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. Expect 3–5 questions directly from this area, with additional overlap in mensuration problems.
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) |
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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.
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**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.
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**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°.
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**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
1. **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.
2. **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.
3. **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.
4. **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.
5. **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.