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CDS Mathematics

Geometry

📐 Geometry · MC09 CDS Elementary Mathematics 🎯 High Priority

Geometry is the single largest individual chapter in CDS, contributing 15–20 questions — nearly a fifth of the entire paper. The syllabus covers lines, angles, triangles, quadrilaterals, circles, and loci. Questions are theorem-based: learn the property, identify where it applies, substitute. Diagrams and conceptual understanding beat rote formula memorisation here.

📌 CDS exam focus: (1) Angle properties — parallel lines, transversals, polygons; (2) Triangle congruence — SSS, SAS, ASA, AAS, RHS; (3) Similar triangles — AA, SSS, SAS similarity; corresponding sides ratio; (4) Circles — chord properties, tangent perpendicular to radius, cyclic quadrilaterals; (5) Quadrilateral properties — parallelogram, rectangle, rhombus; (6) Pythagoras theorem — direct and converse.

Topics at a Glance

① Lines & Angles

Parallel lines, transversal, angle pairs

② Triangles

Properties, congruence, similarity, centres

③ Quadrilaterals

Parallelogram, rectangle, rhombus, trapezium

④ Circles

Chords, tangents, angles in circle

⑤ Pythagoras

Theorem, converse, triples, applications

⑥ Loci

Equidistant from two points/lines

1. Lines & Angles

1.1

Angle Pairs, Parallel Lines & Polygon Angles

These properties underpin every other geometry topic

2. Triangles — Congruence & Similarity

2.1

Congruence Criteria, Similarity Rules & Key Theorems

Triangles account for nearly half of CDS Geometry questions

🔺 Congruence (≅) Criteria

SSS: All 3 sides equal
SAS: 2 sides and included angle equal
ASA: 2 angles and included side equal
AAS: 2 angles and any corresponding side equal
RHS: Right angle, hypotenuse, one side (right triangles only)
Congruent triangles: all sides and angles equal

🔺 Similarity (~) Criteria

AA: Two angles equal → third also equal
SSS: All 3 sides in same ratio
SAS: 2 sides in ratio and included angle equal
Similar triangles: angles equal; sides proportional
Ratio of areas = (ratio of corresponding sides)²
Ratio of perimeters = ratio of corresponding sides

⚡ Key Triangle Properties

Angle sum property: ∠A + ∠B + ∠C = 180° Exterior angle: Ext∠ = sum of two non-adjacent interior angles Triangle inequality: Sum of any two sides > third side Pythagoras Theorem (right triangle, right angle at C): a² + b² = c² (c = hypotenuse) Converse: if a²+b²=c², then angle C is 90° Common Pythagorean triples: 3-4-5 | 5-12-13 | 8-15-17 | 7-24-25 Median: connects vertex to midpoint of opposite side Centroid (G): divides each median in ratio 2:1 from vertex Circumcentre: equidistant from all 3 vertices Incentre: equidistant from all 3 sides Orthocentre: intersection of altitudes Mid-point theorem: Line joining midpoints of 2 sides is parallel to 3rd side and half its length

3. Quadrilaterals

3.1

Properties of All Standard Quadrilaterals

Know which property belongs to which shape — a common CDS question type

Shape	Sides	Diagonals	Angles	Special Property
Parallelogram	Opp sides equal & parallel	Bisect each other	Opp angles equal; adj supplementary	Area = base × height
Rectangle	Opp sides equal; all angles 90°	Equal and bisect each other	All 90°	Diagonal = √(l²+b²)
Rhombus	All sides equal	Perpendicular bisectors of each other	Opp equal; adj supplementary	Area = ½d₁d₂
Square	All sides equal; all angles 90°	Equal, perpendicular bisectors	All 90°	Diagonal = a√2
Trapezium	One pair of parallel sides	Not generally equal	Co-interior angles supplementary	Area = ½(a+b)×h

4. Circles

4.1

Chord, Tangent & Angle Theorems — CDS Essentials

Circle theorems provide 4–6 questions per CDS paper

⚡ Key Circle Theorems

CHORD PROPERTIES: Equal chords are equidistant from centre Perpendicular from centre to chord bisects the chord Equal chords subtend equal angles at centre ANGLE THEOREMS: Angle at centre = 2 × angle at circumference (same arc) Angles in same segment are equal Angle in semicircle = 90° (angle in a semicircle is always a right angle) Opposite angles of cyclic quadrilateral = 180° (supplementary) TANGENT PROPERTIES: Tangent ⊥ radius at point of contact Two tangents from external point are equal in length Tangent-chord angle = angle in alternate segment (Alternate Segment Theorem) INTERSECTING CHORDS (inside circle): PA × PB = PC × PD SECANT-TANGENT (outside): PT² = PA × PB

"Angle in semicircle = 90°" and "opposite angles of cyclic quad = 180°" are the two most tested circle theorems in CDS.

📋 TOPIC-WISE PYQ

Geometry — CDS Questions

Q1. In a triangle, the exterior angle is 110°. If one interior opposite angle is 45°, find the other.

(a) 55° (b) 65° (c) 70° (d) 75°

Answer: (b) 65°
Ext angle = sum of two non-adjacent interior angles. 110° = 45° + x → x=65°.

Q2. Two chords AB and CD of a circle intersect inside at P. If PA=4, PB=9 and PC=6, find PD.

(a) 4 (b) 6 (c) 8 (d) 9

Answer: (b) 6
Intersecting chords: PA×PB=PC×PD → 4×9=6×PD → PD=36/6=6.

Q3. ABCD is a cyclic quadrilateral with ∠A=80°. Find ∠C.

(a) 80° (b) 100° (c) 90° (d) 120°

Answer: (b) 100°
Opposite angles of cyclic quadrilateral are supplementary. ∠A+∠C=180° → ∠C=100°.

Q4. In triangle ABC, D and E are midpoints of AB and AC. If BC=12 cm, find DE.

(a) 4 cm (b) 6 cm (c) 8 cm (d) 3 cm

Answer: (b) 6 cm
Mid-point theorem: DE = BC/2 = 12/2 = 6 cm. DE is also parallel to BC.

Q5. The angle subtended by an arc at the centre is 70°. The angle subtended at the circumference by the same arc is:

(a) 35° (b) 70° (c) 140° (d) 105°

Answer: (a) 35°
Angle at centre = 2 × angle at circumference → circumference angle = 70°/2 = 35°.

Q6. In a right-angled triangle, the hypotenuse is 13 cm and one side is 5 cm. Find the third side.

(a) 8 cm (b) 10 cm (c) 12 cm (d) 11 cm

Answer: (c) 12 cm
Pythagoras: b²=13²−5²=169−25=144 → b=12 cm. (5-12-13 is a Pythagorean triple.)

🔥 TRICKY QUESTIONS

Geometry — Classic CDS Traps

🧩 T1. From a point P outside a circle, two tangents PA and PB are drawn. If AB=8 cm and PA=6 cm, find PO (distance to centre).

Solution: PO = √(6²−4²)... Let me set up properly.
Two tangents equal: PA=PB=6. Let O=centre, r=radius. AB is the chord of contact.
OM⊥AB where M is midpoint of AB (M is midpoint, AM=4). In △OAP: OA=r (radius), PA=6 (tangent).
OM²=OA²−AM²=r²−16. Also PM=PA−AM... Using: PO²=PA²+r²−... actually: PO²=PA²+OA²=36+r² (no — OA⊥PA only if P is on OA extended).
Key: PO²=PA²+r²? No: PA²=PO²−r² (tangent relation). So r²=PO²−36. Also: OM⊥AB, so AM=4 → r²=OM²+16 and OM+MP=PO. Standard result: PO=√(PA²+r²) only if PA⊥OA. Need r to solve fully — question needs more data or r is given indirectly.

🧩 T2. In △ABC, ∠B=90°, D is the foot of altitude from B to AC. Prove BD²=AD×DC.

Solution using similarity.
In △ABC and △ADB: ∠A is common, ∠ADB=∠ABC=90°. So △ADB~△ABC (AA).
Similarly △BDC~△ABC (AA). From these: BD/AD=DC/BD → BD²=AD×DC. ✓
This geometric mean relation frequently gives CDS numerical questions: "If AD=4, DC=9, find BD."
BD = √(4×9) = √36 = 6.

🧩 T3. Three interior angles of a quadrilateral are 75°, 95° and 110°. Find the fourth angle.

Solution: 80°.
Sum of interior angles of quadrilateral = (4−2)×180° = 360°.
Fourth angle = 360°−75°−95°−110° = 360°−280° = 80°.

📐 Formula Sheet — MC09

Angle Properties

Triangle angles: 180°
Polygon: (n−2)×180°
Ext angle = sum of opp interior angles
Parallel lines: alt interior = equal; co-interior = 180°

Triangle Theorems

Midpoint theorem: DE = BC/2, DE∥BC
Pythagoras: a²+b²=c²
Similar: ratio of areas = (ratio of sides)²
Congruence: SSS, SAS, ASA, AAS, RHS

Circle Theorems

Centre angle = 2× circumference angle
Semicircle angle = 90°
Cyclic quad: opp angles = 180°
Tangent ⊥ radius; two tangents equal

Quadrilateral Facts

Rhombus: all sides equal; diag⊥
Rectangle: diag equal
Square: all sides, all angles, diag equal and ⊥
Parallelogram: diag bisect each other

Chord & Tangent

Intersecting chords: PA×PB=PC×PD
Tangent²=secant×external segment
Perp from centre bisects chord

Pythagorean Triples

3-4-5 (×2: 6-8-10)
5-12-13
8-15-17
7-24-25

⚡ Quick Revision Booster — MC09

Triangle

Angles sum 180°
Ext = sum of 2 opp interior
Midpoint: half opposite side
Centroid: 2:1 from vertex

Congruence

SSS, SAS, ASA, AAS, RHS
Congruent: ALL equal
AAA is NOT congruence (it's similarity)

Circle Angles

Centre = 2× circumference
Semicircle = 90°
Same segment = equal
Cyclic quad: opp = 180°

Tangent

Tangent ⊥ radius
Two tangents = equal length
Alternate segment theorem
PT²=PA×PB (external)

Similarity

AA, SSS, SAS
Area ratio = (sides ratio)²
BD²=AD×DC (altitude)

🚨 Traps

AAA → similarity not congruence
Angle in semicircle = 90° always
Cyclic quad: OPPOSITE angles add to 180°

✏️ PRACTICE EXERCISE

Test Yourself — MC09

E1. Two parallel lines are cut by a transversal. If co-interior angles are (3x+20)° and (2x+10)°, find x.

(a) 25 (b) 30 (c) 35 (d) 40

💡 Co-interior angles sum = 180°. Set (3x+20)+(2x+10)=180.

E2. In triangle ABC, DE∥BC, AD=4, DB=6. Find DE if BC=15.

(a) 6 (b) 7 (c) 8 (d) 9

💡 By Basic Proportionality Theorem: DE/BC = AD/AB = 4/10.

E3. A chord is 8 cm long and is 3 cm from the centre. Find the radius.

(a) 4 cm (b) 5 cm (c) 6 cm (d) 7 cm

💡 Perp from centre bisects chord. Half chord = 4. r²=4²+3².

E4. The diagonals of a rhombus are 16 cm and 12 cm. Find its side.

(a) 8 cm (b) 10 cm (c) 12 cm (d) 14 cm

💡 Diagonals of rhombus bisect at 90°. Half diagonals: 8 and 6. Side=√(8²+6²).

E5. An inscribed angle in a circle subtends an arc of 130°. The inscribed angle is:

(a) 65° (b) 130° (c) 260° (d) 45°

💡 Inscribed (circumference) angle = half the central angle (arc measure).

E6. In a right-angled triangle, altitude from right angle to hypotenuse divides it into 4 cm and 9 cm. Find the altitude.

(a) 5 cm (b) 6 cm (c) 13 cm (d) 6.5 cm

💡 Altitude = √(AD×DC) = √(4×9).

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