This MCQ module is based on: 6.4 Criteria for Similarity of Triangles
TOPIC 15 OF 35
6.4 Criteria for Similarity of Triangles
🎓 Class 10
Mathematics
CBSE
Theory
Ch 6 — Triangles
⏱ ~15 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: 6.4 Criteria for Similarity of Triangles
Targeting Class 10 level in Geometry, with Intermediate difficulty.
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6.4 Criteria for Similarity of Triangles
To conclude two triangles are similar, we need not verify all three angles and all three side ratios. Three powerful criteria — each derivable from BPT — let us test similarity from a smaller amount of data.
Theorem 6.3 — AAA (Angle-Angle-Angle) Similarity
If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (proportional), and the triangles are similar.Given \(\triangle ABC\) and \(\triangle DEF\) with \(\angle A=\angle D,\ \angle B=\angle E,\ \angle C=\angle F\), prove \(\tfrac{AB}{DE}=\tfrac{BC}{EF}=\tfrac{CA}{FD}\).
Fig 6.4: Proof of AAA — cut off DP = AB on DE and DQ = AC on DF, then show PQ ∥ EF.
Proof of AAA
Cut off DP = AB on DE, and DQ = AC on DF. Since \(\angle A = \angle D\) (given) and DP = AB, DQ = AC, by SAS congruence, \(\triangle ABC \cong \triangle DPQ\). Hence \(\angle DPQ = \angle B\).But \(\angle DPQ\) and \(\angle DEF\) are corresponding angles formed when line PQ and EF are cut by the transversal DE, so \(PQ \parallel EF\).
By BPT in \(\triangle DEF\): \(\tfrac{DP}{PE} = \tfrac{DQ}{QF}\), which rearranges to \(\tfrac{DP}{DE} = \tfrac{DQ}{DF}\), i.e., \(\tfrac{AB}{DE} = \tfrac{AC}{DF}\).
Similarly, \(\tfrac{AB}{DE} = \tfrac{BC}{EF}\). Hence all three ratios are equal. ∎
AA Similarity (Corollary)
If two angles of one triangle are respectively equal to two angles of another triangle, the third pair is automatically equal (angle sum = 180°). So AA is enough to conclude similarity.
Theorem 6.4 — SSS (Side-Side-Side) Similarity
If in two triangles, the corresponding sides are in the same ratio (proportional), then their corresponding angles are equal, and hence the triangles are similar.
Theorem 6.5 — SAS (Side-Angle-Side) Similarity
If one angle of a triangle is equal to one angle of the other, and the sides including these angles are proportional, then the triangles are similar.
Summary — Three Criteria
AA (or AAA): two angles equal ⇒ similar.SSS: three sides proportional ⇒ similar.
SAS: two sides proportional & included angles equal ⇒ similar.
Example 5 — AA criterion
In the figure, \(\angle ODC = \angle OBA\), \(\angle BOC = 125°\), \(\angle CDO = 70°\). Find \(\angle DOC\), \(\angle DCO\), and \(\angle OAB\).
\(\angle DOC = 180° - 125° = 55°\) (linear pair). In \(\triangle DOC\), \(\angle DCO = 180° - 55° - 70° = 55°\).
\(\triangle ODC \sim \triangle OBA\) by AA ⇒ \(\angle OAB = \angle OCD = 55°\).
Example 6 — Diagonals of a trapezium
Diagonals AC and BD of a trapezium ABCD (AB ∥ DC) intersect at O. Prove \(\triangle OAB \sim \triangle OCD\) and hence \(\tfrac{OA}{OC}=\tfrac{OB}{OD}\).
In \(\triangle OAB\) and \(\triangle OCD\): \(\angle AOB = \angle COD\) (vertically opposite); \(\angle OAB = \angle OCD\) (alternate angles, AB ∥ DC). By AA, \(\triangle OAB \sim \triangle OCD\). Hence the ratio result follows.
Example 7 — SAS criterion
In \(\triangle ABC\), \(\tfrac{AB}{PQ} = \tfrac{AC}{PR} = 2\) and \(\angle A = \angle P\). Are \(\triangle ABC\) and \(\triangle PQR\) similar?
Yes, by SAS. Hence \(\tfrac{BC}{QR} = 2\) also.
Example 8 — SSS criterion
In \(\triangle ABC\), AB = 5 cm, BC = 7 cm, CA = 8 cm; in \(\triangle DEF\), DE = 10, EF = 14, FD = 16. Are they similar?
\(\tfrac{AB}{DE}=\tfrac{5}{10}=\tfrac12,\ \tfrac{BC}{EF}=\tfrac{7}{14}=\tfrac12,\ \tfrac{CA}{FD}=\tfrac{8}{16}=\tfrac12\). All equal ⇒ similar by SSS with scale factor \(\tfrac12\).
Example 9 — Altitude from right-angle vertex
In a right triangle ABC right-angled at B, let BD be the altitude to AC. Prove that \(\triangle ADB \sim \triangle ABC\), \(\triangle BDC \sim \triangle ABC\), and \(\triangle ADB \sim \triangle BDC\).
In \(\triangle ADB\) and \(\triangle ABC\): \(\angle A = \angle A\) (common); \(\angle ADB = \angle ABC = 90°\). By AA, \(\triangle ADB \sim \triangle ABC\). Similarly, \(\triangle BDC \sim \triangle ABC\) (common \(\angle C\), right angles). Transitively, \(\triangle ADB \sim \triangle BDC\). This is the famous "altitude-on-hypotenuse" result used in the proof of Pythagoras' theorem.
Fig 6.5: Right triangle ABC with right angle at B; altitude BD splits it into two triangles similar to the original.
Example 10 — Finding lengths in similar triangles
\(\triangle ABC \sim \triangle PQR\) with \(\tfrac{AB}{PQ}=\tfrac{2}{3}\). If PQ = 9, QR = 12, PR = 15, find AB, BC, CA.
Scale = \(\tfrac23\). AB = \(\tfrac23\cdot 9 = 6\); BC = \(\tfrac23\cdot 12 = 8\); CA = \(\tfrac23\cdot 15 = 10\).
Example 11 — Prove AB² = AD · AC
In a right triangle with right angle at B and altitude BD on AC, prove \(AB^2 = AD \cdot AC\).
From Example 9, \(\triangle ADB \sim \triangle ABC\). So \(\tfrac{AD}{AB} = \tfrac{AB}{AC}\Rightarrow AB^2 = AD\cdot AC\). (Similarly \(BC^2 = CD\cdot CA\).)
Activity: Similarity with Two Triangular Cut-outs
L3 ApplyMaterials: Two sheets of paper, ruler, protractor, scissors.
Predict: If you cut two triangles whose angles are 40°, 60°, 80°, will they always be similar even if side lengths differ?
- On one sheet draw a triangle with angles 40°, 60°, 80° and sides measuring 5 cm, x, y. Cut it out.
- On the second sheet draw another triangle with the same angles but sides 7.5 cm, ..., .... Cut it out.
- Measure all sides of both triangles. Compute the ratios.
- Observe: the two sets of three ratios are equal — the scale factor is the same on all three sides.
This directly confirms the AAA (AA) similarity theorem: equal angles force proportional sides.
Exercise 6.3 (Selected)
Q1. State which pairs of triangles are similar and by which criterion. (a) \(\triangle ABC\): 60°, 80°, 40°; \(\triangle PQR\): 60°, 80°, 40°. (b) Sides 2, 3, 4 and 4, 6, 8. (c) Sides 3, 5, 7 and 6, 10, 14.
(a) Similar by AAA (or AA).
(b) Similar by SSS: 2/4 = 3/6 = 4/8 = 1/2.
(c) Similar by SSS: 3/6 = 5/10 = 7/14 = 1/2.
(b) Similar by SSS: 2/4 = 3/6 = 4/8 = 1/2.
(c) Similar by SSS: 3/6 = 5/10 = 7/14 = 1/2.
Q2. Diagonals AC and BD of a trapezium ABCD (AB ∥ DC) intersect at O. Using similarity, prove AO/OC = BO/OD.
See Example 6 — \(\triangle AOB \sim \triangle COD\) by AA (vertically opposite + alternate). Hence the ratio.
Q3. If \(\triangle ABC \sim \triangle DEF\) and scale factor is 3, and EF = 4 cm, find BC.
\(\tfrac{BC}{EF}=3\Rightarrow BC = 12\) cm.
Q4. CM and RN are respectively medians of similar triangles \(\triangle ABC \sim \triangle PQR\). Prove that \(\triangle AMC \sim \triangle PNR\).
\(\triangle ABC \sim \triangle PQR\) ⇒ \(\tfrac{AB}{PQ}=\tfrac{BC}{QR}=\tfrac{CA}{RP}=k\) and \(\angle A=\angle P\). Medians bisect the opposite sides: \(AM=\tfrac12 AB,\ PN=\tfrac12 PQ\), so \(\tfrac{AM}{PN}=k\). In \(\triangle AMC\) and \(\triangle PNR\): \(\angle A=\angle P\); \(\tfrac{AM}{PN}=\tfrac{AC}{PR}=k\). By SAS, similar. ∎
Q5. In a right triangle ABC right-angled at B, BD ⊥ AC with D on AC. If AB = 6 cm, AC = 10 cm, find AD, DC, BD.
BC = \(\sqrt{10^2-6^2}=8\). \(AB^2=AD\cdot AC\Rightarrow 36=10 AD\Rightarrow AD=3.6\). Then DC = 10−3.6 = 6.4. \(BD^2 = AD\cdot DC = 3.6\cdot 6.4 = 23.04\Rightarrow BD = 4.8\) cm.
Competency-Based Questions
Scenario: A surveyor wants to find the width AB of a river without crossing it. From a point C on her side, she measures AC = 50 m along the bank and CB = 30 m directly across (using a known right angle at C). Unfortunately, only part of the across-line is accessible; she uses a smaller similar triangle to estimate.
Q1. She sets up a small triangle with AC' = 5 m and C'B' = 3 m, right-angled at C'. By similarity, what is AB if \(\triangle ABC \sim \triangle AB'C'\) with a common angle A?
L3 Apply(c) ≈ 58.3 m. AB is the hypotenuse of a right triangle with legs 50, 30. Scale 10: small hyp = \(\sqrt{25+9}=\sqrt{34}\approx 5.83\), so AB ≈ 58.3 m.
Q2. Analyse which similarity criterion justifies the surveyor's method.
L4 AnalyseBoth triangles share angle A, and both have a right angle (C and C′). Two pairs of equal angles ⇒ AA similarity.
Q3. Evaluate: if the smaller triangle's right angle is off by 2°, roughly how much error is introduced in AB?
L5 EvaluateSimilarity depends on exact angle equality. A 2° error means the triangles are not truly similar; the computed ratio is wrong by up to ~tan(2°) ≈ 3.5%, introducing ≈ 2 m of error in AB. Surveyors repeat and average to minimise this.
Q4. Design an alternative method to find AB using only a tape and a mirror placed on the ground.
L6 CreatePlace a mirror flat on the ground between her and a landmark on the far bank. Move until she sees the landmark's reflection in the mirror. By equal angles of incidence/reflection, the two right-triangles formed (her eye-mirror-landmark) are similar. Measure her eye height, her distance to the mirror, and the mirror-to-landmark distance; then AB is found by similar-triangle ratio.
Assertion–Reason Questions
Assertion (A): Two triangles with angles 30°, 60°, 90° and 30°, 60°, 90° are always similar.
Reason (R): AA similarity.
Reason (R): AA similarity.
Answer: (a).
Assertion (A): If the sides of two triangles are 2, 3, 4 and 4, 6, 9, the triangles are similar by SSS.
Reason (R): SSS requires all three side-ratios to be equal.
Reason (R): SSS requires all three side-ratios to be equal.
Answer: (d). \(\tfrac24=\tfrac12,\ \tfrac36=\tfrac12,\ \tfrac49\ne\tfrac12\). Not similar. R is true.
Assertion (A): In a right triangle, the altitude to the hypotenuse divides it into two triangles similar to the original.
Reason (R): Each smaller triangle shares an acute angle with the original and has a right angle.
Reason (R): Each smaller triangle shares an acute angle with the original and has a right angle.
Answer: (a).
Frequently Asked Questions — Triangles
What is Part 2 — Criteria for Similarity of Triangles | Class 10 Maths Ch 6 | MyAiSchool in NCERT Class 10 Mathematics?
Part 2 — Criteria for Similarity of Triangles | Class 10 Maths Ch 6 | MyAiSchool is a key concept covered in NCERT Class 10 Mathematics, Chapter 6: Triangles. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Part 2 — Criteria for Similarity of Triangles | Class 10 Maths Ch 6 | MyAiSchool step by step?
To solve problems on Part 2 — Criteria for Similarity of Triangles | Class 10 Maths Ch 6 | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 10 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 6: Triangles?
The essential formulas of Chapter 6 (Triangles) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Part 2 — Criteria for Similarity of Triangles | Class 10 Maths Ch 6 | MyAiSchool important for the Class 10 board exam?
Part 2 — Criteria for Similarity of Triangles | Class 10 Maths Ch 6 | MyAiSchool is part of the NCERT Class 10 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Part 2 — Criteria for Similarity of Triangles | Class 10 Maths Ch 6 | MyAiSchool?
Common mistakes in Part 2 — Criteria for Similarity of Triangles | Class 10 Maths Ch 6 | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Part 2 — Criteria for Similarity of Triangles | Class 10 Maths Ch 6 | MyAiSchool?
End-of-chapter NCERT exercises for Part 2 — Criteria for Similarity of Triangles | Class 10 Maths Ch 6 | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 6, and solve at least one previous-year board paper to consolidate your understanding.
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