Triangles, Quadrilaterals and Circles
Overview
This topic forms the geometric backbone of upper-primary mathematics in KAR TET Paper II. Questions typically test your understanding of triangle properties, congruence and similarity criteria, quadrilateral theorems, and fundamental circle results.
Mastery here requires two skills: recognising which theorem or criterion applies to a given figure, and executing short logical proofs or calculations. The topic connects strongly with coordinate geometry (distance formula proofs) and mensuration (area and perimeter). Focus on the five congruence/similarity criteria, the angle-sum properties, and the three major circle theorems—these appear repeatedly.
Key Concepts
- Angle-sum property: Interior angles of a triangle sum to 180°; for a quadrilateral, the sum is 360°. Exterior angle of a triangle equals the sum of the two non-adjacent interior angles.
- Congruence of triangles: Two triangles are congruent if all corresponding sides and angles are equal. Five criteria establish congruence without checking all six parts: SSS, SAS, ASA, AAS, and RHS (right-angle–hypotenuse–side).
- Similarity of triangles: Triangles are similar if corresponding angles are equal (and hence sides are proportional). Three criteria: AAA (or AA), SSS (ratio), and SAS (ratio with included angle).
- Basic Proportionality Theorem (Thales' theorem): A line drawn parallel to one side of a triangle divides the other two sides proportionally. Converse also holds.
- Pythagoras theorem: In a right triangle, hypotenuse² = sum of squares of other two sides. Converse: if c² = a² + b², the triangle is right-angled.
- Quadrilateral hierarchy: Parallelogram → Rectangle / Rhombus → Square. Each inherits properties of the previous and adds constraints (right angles or equal sides).
- Circle angle theorems: Angle subtended by an arc at the centre is twice that at any point on the remaining circle. Angles in the same segment are equal. Angle in a semicircle is 90°.
- Tangent properties: A tangent is perpendicular to the radius at the point of contact. Tangents drawn from an external point are equal in length.
Formulas / Key Facts
| Concept | Formula / Fact |
|---|---|
| Triangle angle sum | ∠A + ∠B + ∠C = 180° |
| Exterior angle | Exterior ∠ = sum of two remote interior angles |
| Pythagoras | c² = a² + b² (right triangle) |
| Area of triangle | ½ × base × height; also √[s(s−a)(s−b)(s−c)] where s = (a+b+c)/2 |
| Basic Proportionality | If DE ‖ BC in △ABC, then AD/DB = AE/EC |
| Ratio of areas (similar △) | (Area₁)/(Area₂) = (side₁/side₂)² |
| Parallelogram diagonals | Bisect each other |
| Rectangle diagonals | Equal and bisect each other |
| Rhombus diagonals | Perpendicular bisectors of each other |
| Central angle theorem | ∠ at centre = 2 × ∠ at circumference (same arc) |
| Angle in semicircle | 90° |
| Tangent-radius | Tangent ⊥ radius at contact point |
| Tangent lengths | PA = PB (tangents from external point P) |
Worked Examples
Example 1 — Congruence (SAS)
In △ABC and △DEF, AB = DE = 5 cm, AC = DF = 7 cm, and ∠A = ∠D = 50°. Prove the triangles are congruent.
Solution: The given information shows two sides and the included angle of one triangle equal to the corresponding parts of the other. By SAS criterion, △ABC ≅ △DEF.
Example 2 — Basic Proportionality Theorem
In △PQR, a line parallel to QR meets PQ at X and PR at Y such that PX = 4 cm, XQ = 6 cm. If PY = 5 cm, find YR.
Solution: By BPT, PX/XQ = PY/YR ⇒ 4/6 = 5/YR ⇒ YR = (5 × 6)/4 = 7.5 cm.
Example 3 — Circle theorem (Central and inscribed angle)
An arc AB subtends an angle of 70° at point C on the circle. Find the angle subtended at the centre O.
Solution: Central angle = 2 × angle at circumference = 2 × 70° = 140°.
Example 4 — Tangent lengths
From an external point P, tangents PA and PB are drawn to a circle with centre O. If PA = 12 cm and OP = 13 cm, find the radius.
Solution: Tangent ⊥ radius ⇒ △OAP is right-angled at A. OA² + PA² = OP² ⇒ OA² = 169 − 144 = 25 ⇒ OA = 5 cm.
Common Mistakes
- Confusing congruence with similarity — Students write "△ABC ~ △DEF" when they mean congruent (≅). Similarity requires only proportional sides; congruence requires equal sides. Fix: Use ≅ for congruence, ~ for similarity.
- Wrong order of vertices in similarity — Writing △ABC ~ △EDF when the correct correspondence is △ABC ~ △DEF scrambles ratios. Fix: Always list vertices so that corresponding angles match.
- Applying BPT when line is not parallel — The theorem holds only when the transversal is parallel to one side. Fix: Verify parallelism before using the proportionality result.
- Forgetting the "included angle" in SAS — Two sides and any angle are not enough; the angle must be between the two given sides. Fix: Check that the angle is formed by the two sides being compared.
- Mixing central and inscribed angles — Using the inscribed angle value as the central angle (or vice versa) doubles or halves the correct answer. Fix: Remember central angle = 2 × inscribed angle for the same arc.
Quick Reference
- Triangle interior angles = 180°; quadrilateral = 360°.
- Congruence criteria: SSS, SAS, ASA, AAS, RHS.
- Similarity criteria: AA, SSS (ratio), SAS (ratio).
- BPT: Parallel line divides two sides proportionally.
- Central ∠ = 2 × inscribed ∠ (same arc); semicircle ∠ = 90°.
- Tangent ⊥ radius; tangents from external point are equal.