Class 10 Mathematics Chapter 6 Triangles — Formulas & Key Points

Before diving into theorems, clarity on terminology prevents half the errors in board exams. A figure is a closed shape; a triangle is a polygon with three sides and three angles summing to 180°. Congruent triangles are identical in shape and size (all corresponding sides and angles equal). Similar triangles have the same shape but not necessarily the same size; corresponding angles are equal and corresponding sides are in proportion. The symbol for similarity is '∼' and for congruence is '≅'. In NCERT Class 10 Mathematics, the focus shifts from congruence (Class 9) to similarity. The term 'corresponding' is critical: when writing ΔABC ∼ ΔPQR, A corresponds to P, B to Q, and C to R; mixing this order in proofs loses marks. Always write the correspondence in the same sequence as given in the question or as you establish through angle-chasing.

The Basic Proportionality Theorem, attributed to the Greek mathematician Thales, is the gateway to similarity. It states: 'If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides those sides in the same ratio.' Formally, in ΔABC, if DE ∥ BC and D lies on AB, E lies on AC, then AD/DB = AE/EC. The converse is equally important: if a line divides two sides of a triangle in the same ratio, then that line is parallel to the third side. This converse is tested in 'prove that DE ∥ BC' type questions. Both statements appear verbatim in CBSE papers; write the theorem name in your proof to secure method marks. BPT underpins every similarity proof and is worth 3-4 marks when asked standalone.

CBSE Class 10 Mathematics Chapter 6 introduces four criteria to prove two triangles similar. Unlike congruence (which has five criteria), similarity focuses on angles and side ratios. The AA (Angle-Angle) criterion is the most frequently tested because measuring two angles automatically fixes the third (sum = 180°). The SSS (Side-Side-Side) criterion requires all three pairs of corresponding sides to be proportional. The SAS (Side-Angle-Side) criterion needs two sides proportional and the included angle equal. The AAA criterion is theoretically valid but reduces to AA in practice. Boards often ask: 'State the criterion by which ΔABC ∼ ΔPQR' – answer with the exact acronym (AA, SSS, or SAS) and a one-line reason. Mixing up the order of sides or forgetting the included-angle condition in SAS costs marks.

The table below lists each similarity criterion, its formal statement, and the typical scenario in which you apply it. Use this table as a flowchart: identify what information the question gives (angles? side lengths? both?) and pick the matching criterion. In MCQs, eliminate options by checking if the given data satisfies the conditions. For instance, if only two angles are mentioned, AA is your criterion; if three side ratios are given, SSS applies. The included-angle requirement in SAS is a common trap: the equal angle must lie between the two proportional sides. If the angle is opposite or adjacent but not between, SAS does not apply. Boards sometimes present a diagram with tick marks and arc symbols; decode them into mathematical statements before choosing a criterion. Writing the correct criterion name earns you 1 mark even if the subsequent working has minor slips.

One of the most examined results in Triangles is the area-ratio theorem: 'The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.' Formally, if ΔABC ∼ ΔPQR, then (Area of ΔABC) / (Area of ΔPQR) = (AB/PQ)² = (BC/QR)² = (CA/RP)². Notice the square: a common error is writing the ratio of areas equal to the ratio of sides (linear), but area is two-dimensional, hence the square. This theorem is tested in word problems involving scale models, maps, or similar figures. Boards may give one area and one side ratio, asking for the other area; always square the side ratio before multiplying. The theorem also extends to other polygons, but NCERT Class 10 restricts it to triangles. Proof of this theorem uses the formula for area (½ × base × height) and the fact that corresponding altitudes are in the same ratio as corresponding sides. Write this theorem statement verbatim in your answer to secure full marks, especially in 4-mark proofs where method marks are awarded for correct citations.

Although Pythagoras theorem is formally in Chapter 6 (sometimes shifted to Chapter 7 depending on edition), it is integral to Triangles problems. The theorem states: 'In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.' If ∠C = 90°, then AB² = AC² + BC². The converse is equally important for proof questions: 'If in a triangle, the square of one side equals the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.' This converse is used to prove a triangle is right-angled when side lengths are given. Boards love combining Pythagoras with similarity: for example, proving two right triangles similar via AA criterion (both have a 90° angle plus one more common angle). Pythagorean triplets like (3,4,5), (5,12,13), (8,15,17) and their multiples appear frequently; recognizing them saves calculation time. Always label the hypotenuse (longest side opposite the right angle) clearly in your diagram to avoid substituting sides incorrectly in the formula, a mistake that leads to negative or illogical results.

Beyond the main theorems, NCERT Class 10 Mathematics Chapter 6 Triangles presents several corollaries that streamline problem-solving. One key corollary of BPT: if a line divides two sides of a triangle proportionally, it is parallel to the third side (converse BPT). Another useful result: the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length (mid-point theorem, often recalled from Class 9). In the context of similar triangles, if a perpendicular is drawn from the right angle to the hypotenuse in a right triangle, it divides the triangle into two smaller triangles each similar to the original and to each other. This result is exploited in altitude-on-hypotenuse problems. Boards also test the angle-bisector theorem (internal bisector divides opposite side in the ratio of adjacent sides), though it is not always in every state syllabus; verify your NCERT edition. Memorize these corollaries as one-liners; they are perfect for filling gaps in proofs or justifying a step when the main theorem does not directly apply. CBSETUTOR.ai offers real-time doubt clearing with photo upload at a flat ₹999/month for all classes 6-12, and students can practice these corollaries through personalized problem sets with instant feedback during the 3-day free trial, making last-minute revision seamless.

Formulas stick when you attach a story or acronym. For similarity criteria, remember 'A-S-S' order: AA (Angle-Angle), SSS (Side-Side-Side), SAS (Side-Angle-Side) – but note there is no ASS in similarity (unlike congruence's ASA). For BPT, think 'Parallel ⇒ Proportional' – a parallel line slices sides in equal ratios. For the area theorem, chant 'Area is two-dimensional, square the ratio!' before every problem. To recall Pythagoras triplets, note 3-4-5 is the smallest, then double/triple them (6-8-10, 9-12-15); or remember 5-12-13 and 8-15-17 separately. When proving similarity, draw both triangles side by side, mark equal angles with arcs of the same color, and proportional sides with the same number of ticks; visual symmetry reduces mistakes. For the converse of Pythagoras, the mnemonic 'If the square-sum fits, the angle sits at ninety' helps you remember that equality of a² + b² = c² forces a 90° angle. Finally, in proofs, write the theorem name in parentheses after each assertion, e.g., '∠A = ∠P (given), ∠B = ∠Q (given), hence ΔABC ∼ ΔPQR (AA criterion).' This structured style earns full method marks even if arithmetic slips occur later in the solution.

Half of the marks lost in Triangles come from notation slips, not conceptual gaps. Writing ΔABC ∼ ΔQPR when the correct correspondence is ΔABC ∼ ΔPQR inverts ratios and leads to wrong answers; always match vertices in the same order. Confusing the symbols ∼ (similar) and ≅ (congruent) signals a fundamental misunderstanding to the examiner. Another frequent error: writing AD/DB when you mean AD/AB (part versus whole). The ratio AD/DB is 'part to remaining part' while AD/AB is 'part to total'; both appear in BPT but in different forms – clarify which the question uses. In the area-ratio theorem, forgetting to square the side ratio is the single most common mistake; double-check your exponent before substituting numbers. Units matter less in pure geometry (most lengths are unitless or in cm), but if a question mixes meters and centimeters, convert before computing ratios. In proofs, jumping steps without justification (e.g., stating ΔABC ∼ ΔPQR without naming the criterion) loses method marks even if the final answer is correct. Finally, when drawing additional construction lines (common in BPT or altitude proofs), label them immediately and state 'Construction: Draw DE ∥ BC' so the examiner knows it is intentional, not a diagram error. Clean notation and explicit reasoning can add 2-3 marks to an otherwise mediocre answer.

Example 1 (BPT application): In ΔABC, D is on AB such that AD = 3 cm, DB = 5 cm. A line through D parallel to BC intersects AC at E. If AC = 12 cm, find AE. Solution: By BPT, AD/AB = AE/AC. AB = AD + DB = 8 cm. So 3/8 = AE/12 ⇒ AE = 3×12/8 = 4.5 cm. Example 2 (Similarity via AA): In ΔPQR and ΔXYZ, ∠P = 50°, ∠Q = 60°, ∠X = 50°, ∠Y = 60°. Are the triangles similar? Solution: Two pairs of corresponding angles are equal. By AA criterion, ΔPQR ∼ ΔXYZ. (Third angles are 70° each automatically.) Example 3 (Area ratio): ΔABC ∼ ΔDEF. AB = 6 cm, DE = 9 cm. If area of ΔABC = 20 cm², find area of ΔDEF. Solution: Side ratio AB/DE = 6/9 = 2/3. Area ratio = (2/3)² = 4/9. So 20 / Area(ΔDEF) = 4/9 ⇒ Area(ΔDEF) = 20×9/4 = 45 cm². These mini-examples mirror the 2-3 mark short-answer questions in CBSE board papers; practice them until you can solve each in under two minutes without referring to notes.

Use this box 30 minutes before your exam. BPT: Line ∥ to one side ⇒ divides other two sides proportionally (AD/DB = AE/EC). Converse: Proportional division ⇒ line is parallel. Similarity criteria: AA (two ∠s equal), SSS (three sides proportional), SAS (two sides proportional + included ∠ equal). Area ratio: (Area₁/Area₂) = (side₁/side₂)². Pythagoras: c² = a² + b² in right triangle. Converse: if c² = a² + b², then triangle is right-angled at the vertex opposite c. Triplets: 3-4-5, 5-12-13, 8-15-17 (and multiples). Mid-point theorem: Line joining midpoints ∥ third side and = ½ third side. Notation: ΔABC ∼ ΔPQR means exact vertex correspondence; ∼ is similarity, ≅ is congruence. Proof checklist: (1) Mark given data on diagram, (2) State the criterion name, (3) Write 'Hence proved' at the end. Marks distribution: BPT 3-4 marks, similarity proof 4-6 marks, area-ratio problem 3 marks, Pythagoras 2-3 marks. Total chapter weight: 8-10 marks in the 80-mark paper. Revise theorem statements verbatim from NCERT pages 121-141 (most editions). Good luck!