Chapter 5 of 24

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ICSE Class 10 Mathematics★ ICSE Class 10 Mathematics · ~28 marks · Board: ICSE

Geometry, Mensuration, Trigonometry, Statistics & Probability

The FINAL ICSE Class 10 Mathematics toolkit: SIMILARITY of triangles, CIRCLE theorems, LOCI, 3D MENSURATION, advanced TRIGONOMETRY (heights and distances, trig identities), STATISTICS (mean, median, mode, ogive), and PROBABILITY. ICSE Class 10 Mathematics.

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By the end of this chapter you'll be able to…

1Apply AA, SAS, SSS criteria to prove triangles similar; use the ratio of areas theorem
2State and apply all seven ICSE circle theorems: angle at centre, angle in semicircle, angles in same segment, cyclic quadrilateral, tangent-radius, equal tangents, Alternate Segment Theorem
3Find the locus of a point satisfying given geometric conditions
4Calculate surface area and volume of cylinders, cones, spheres, hemispheres, and combinations thereof
5Prove trigonometric identities using sin, cos, tan transformations; solve heights-and-distances problems using angles of elevation and depression
6Find mean, mode, and median of grouped data; draw ogive and estimate median from it
7Calculate theoretical and experimental probability; solve problems on combined events

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Why this chapter matters

Geometry and Trigonometry together dominate the second half of the ICSE paper. Circle theorems are a guaranteed 8–10 marks every year — the Alternate Segment Theorem is the most-tested theorem in ICSE. Mensuration (3D solids) is 6–8 marks and highly formula-dependent. Statistics (ogive, mean, median) is 6 marks of reliable scoring if you master the cumulative frequency table. Trigonometric identities and heights-and-distances appear in every ICSE paper. This file covers the 'hard' portion of the paper — but with practice, these become manageable.

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Geometry, Mensuration, Trig, Statistics & Probability

1. Similarity of Triangles

Criteria

AA (Angle-Angle): Two angles equal → triangles SIMILAR.
SAS (Side-Angle-Side): Two sides PROPORTIONAL, included angle EQUAL.
SSS (Side-Side-Side): Three sides PROPORTIONAL.

Key Theorems

Ratio of areas of similar triangles = (Ratio of corresponding sides)²
Basic Proportionality Theorem (Thales) : A line parallel to one side divides the other two sides PROPORTIONALLY.
Converse: If a line divides two sides proportionally, it is PARALLEL to the third side.

2. Circle Theorems

Theorem	Statement
Angle at centre	= 2 × Angle at circumference (subtended by same arc).
Angle in semicircle	= 90°.
Angles in same segment	Are EQUAL.
Cyclic Quadrilateral	Opposite angles SUM to 180°.
Tangent-Radius	Tangent is PERPENDICULAR to radius at point of contact.
Tangents from external point	EQUAL in length.
Alternate Segment Theorem	Angle between tangent and chord = angle in ALTERNATE segment.

3. Loci (Constructions)

The LOCUS is the SET OF ALL POINTS satisfying a given condition. Examples: Locus of points equidistant from two points → perpendicular bisector. Equidistant from two intersecting lines → angle bisectors. At a fixed distance from a point → circle.

4. Mensuration — 3D Solids

Solid	CSA	TSA	Volume
Cylinder	2πrh	2πr(r+h)	πr²h
Cone	πrl [l=√(r²+h²)]	πr(r+l)	⅓πr²h
Sphere	4πr²	4πr²	(4/3)πr³
Hemisphere	2πr²	3πr²	(2/3)πr³

Frustum of a Cone

When the top of a cone is cut off by a plane parallel to the base.

5. Trigonometry — Advanced

Identities (Must Memorise)

sin²θ + cos²θ = 1 → sec²θ — tan²θ = 1 → cosec²θ — cot²θ = 1
sin(90°—θ) = cos θ. cos(90°—θ) = sin θ. tan(90°—θ) = cot θ.

Heights and Distances

Angle of Elevation: Looking UP from horizontal.
Angle of Depression: Looking DOWN from horizontal.
Two common problem types:
1. Single observation: One triangle. Use appropriate trig ratio.
2. Two observations: Two right triangles. Often from different positions/heights. Use simultaneous equations.

Proving Trigonometric Identities

Strategy: Start with the MORE COMPLICATED side. Express everything in terms of sin and cos. Use the fundamental identity. 'ICSE exams FREQUENTLY ask: Prove that LHS = RHS. Practice identifying which side to simplify.'

6. Statistics

Mean (Average)

Direct: X̄ = Σfx / Σf
Assumed Mean (Shortcut) : X̄ = A + (Σfd / Σf), where d = x — A
Step Deviation: X̄ = A + h(Σfu / Σf), where u = (x — A)/h

Median (Grouped Data)

Median = L + [(N/2 — CF) / f] × h L = lower limit of median class. N = total frequency. CF = cumulative frequency BEFORE median class. f = frequency of median class. h = class width.

Mode

Mode = L + [(f₁ — f₀) / (2f₁ — f₀ — f₂)] × h f₁ = frequency of modal class. f₀ = frequency before. f₂ = frequency after.

Empirical Relationship (for Moderately Skewed Data)

Mode ≈ 3 Median — 2 Mean

Ogive (Cumulative Frequency Curve)

Plot: Upper class limits vs. cumulative frequencies. 'Less than' ogive. FIND MEDIAN GRAPHICALLY: on y-axis at N/2. Draw horizontal line to curve. Drop vertical to x-axis. The x-value is the MEDIAN.

7. Probability

Classical Definition

P(E) = n(E) / n(S) = Number of favourable outcomes / Total number of outcomes. 0 ≤ P(E) ≤ 1.

Key Concepts

Sample Space (S) : All possible outcomes. Two dice: 36 outcomes. Deck of cards: 52.
Event (E) : A subset of S. 'Getting a sum of 7 on two dice.'
Complement: P(not E) = 1 — P(E).
Addition Rule: P(A ∪ B) = P(A) + P(B) — P(A ∩ B).
Mutually Exclusive: P(A ∩ B) = 0. Cannot happen together.

Playing Cards (Standard 52-Card Deck)

4 suits (Spades ♠, Hearts ♥, Diamonds ♦, Clubs ♣). 13 cards per suit. 26 RED (♥♦). 26 BLACK (♠♣).
Face cards: Jack, Queen, King (12 total). Aces (4 total).

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Circle Theorems

ANGLE AT CENTRE: ∠AOB = 2 × ∠ACB (A, B on circle, O = centre, C = any point on major arc). ANGLE IN SEMICIRCLE: If AB is diameter, ∠ACB = 90° for any C on circle. ANGLES IN SAME SEGMENT: ∠ACB = ∠ADB (C, D on same side of chord AB). CYCLIC QUADRILATERAL: ∠A + ∠C = 180°, ∠B + ∠D = 180° (opposite angles supplementary). TANGENT-RADIUS: The tangent at any point is perpendicular to the radius at that point. EQUAL TANGENTS: Two tangents from external point are equal. ALTERNATE SEGMENT THEOREM: Angle between tangent and chord = angle in alternate segment.

MOST TESTED: Alternate Segment Theorem. If PT is a tangent at P and PQ is a chord, then ∠TPQ = ∠PRQ where R is any point in the segment on the OTHER side of PQ. ICSE examiners test this in both 'prove' and 'find angle' forms. Strategy for all circle problems: (1) Mark all known angles on the diagram. (2) Look for which theorem connects the known to the unknown. (3) Work step by step, naming each theorem.

Mensuration — 3D

CYLINDER: CSA = 2πrh, TSA = 2πr(r+h), V = πr²h. CONE: Slant height l = √(r²+h²). CSA = πrl, TSA = πr(r+l), V = (1/3)πr²h. SPHERE: SA = 4πr², V = (4/3)πr³. HEMISPHERE: CSA = 2πr², TSA = 3πr², V = (2/3)πr³. HOLLOW CYLINDER: V = π(R²−r²)h. COMBINATION: Add or subtract volumes and areas as appropriate (e.g., a cone on top of a cylinder: V = V_cylinder + V_cone; TSA = base circle + CSA cylinder + CSA cone).

COMMON ICSE TYPES: (1) Melting a solid and recasting into another shape — volumes are EQUAL. (2) A cone/cylinder dipped into water — water displaced = volume of solid. (3) Hollow sphere/cylinder — use R (outer) and r (inner) radius. For combinations, SKETCH the solid first. Identify which surfaces are exposed (form TSA) and which are internal (not counted).

Trigonometric Identities

FUNDAMENTAL: sin²θ + cos²θ = 1. Derived: 1 + tan²θ = sec²θ. 1 + cot²θ = cosec²θ. CO-RATIO: sin(90°−θ) = cosθ, cos(90°−θ) = sinθ, tan(90°−θ) = cotθ. HEIGHTS: tan(angle of elevation) = height / distance. tan(angle of depression) = height / distance (same angle, measured from horizontal). RECIPROCAL IDENTITIES: cosecθ = 1/sinθ, secθ = 1/cosθ, cotθ = 1/tanθ.

PROVING IDENTITIES STRATEGY: (1) Start with the MORE COMPLEX side. (2) Convert everything to sin and cos. (3) Use sin²θ + cos²θ = 1 and its derived forms. (4) Factorise or expand as needed. (5) Never transfer terms across the '=' sign — only simplify each side independently. HEIGHTS AND DISTANCES: Always draw a clear diagram. Label angles of elevation/depression precisely. Two-triangle problems → set up two equations in two unknowns (height h and horizontal distance d).

Statistics

MEAN (grouped): X̄ = Σfx / Σf (direct), or X̄ = A + Σfd/Σf (assumed mean, d = x−A). MEDIAN (grouped): M = L + [(n/2 − CF) / f] × h, where L = lower class boundary, CF = cumulative freq before median class, f = freq of median class, h = class width. MODE: Class with highest frequency (modal class). Modal value = L + [(f₁−f₀)/(2f₁−f₀−f₂)] × h. OGIVE: Plot cumulative frequency against upper class boundary. Median = x-value at n/2 cumulative frequency. Quartiles: Q₁ at n/4, Q₃ at 3n/4.

OGIVE QUESTION: ICSE asks to draw an ogive and estimate the median, lower quartile, and upper quartile graphically. Steps: (1) Make a cumulative frequency table (use LESS THAN type — upper boundaries). (2) Plot points (upper boundary, cumulative freq). (3) Join with a smooth curve. (4) Draw a horizontal line at n/2 from y-axis, where it meets the ogive, drop a perpendicular to x-axis → that x-value is the MEDIAN.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Confusing the Alternate Segment Theorem with angles in the same segment

✓ Alternate Segment Theorem: The angle between the TANGENT and a CHORD at the point of contact equals the INSCRIBED ANGLE subtending the same chord from the OPPOSITE segment. If TP is tangent at P and PQ is a chord: ∠TPQ = ∠PRQ (where R is in the segment on the OTHER side of PQ — the ALTERNATE segment). 'Alternate' means 'the other side.' Draw carefully: the angle is between tangent and chord on one side; the equal angle is the inscribed angle from the other side.

WATCH OUT

✗ Using diameter instead of radius in mensuration formulas

✓ ALL mensuration formulas use RADIUS (r), not diameter. If the problem gives diameter, HALVE it first. Example: 'A sphere of diameter 14 cm' → r = 7 cm → V = (4/3)π(7)³, not (4/3)π(14)³. This is the most common mensuration error — it multiplies the volume by 8.

WATCH OUT

✗ Drawing the angle of depression incorrectly

✓ Angle of DEPRESSION is measured DOWNWARD from the HORIZONTAL, not from the vertical. In a heights-and-distances diagram: stand at the top of the tower, draw a horizontal line from your eye level going OUTWARD. The angle DOWN to the object below is the angle of depression. IMPORTANT: Angle of depression from top of tower to base = Angle of elevation from base to top of tower (alternate interior angles). This equality is used in most two-triangle problems.

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· circle-theorem

In a circle with centre O, AB is a chord. C is a point on the major arc. If ∠AOB = 140°, find ∠ACB.

▶ Show solution

By the Angle at Centre Theorem: ∠AOB = 2 × ∠ACB (angle at centre = twice the angle at circumference, subtended by the same arc AB). Therefore ∠ACB = ∠AOB / 2 = 140° / 2 = 70°.

Q2MEDIUM· mensuration-combination

A toy is made by mounting a cone of radius 5 cm and height 12 cm on a hemisphere of the same radius. Find the total surface area and volume of the toy.

▶ Show solution

Cone: r = 5, h = 12. Slant height l = √(5² + 12²) = √(25 + 144) = √169 = 13 cm. Hemisphere: r = 5 cm. TOTAL SURFACE AREA: Only the CURVED surface of the cone + CURVED surface of hemisphere (the flat base of cone and flat face of hemisphere are joined — hidden). TSA = πrl + 2πr² = π(5)(13) + 2π(25) = 65π + 50π = 115π = 115 × 22/7 ≈ 361.4 cm². VOLUME = V_cone + V_hemisphere = (1/3)πr²h + (2/3)πr³ = (1/3)π(25)(12) + (2/3)π(125) = 100π + (250/3)π = (300π + 250π)/3 = 550π/3 ≈ 576.2 cm³.

Q3HARD· heights-distances-two-observations

From the top of a cliff 100 m high, the angles of depression of the top and bottom of a tower are 30° and 60° respectively. Find the height of the tower and its distance from the base of the cliff.

▶ Show solution

Let tower height = h, distance from cliff to tower = d. From top of cliff (height 100m): angle of depression to BOTTOM of tower = 60°. So tan60° = 100/d → √3 = 100/d → d = 100/√3 metres. Angle of depression to TOP of tower = 30°. Vertical distance from top of cliff to top of tower = 100 − h. tan30° = (100 − h)/d → 1/√3 = (100 − h) / (100/√3) → 1/√3 × 100/√3 = 100 − h → 100/3 = 100 − h → h = 100 − 100/3 = 200/3 ≈ 66.67 m. Distance d = 100/√3 = 100√3/3 ≈ 57.74 m. Tower height = 200/3 m ≈ 66.7 m.

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