Unit 3: Mensuration, Calendar and Clocks

I. Foundations — Measurement of Geometrical Figures

Mensuration applies geometrical formulae to calculate lengths, areas, surface areas, and volumes; calendar and clock methods similarly convert recurring cycles into numerical relationships.

Defining properties:

• Dimensions: Length is one-dimensional, area is two-dimensional, and volume is three-dimensional.
• Units: Length uses units such as cm; area uses square units such as cm²; volume uses cubic units such as cm³.
• Surface classification: A solid may have lateral or curved surface area and total surface area.
• Calendar cycle: Days repeat in a seven-day cycle, so calculations use remainders modulo 7.
• Clock cycle: A clock represents 12 hours by 360°, enabling time–angle conversions.

A. Mensuration

Mensuration determines the measurements of plane figures and three-dimensional solids.

• Perimeter: The total length around a plane figure; for a rectangle, it is 2(l + b).
• Area: The region enclosed by a plane figure; for a circle, it is πr².
• Lateral or curved surface area: The area of side surfaces only, excluding bases.
• Total surface area: The sum of all exposed surfaces, including bases.
• Volume: The space occupied by a solid.
• Capacity: The quantity a container can hold; 1 cm³ = 1 mL and 1000 cm³ = 1 L.
• Conventions:
  • Use consistent units before substitution.
  • Take π ≈ 22/7 or 3.1416 as directed.
  • Square the scale factor for area and cube it for volume.

II. Cube — Solid with Six Equal Square Faces

A. Surface area and volume of cube

A cube has six congruent square faces, twelve equal edges, and eight vertices.

LSA = 4a²
TSA = 6a²
V = a³
d = a√3

Here, a is the edge, LSA is lateral surface area, TSA is total surface area, V is volume, and d is the space diagonal.

• Lateral area: Four side faces each have area a², producing 4a².
• Total area: All six faces produce 6a².
• Volume: Multiplying length, breadth, and height gives a × a × a = a³.
• Example: For a = 5 cm, TSA = 150 cm² and V = 125 cm³.

III. Cuboid — Solid with Rectangular Faces

A. Surface area and volume of cuboid

A cuboid has length, breadth, and height, with opposite rectangular faces equal and parallel.

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

Here, l is length, b is breadth, h is height, and d is the space diagonal.

• Lateral area: The four walls consist of two lh faces and two bh faces.
• Total area: The three pairs of faces have areas lb, bh, and hl.
• Volume: Base area lb multiplied by perpendicular height h gives lbh.
• Example: For dimensions 6 cm × 4 cm × 3 cm, TSA = 108 cm² and V = 72 cm³.

IV. Sphere — Solid Whose Surface Is Equidistant from Its Centre

A. Surface area and volume of sphere

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

Surface area = 4πr²
Volume = (4/3)πr³

Here, r is the radius and π is the circle constant.

• Surface: A sphere has one continuous curved surface and no edge, vertex, or base.
• Diameter: d = 2r; therefore, formulas given in diameter require substituting r = d/2.
• Great circle: A plane through the centre forms the largest circular cross-section, of area πr².
• Example: For r = 3 cm, surface area is 36π cm² and volume is 36π cm³.

V. Hemisphere — Half of a Sphere

A. Surface area and volume of hemisphere

A hemisphere is formed by dividing a sphere into two equal parts through its centre.

CSA = 2πr²
TSA = 3πr²
V = (2/3)πr³

Here, r is radius and CSA is curved surface area.

• Curved area: Half the surface area of a sphere is 2πr².
• Total area: Adding the circular base πr² gives 3πr².
• Volume: Half the sphere’s volume is (2/3)πr³.
• Distinction: An open hemispherical bowl uses CSA; a closed solid hemisphere uses TSA.

VI. Cone — Solid with a Circular Base and One Vertex

A. Surface area and volume of cone

A right circular cone narrows from a circular base to a vertex directly above the base’s centre.

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

Here, r is base radius, h is perpendicular height, and s is slant height.

• Slant height: The radius, height, and slant height form a right triangle.
• Curved area: The curved surface unfolds into a circular sector with area πrs.
• Total area: Adding base area πr² produces πr(s + r).
• Volume relation: A cone has one-third the volume of a cylinder with the same base and height.
• Example: For r = 3 cm and h = 4 cm, s = 5 cm and V = 12π cm³.

VII. Cylinder — Solid with Two Parallel Circular Bases

A. Surface area and volume of cylinder

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

CSA = 2πrh
TSA = 2πr(h + r)
V = πr²h

Here, r is base radius and h is perpendicular height.

• Curved area: Unrolling the curved surface forms a rectangle of length 2πr and breadth h.
• Total area: Adding two circular bases, with combined area 2πr², gives 2πr(h + r).
• Volume: Base area πr² multiplied by height gives πr²h.
• Example: For r = 2 cm and h = 7 cm, CSA = 28π cm² and V = 28π cm³.

VIII. Calendar — Recurring Organization of Days

A. Calendar

A calendar arranges days into weeks, months, and years according to a repeating seven-day cycle.

• Week: Seven days run from Monday through Sunday and then repeat.
• Ordinary year: Contains 365 days, or 52 weeks + 1 day.
• Leap year: Contains 366 days, or 52 weeks + 2 days.
• Odd days: Days remaining after complete weeks; an ordinary year contributes one odd day and a leap year contributes two.
• Cycle: The Gregorian calendar repeats over 400 years because 400 years contain exactly 20,871 weeks.

B. Basic concept of calendar

Calendar calculations depend on month lengths, leap-year rules, and remainders after division by seven.

• Month lengths:
  • 31 days: January, March, May, July, August, October, December.
  • 30 days: April, June, September, November.
  • February has 28 days ordinarily and 29 in a leap year.
• Leap-year rule: A year divisible by 4 is normally a leap year; a century year must also be divisible by 400.
• Examples: 2024 and 2000 are leap years, whereas 1900 and 2100 are not.
• Day shift: A date advances by one weekday after an ordinary year and two after a leap year, provided February 29 falls within the interval.

C. Dates and days

The weekday difference between two dates equals the total elapsed days reduced modulo seven.

Weekday shift = Total elapsed days mod 7

• Inclusive counting: To find elapsed days from the first date to the second, exclude the starting date and count up to the target date.
• Same month: Subtract date numbers; from May 8 to May 20, 20 − 8 = 12 days.
• Successive months: Add the remaining days of the first month, complete intervening months, and elapsed days in the final month.
• Equivalent dates: Two dates fall on the same weekday when their elapsed-day difference is divisible by 7.

D. Finding the exact day

An exact weekday can be found by counting odd days from a known reference date.

Total shift =
(year contributions + completed-month days + elapsed date days) mod 7

• Reference: January 1, 1900 was Monday.
• Year contribution: Count one odd day for each ordinary year and two for each leap year before the target year.
• Month contribution: Add days in completed months of the target year.
• Date contribution: For day D, add D − 1 because no day has elapsed at the start of day 1.
• Example: Before August 15, 1947, years 1900–1946 contribute 47 + 11 = 58 odd-day units; 226 days elapsed in 1947. Since (58 + 226) mod 7 = 4, Monday advances four places to Friday.

IX. Clocks — Angular Measurement of Time

A. Clocks

A clock converts the passage of time into the circular motion of hour, minute, and second hands.

• Dial division: The 360° dial has 12 hour spaces of 30° each and 60 minute spaces of 6° each.
• Continuous movement: The hour hand moves gradually between hour marks rather than jumping once per hour.
• Relative motion: Angle problems compare the hands’ positions or their relative angular speed.
• Smaller angle: If the direct difference exceeds 180°, subtract it from 360°.

B. Concept of clock

Clock-hand positions are determined by multiplying elapsed time by each hand’s angular speed.

Minute hand speed = 6° per minute
Hour hand speed = 0.5° per minute
Second hand speed = 6° per second

• Hour-hand position: At H:M, it is 30H + 0.5M degrees from 12.
• Minute-hand position: At H:M, it is 6M degrees from 12.
• Example: At 3:20, the hour hand is at 100° and the minute hand at 120°; the smaller angle is 20°.

C. Facts and formulae

Standard clock results follow from the hands’ relative speed of 6 − 0.5 = 5.5° per minute.

θ = |30H − 5.5M|
Smaller angle = min(θ, 360° − θ)
Coincidence after H o’clock: M = 60H/11

Here, H is the completed hour from 0 to 11, M is minutes after that hour, and θ is the direct angular difference.

• Coincidence: The hands overlap 11 times in 12 hours, approximately every 65 5/11 minutes.
• Opposition: They form 180° eleven times in 12 hours.
• Right angles: They form 90° or 270° twenty-two times in 12 hours.
• Faulty clock: A clock gaining g minutes daily runs at (1440 + g)/1440 of correct elapsed time; replace +g with −g for a losing clock.

D. Practice problems

Practice is most effective when each problem is reduced to the governing measurement or cycle before calculation.

• Mensuration method: Identify the solid, mark dimensions, distinguish curved from total area, standardize units, and then substitute into the appropriate formula.
• Calendar method: Select a known weekday, count leap and ordinary years, add completed-month and date contributions, and reduce the result modulo 7.
• Clock method: Convert each hand’s position into degrees, take the absolute difference, and choose the smaller angle when required.
• Verification: Check dimensional units—square units for surface area, cubic units for volume, weekdays for calendar remainders, and degrees for clock angles.