Unit 3: Mensuration, Calendar and Clocks - Subjective Questions

Mensuration is the branch of mathematics concerned with measuring the dimensions, boundaries, areas, surface areas, and volumes of geometric figures.

• Perimeter: The total length of the boundary of a two-dimensional figure. It is measured in linear units such as or .
• Area: The region occupied by a two-dimensional figure. It is measured in square units such as or .
• Surface area: The total area of the exposed surfaces of a three-dimensional solid. It is also measured in square units.
• Volume: The amount of space occupied by a three-dimensional solid. It is measured in cubic units such as or .

Unit relationships:

A cube has six identical square faces, each having side .

1. Lateral surface area:
The four vertical faces form the lateral surface. The area of each face is .

2. Total surface area:
There are six square faces.

3. Volume:
Volume is the product of length, breadth, and height. All three dimensions are equal to .

Thus, the required formulas are:

• Lateral surface area:
• Total surface area:
• Volume:

Let the side of the cube be .

Step 1: Find the side

Step 2: Find the volume

Step 3: Find the body diagonal
The body diagonal of a cube is .

Therefore:

• Side:
• Volume:
• Body diagonal:

A cuboid has three pairs of congruent rectangular faces.

1. Lateral surface area:
The lateral faces have areas , , , and .

2. Total surface area:
The three pairs of faces have areas , , and .

3. Volume:

4. Body diagonal:
First, the diagonal of the base is

Using the Pythagorean theorem again with height :

Hence:

Given , , and :

Lateral surface area:

Total surface area:

Volume:

Body diagonal:

Therefore, the required values are , , , and , respectively.

Surface area of a sphere:
The surface area of a sphere is four times the area of a great circular cross-section:

Derivation of volume using integration:
Consider the sphere

At a distance from the center, the radius of a circular cross-section is

Its area is

Therefore,

Thus:

• Surface area:
• Volume:

Surface area is measured in square units, whereas volume is measured in cubic units.

Given :

Surface area:

Volume:

Therefore:

• Surface area:
• Volume:

A hemisphere is one-half of a sphere.

Formulas for a hemisphere:

• Curved surface area:
  
• Total surface area, including the circular base:
  
• Volume:
  

Comparison with a sphere:

• A sphere has surface area , so the hemisphere's curved surface area is half of it.
• The ratio of the sphere's surface area to the hemisphere's total surface area is
  
• A sphere has volume , which is twice the volume of a hemisphere.

The circular base must be included only when calculating the total surface area of a hemisphere.

For a right circular cone, the radius, height, and slant height satisfy

1. Curved surface area:
When the curved surface is opened, it forms a sector of radius . Its arc length equals the circumference of the cone's base. Since the area of a sector is half the product of its radius and arc length:

2. Total surface area:
Adding the base area gives

3. Volume:
At distance from the vertex, the cross-sectional radius is . Therefore,

Thus, , , and .

Given and :

Slant height:

Curved surface area:

Total surface area:

Volume:

The required values are , , , and .

Consider a cylinder of radius and height .

When its curved surface is opened, it forms a rectangle with:

• Length equal to the circumference of the base,
• Breadth equal to the height,

Therefore, the curved surface area is

A closed cylinder has two circular bases, each of area . Hence,

Different cases:

• Open at both ends:
• Open at one end:
• Closed at both ends:

The volume is unaffected by whether the ends are physically covered:

Given and :

Curved surface area:

Total surface area of a closed cylinder:

Volume:

Surface area when open at the top:
Only the curved surface and bottom base are included.

Thus, the required values are , , , and .

The two hemispheres together form one complete sphere of radius . The circular joining faces are internal and are not included in the external surface area.

External surface area:

Volume:

Therefore:

• External surface area:
• Volume:

The Gregorian calendar is the commonly used civil calendar.

• An ordinary year contains days.
  
  Therefore, an ordinary year has complete weeks and 1 odd day.

• A leap year contains days.
  
  Therefore, a leap year has complete weeks and 2 odd days.

Leap-year rule:

• A year divisible by is generally a leap year.
• A century year must be divisible by to be a leap year.
• Thus, was a leap year, whereas and are not.

An odd day is a day remaining after complete weeks have been removed. Odd days are calculated modulo and are used to determine the weekday of a given date.

The number of odd days is the total number of days taken modulo .

In years:
There are leap years because the hundredth year is not a leap year unless divisible by .

Thus, years have 5 odd days.

In years:
There are leap years.

Thus, years have 3 odd days.

In years:
There are leap years.

Thus, years have 1 odd day.

In years:
There are

leap years. Hence,

Thus, years have 0 odd days, and the Gregorian calendar repeats its weekday cycle every years.

Step 1: Count completed years
From through , there are completed years.

The leap years are , giving leap years. The year is not a leap year.

Total odd days contributed by these years:

Therefore, January was two days after Monday, namely Wednesday.

Step 2: Count days before August
Days in the completed months are

There are also completed days in August before August.

Two days after Wednesday is Friday.

Therefore, August was a Friday.

Two years have identical calendars when:

• They begin on the same day of the week.
• They are of the same type: both ordinary years or both leap years.

From the beginning of to the beginning of , the ordinary years contribute odd day each, while leap years contribute odd days each.

The leap years in this interval are , , and . Thus, there are ordinary years and leap years.

Therefore, January falls on the same weekday as January .

Both and are ordinary years. Hence, the calendar for repeats in .

Minute hand:
The minute hand completes in minutes. Its speed is

Hour hand:
The hour hand completes in hours or minutes. Its speed is

At hours and minutes:

• Hour-hand angle from is .
• Minute-hand angle from is .

Therefore, their angular difference is

The smaller angle is .

At :

Thus, the smaller angle between the hands at is .

Let be the number of minutes after o'clock. The angular difference is

Coincidence:
For coincidence, the difference is zero.


Thus, the hands coincide at

or approximately .

Right angles:
For a right angle,

First solution:

Second solution:

Therefore, the hands are at right angles at

• , approximately
• , approximately

The faulty clock gains minutes in every minutes of actual time. Therefore, it records minutes while actual minutes pass.

From a.m. to the displayed p.m., the clock records

Let the actual elapsed time be minutes. Then

This equals

Adding this to a.m., the actual time is approximately

Therefore, when the faulty clock displays p.m., the actual time is approximately p.m.