Unit 3: Mensuration, Calendar, and Clocks

I. Foundations and Conventions

Mensuration measures the boundaries and occupied space of three-dimensional solids; calendar reasoning identifies weekdays through repeating day cycles; clock reasoning studies the relative angular motion of clock hands.

• Mensuration conventions:
  • Surface area is measured in square units, such as cm² or m².
  • Volume is measured in cubic units, such as cm³ or m³.
  • Curved surface area excludes flat bases; total surface area includes every exposed surface.
  • All dimensions must be converted to the same unit before calculation.
• Calendar conventions:
  • One ordinary year has 365 days; one leap year has 366 days.
  • A week contains 7 days, so weekday calculations use remainders modulo 7.
  • Under the Gregorian calendar, leap years are divisible by 4, except century years, which must be divisible by 400.
• Clock conventions:
  • A standard clock face contains 12 hour divisions and 60 minute divisions.
  • Angles are normally measured clockwise from 12.
  • Unless stated otherwise, clock-hand motion is treated as continuous.

II. Mensuration — Surface Area and Volume of Solids

A. Surface area and volume of cube

A cube is a solid with six congruent square faces and twelve equal edges.

• Dimensions: Let a be the length of each edge.
• Areas: Four side faces form the lateral surface, while all six faces form the total surface.
• Volume: The occupied space equals length × breadth × height, with all three dimensions equal.

LSA = 4a²
TSA = 6a²
V = a³

Here, LSA is lateral surface area, TSA is total surface area, and V is volume.

• Example: For a = 5 cm, TSA = 6(5²) = 150 cm² and V = 5³ = 125 cm³.
• Useful relation: If the volume is known, the edge is a = ∛V.

B. Surface area and volume of cuboid

A cuboid is a rectangular solid whose opposite faces are congruent rectangles.

• Dimensions: Let l, b, and h denote length, breadth, and height.
• Lateral area: The four vertical faces have combined area 2lh + 2bh.
• Total area: The three distinct face areas, lb, bh, and hl, each occur twice.

LSA = 2h(l + b)
TSA = 2(lb + bh + hl)
V = lbh
Diagonal = √(l² + b² + h²)

Here, LSA, TSA, and V have their usual meanings, and the diagonal joins opposite vertices.

• Example: For l = 8 cm, b = 5 cm, and h = 3 cm, TSA = 2(40 + 15 + 24) = 158 cm², while V = 8 × 5 × 3 = 120 cm³.
• Special case: A cube is a cuboid for which l = b = h.

C. Surface area and volume of sphere

A sphere consists of all points in space at a fixed distance from its centre.

• Dimension: Let r be the radius and d = 2r the diameter.
• Surface: A sphere has one continuous curved surface and no flat base.
• Measurement: Its curved surface area and total surface area are therefore identical.

Surface area = 4πr²
Volume = (4/3)πr³

Here, π ≈ 3.1416 or may be taken as 22/7 when appropriate.

• Example: For r = 3 cm, surface area is 4π(3²) = 36π cm², and volume is (4/3)π(3³) = 36π cm³.
• Scaling: If the radius is doubled, surface area becomes four times and volume becomes eight times the original.

D. Surface area and volume of hemisphere

A hemisphere is one-half of a sphere cut through its centre.

• Curved surface area: Half the surface area of a sphere gives 2πr².
• Base area: The circular base contributes πr².
• Total surface area: Adding the curved surface and base gives 3πr².

CSA = 2πr²
TSA = 3πr²
V = (2/3)πr³

Here, CSA is curved surface area, TSA includes the circular base, and r is the radius.

• Example: For r = 6 cm, CSA = 72π cm², TSA = 108π cm², and V = 144π cm³.
• Interpretation: An open hemispherical bowl uses CSA; a closed solid hemisphere uses TSA.

E. Surface area and volume of cone

A right circular cone has a circular base and a curved surface tapering to a vertex directly above the base centre.

• Dimensions: Let r be base radius, h vertical height, and s slant height.
• Slant relation: The radius, height, and slant height form a right triangle.
• Area distinction: Total surface area includes both the curved surface and circular base.

s = √(r² + h²)
CSA = πrs
TSA = πr(s + r)
V = (1/3)πr²h

Here, CSA is curved surface area and V is volume.

• Example: If r = 3 cm and h = 4 cm, then s = 5 cm, TSA = π(3)(5 + 3) = 24π cm², and V = 12π cm³.
• Comparison: A cone has one-third the volume of a cylinder with the same base radius and perpendicular height.

F. Surface area and volume of cylinder

A right circular cylinder has two congruent circular bases connected by a curved surface.

• Dimensions: Let r be the base radius and h the perpendicular height.
• Curved surface: Unrolling it produces a rectangle of length 2πr and height h.
• Total surface: Two circular bases add 2πr² to the curved area.

CSA = 2πrh
TSA = 2πr(h + r)
V = πr²h

Here, CSA excludes the bases, whereas TSA includes both bases.

• Example: For r = 7 cm and h = 10 cm, using π = 22/7, CSA = 440 cm², TSA = 748 cm², and V = 1540 cm³.
• Capacity: The internal volume of a cylindrical vessel represents its capacity after unit conversion; 1000 cm³ = 1 litre.

III. Calendar — Weekday and Date Calculations

A. Basic concept of calendar

A calendar arranges days into weeks, months, and years, with weekday patterns governed by complete weeks and leftover days.

• Ordinary year: 365 = 52 × 7 + 1, so it contributes one odd day.
• Leap year: 366 = 52 × 7 + 2, so it contributes two odd days.
• Month lengths:
  • January, March, May, July, August, October, and December have 31 days.
  • April, June, September, and November have 30 days.
  • February has 28 days ordinarily and 29 in a leap year.
• Leap-year rule: 2024 is a leap year because it is divisible by 4; 1900 is not because it is a century year not divisible by 400; 2000 is a leap year.
• Repetition: Calendar patterns usually recur after 28 years, although Gregorian century rules can interrupt this cycle.

B. Dates and days

The weekday of a later or earlier date is obtained by counting elapsed days and reducing the total modulo seven.

• Odd days: The remainder after division by 7 determines the weekday shift.

Weekday shift = Total elapsed days mod 7

Here, mod 7 means the remainder after division by 7.

• Direction: A positive remainder moves forward through the week; backward counting moves in the reverse direction.
• Same date next year:
  • It advances by one weekday after an ordinary year.
  • It advances by two weekdays when February 29 lies within the interval.
• Example: If 1 January is Monday, then 1 February is Thursday because January contributes 31 mod 7 = 3 odd days.

C. Finding the exact day

An exact weekday can be found by adding the day contributions of completed years, completed months, and the date offset.

• Reference method: Begin from a known date, count all intervening days, and take the remainder modulo 7.
• Date offset: For the dth day of a month, count d - 1 completed days because the first day has offset zero.
• Month adjustment: Include February 29 only when the target year is leap and the target date falls after February.
• Worked determination: To find 15 August 2024, use the known fact that 1 January 2024 was Monday.
  • Completed months contain 31 + 29 + 31 + 30 + 31 + 30 + 31 = 213 days.
  • The date offset is 15 - 1 = 14.
  • Total offset is 227, and 227 mod 7 = 3.
  • Three days after Monday is Thursday; therefore, 15 August 2024 was Thursday.
• Verification principle: Counting must exclude the starting date but include each fully elapsed day.

IV. Clocks — Angular Motion and Time Relations

A. Concept of clock

Clock questions treat the hour and minute hands as objects moving at constant but different angular speeds.

• Clock face: The full circle is 360°, divided into 12 hour spaces of 30° each.
• Minute divisions: Each of the 60 minute marks represents 6°.
• Minute hand: It completes 360° in 60 minutes.
• Hour hand: It completes 360° in 12 hours and moves continuously as minutes pass.
• Relative motion: The minute hand gains 5.5° per minute on the hour hand.
• Smaller angle: If the calculated angle exceeds 180°, subtract it from 360°.

B. Clock facts and formulae

Clock formulae compare the angular positions of the two hands after a stated number of hours and minutes.

• Angular speeds:

Minute-hand speed = 6° per minute
Hour-hand speed = 0.5° per minute
Relative speed = 6° - 0.5° = 5.5° per minute

• Hand positions: At H:M, where H is the hour number and M is minutes after that hour:

Hour-hand angle = 30H + 0.5M
Minute-hand angle = 6M
Angle between hands = |30H - 5.5M|
Smaller angle = min(θ, 360° - θ)

Here, θ = |30H - 5.5M| is the absolute angular difference.

• Coincidence: The hands overlap when 30H = 5.5M, giving M = 60H/11.
• Right angle: The hands are perpendicular when |30H - 5.5M| = 90° or 270°.
• Opposition: The hands are in a straight line when their angular difference is 180°.
• Frequency: In 12 hours, the hands coincide 11 times, form opposite straight lines 11 times, and form right angles 22 times.

C. Clock practice problems

Standard clock problems are solved by translating the stated hand position into an angular equation.

• Angle at a stated time: At 4:20, the hour-hand angle is 30(4) + 0.5(20) = 130°, while the minute-hand angle is 6(20) = 120°; therefore, the angle is 10°.
• Coincidence after an hour: Between 5 and 6, apply the overlap formula:

M = 60H/11 = 60(5)/11 = 300/11 minutes

Thus, the hands coincide at approximately 5:27:16.

• Mirror-image time: For a standard vertical mirror, subtract the shown time from 11:60.

Mirror time = 11:60 - Actual time

For 7:25, the mirror reading is 11:60 - 7:25 = 4:35.

• Fast or slow clock: Compare indicated elapsed time with actual elapsed time. A clock gaining 5 minutes every hour shows 65 minutes for each actual 60, so its indicated-to-actual ratio is 65:60.
• Method check: Use continuous hour-hand movement; treating the hour hand as fixed exactly on an hour numeral produces incorrect angles whenever M ≠ 0.