7, so write the sides of $△ A B C$ as $3 s$ , $5 s$ , $7 s$ .
Answer
(i) $3 t + 5 t + 7 t = 15 t = 60$ , so $t = 4$ : sides 12 cm, 20 cm, 28 cm. (1)
(ii) Sides of $△ A B C$ are $3 s$ , $5 s$ , $7 s$ with $7 s - 3 s = 4 s = 24$ , so $s = 6$ : sides 18 cm, 30 cm, 42 cm, perimeter 90 cm. (1)
(iii) $k = \frac{18}{12} = \frac{30}{20} = \frac{42}{28} = \frac{3}{2}$ , and $\frac{90}{60} = \frac{3}{2}$ as well. (1)
Q4 · 3 marks · Short answerHarder board level
In $△ A B C$ , $A B = x$ cm, $B C = 8$ cm, $C A = 8$ cm. In $△ D E F$ , $D E = 3$ cm, $E F = (x + 2)$ cm, $F D = 6$ cm. The triangles are similar with $\frac{A B}{D E} = \frac{B C}{E F} = \frac{C A}{F D}$ . (i) Find $x$ , giving a reason for rejecting any value. (ii) Verify all three ratios. (iii) Find the ratio of the perimeters.
Hint
Equate the first two ratios and cross-multiply; you will get a quadratic.
Answer
(i) $\frac{x}{3} = \frac{8}{x + 2}$ gives $x^{2} + 2 x - 24 = 0$ , so $(x + 6) (x - 4) = 0$ . A length cannot be negative, so $x = 4$ . (ii) $\frac{4}{3}$ , $\frac{8}{6} = \frac{4}{3}$ , $\frac{8}{6} = \frac{4}{3}$ : all equal, so similar by SSS. (iii) Perimeters 20 cm and 15 cm, ratio 4 : 3.
Q5 · 5 marks · Long answerHarder board level
In quadrilateral ABCD, $A B = 4$ cm, $B C = 6$ cm, $C A = 8$ cm, $C D = 12$ cm and $D A = 16$ cm. The diagonal AC divides it into $△ A B C$ and $△ A C D$ , with B and D on opposite sides of AC.
(i) Prove that $△ A B C \sim △ A C D$ .
(ii) Name the three pairs of equal angles.
(iii) Show that AC bisects $∠ B A D$ .
(iv) Find the perimeter of ABCD.
Hint
Sorted sides: 4, 6, 8 and 8, 12, 16. Match each vertex by the two sides that meet there.
Answer
(i) $\frac{A B}{A C} = \frac{4}{8}$ , $\frac{B C}{C D} = \frac{6}{12}$ , $\frac{C A}{D A} = \frac{8}{16}$ , all equal to $\frac{1}{2}$ . A (sides AB, CA) matches A (sides AC, DA); B (sides AB, BC) matches C (sides AC, CD); C (sides BC, CA) matches D (sides CD, DA). So $△ A B C \sim △ A C D$ by SSS. (2 marks)
(ii) $∠ B A C = ∠ C A D$ , $∠ A B C = ∠ A C D$ , $∠ B C A = ∠ C D A$ . (1½ marks)
(iii) From (ii), $∠ B A C = ∠ C A D$ . Since B and D lie on opposite sides of AC, $∠ B A D = ∠ B A C + ∠ C A D = 2∠ B A C$ , so AC divides $∠ B A D$ into two equal parts: AC bisects $∠ B A D$ . (1 mark)
(iv) $4 + 6 + 12 + 16 = 38$ cm. (½ mark)
Q6 · 4 marks · Long answerHarder board level
ABCD is a quadrilateral in which A and C lie on opposite sides of the diagonal BD. $A B = 9$ cm, $B D = 12$ cm, $D A = 6$ cm, $D C = 16$ cm and $C B = 8$ cm.
(i) Prove that $△ A B D \sim △ B D C$ , stating the correspondence.
(ii) Hence prove that $A B ∥ D C$ .
(iii) Find the ratio of the perimeters of $△ A B D$ and $△ B D C$ .
Hint
Sort the sides of each triangle: 6, 9, 12 and 8, 12, 16. Then look for a pair of alternate angles at the diagonal BD.
Answer
(i) $\frac{D A}{C B} = \frac{6}{8}$ , $\frac{A B}{B D} = \frac{9}{12}$ , $\frac{B D}{D C} = \frac{12}{16}$ , all equal to $\frac{3}{4}$ . A (sides AB, DA) ↔ B (sides BD, CB); B (sides AB, BD) ↔ D (sides BD, DC); D ↔ C. So $△ A B D \sim △ B D C$ by SSS. (2)
(ii) Since B ↔ D, $∠ A B D = ∠ B D C$ . These are alternate angles made by the transversal BD with AB and DC, so $A B ∥ D C$ (ABCD is a trapezium). (1)
(iii) Perimeters $6 + 9 + 12 = 27$ cm and $8 + 12 + 16 = 36$ cm, ratio $27 : 36 = 3 : 4$ , equal to the side ratio. (1)
Q7 · 4 marks · Short answerHOTS / competency
In $△ A B C$ , D is a point on BC with $B D = 8$ cm and $B C = 18$ cm. Also $A B = 12$ cm, $A D = 10$ cm and $A C = 15$ cm.
(i) Prove that $△ A B D \sim △ C B A$ .
(ii) Show that $A B^{2} = B D \times B C$ .
(iii) Prove that AD bisects $∠ B A C$ .
Hint
The two triangles share side AB. Compare AB with CB, BD with BA, DA with AC. For (iii), find DC and compare it with AD.
Answer
(i) $\frac{A B}{C B} = \frac{12}{18} = \frac{2}{3}$ , $\frac{B D}{B A} = \frac{8}{12} = \frac{2}{3}$ , $\frac{D A}{A C} = \frac{10}{15} = \frac{2}{3}$ . All equal, so $△ A B D \sim △ C B A$ by SSS (A ↔ C, B ↔ B, D ↔ A). (1½ marks)
(ii) From $\frac{A B}{C B} = \frac{B D}{B A}$ , $A B^{2} = B D \times B C$ . Check: $144 = 8 \times 18$ . (1 mark)
(iii) From the similarity, $∠ B A D = ∠ B C A$ . $D C = 18 - 8 = 10$ cm $= A D$ , so in $△ A D C$ , $∠ D A C = ∠ D C A$ (angles opposite equal sides). Hence $∠ B A D = ∠ D A C$ , so AD bisects $∠ B A C$ . (1½ marks)
Q8 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a stage designer cuts triangular wooden panels for a school play. Panel X has $X Y = 8$ cm, $Y Z = 12$ cm, $Z X = 10$ cm on the drawing. Panel P has $P Q = 15$ cm, $QR = 18$ cm, $R P = 12$ cm.
(i) Find the scale factor from Panel X to Panel P. (1 mark)
(ii) Which statement is correct: $△ X Y Z \sim △ P QR$ , $△ X Y Z \sim △ P R Q$ or $△ X Y Z \sim △ R QP$ ? (1 mark)
(iii) A third panel similar to Panel X has perimeter 75 cm. Find its longest side, and name the angle of Panel P equal to $∠ Y$ . (2 marks)
Hint
Sort both lists: 8, 10, 12 and 12, 15, 18. Then match each vertex by the two sides meeting there.
Answer
(i) $\frac{12}{8} = \frac{15}{10} = \frac{18}{12} = \frac{3}{2}$ , so $k = \frac{3}{2}$ . (ii) XY ↔ RP, ZX ↔ PQ, YZ ↔ QR. X (XY, ZX) ↔ P (RP, PQ); Y (XY, YZ) ↔ R (RP, QR); Z ↔ Q. So $△ X Y Z \sim △ P R Q$ . (iii) Perimeter of Panel X is 30 cm, so the scale factor is $\frac{75}{30} = \frac{5}{2}$ and the longest side is $\frac{5}{2} \times 12 = 30$ cm. Since Y ↔ R, $∠ R = ∠ Y$ .

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.State the SSS similarity criterion.Show answer
If corresponding sides of two triangles are proportional, the triangles are similar and corresponding angles are equal.
2.Is SSS similarity proved in your syllabus?Show answer
No, it is stated without proof.
3.Ratio of perimeters of similar triangles equals?Show answer
The ratio of corresponding sides.
4.Sides 2, 3, 4 and 6, 9, 12: similar?Show answer
Yes, every ratio is 1 : 3.
5.Which angle matches in a similar triangle?Show answer
The angles opposite corresponding (matching) sides are equal; match each vertex by the pair of sides that meet there.

SSS Similarity Criterion: frequently asked questions

What is the difference between SSS similarity and SSS congruence?

SSS congruence needs the sides to be equal. SSS similarity needs them only to be in the same ratio. Congruence is the special case where the ratio is 1.

Do I have to prove the SSS criterion in the exam?

No. In the 2026-27 syllabus it is stated without proof. Only the Basic Proportionality Theorem must be proved.

Why does SSS not work for quadrilaterals?

A quadrilateral can change its angles while keeping its sides, like a square becoming a rhombus. A triangle is rigid, so proportional sides force equal angles.