Class 8 Mathematics: Mensuration – Area, Surface Area, and Volume

Mensuration is the branch of mathematics concerned with measuring geometric figures and their parameters, such as perimeter, area, surface area, and volume. In earlier classes, foundational concepts of two-dimensional (2D) shapes—including rectangles, squares, triangles, and circles—were established. At the Class 8 level, this knowledge expands in two distinct directions:

1. Complex 2D Plane Figures: Determining the area of non-standard plane shapes, such as trapeziums, rhombuses, general quadrilaterals, and irregular polygons, by decomposing them into simpler geometric components.
2. Three-Dimensional (3D) Solid Figures: Transitioning from flat shapes to spatial objects, specifically evaluating the total surface area, lateral (or curved) surface area, and volume (or capacity) of cubes, cuboids, and cylinders.

Mastering these principles is fundamental across many fields. Architects, civil engineers, packaging designers, and urban planners rely daily on these formulas to estimate raw material quantities, compute storage capacities, and optimize manufacturing processes.

In-Depth Conceptual Breakdown

Part 1: Area of Plane Figures (2D Geometry)

1. Area of a Trapezium

A trapezium is a quadrilateral with at least one pair of parallel sides.

• Let the parallel sides be of lengths $a$ and $b$.
• Let the perpendicular distance between these parallel sides (height) be $h$.

Derivation of the Formula

Consider a trapezium $ABCD$ with parallel sides $AB = a$ and $CD = b$. Draw a diagonal $AC$, dividing the trapezium into two triangles: $\Delta ABC$ and $\Delta ADC$.

$\text{Area of Trapezium } ABCD = \text{Area}(\Delta ABC) + \text{Area}(\Delta ADC)$

• For $\Delta ABC$: $\text{Base} = a$, $\text{Height} = h \implies \text{Area} = \frac{1}{2} \times a \times h$
• For $\Delta ADC$: $\text{Base} = b$, $\text{Height} = h \implies \text{Area} = \frac{1}{2} \times b \times h$

Adding the two areas together:

$\text{Area} = \frac{1}{2} ah + \frac{1}{2} bh = \frac{1}{2} h (a + b)$

$\text{Area of a Trapezium} = \frac{1}{2} \times (\text{Sum of parallel sides}) \times (\text{Distance between them})$

$\text{Area} = \frac{1}{2}(a + b)h$

2. Area of a General Quadrilateral

A general quadrilateral with no special properties (such as parallel sides or equal angles) can be evaluated by dividing it into two triangles along one of its diagonals.

Given a quadrilateral $ABCD$:

1. Draw diagonal $AC$ of length $d$.
2. Draw perpendicular offsets $h_1$ from vertex $B$ to $AC$, and $h_2$ from vertex $D$ to $AC$.

$\text{Area of Quadrilateral } ABCD = \text{Area}(\Delta ABC) + \text{Area}(\Delta ADC)$

$\text{Area} = \left(\frac{1}{2} \times d \times h_1\right) + \left(\frac{1}{2} \times d \times h_2\right)$

$\text{Area of a General Quadrilateral} = \frac{1}{2} d (h_1 + h_2)$

3. Area of Special Quadrilaterals: Rhombus

A rhombus is a quadrilateral in which all four sides are equal, and its diagonals bisect each other at right angles ($90^\circ$).

Let the lengths of the two diagonals be $d_1$ and $d_2$.

$\text{Area of a Rhombus} = \frac{1}{2} \times d_1 \times d_2$

Derivation

A rhombus $ABCD$ can be split into two triangles by diagonal $AC$ ($d_1$). The diagonal $BD$ ($d_2$) is perpendicular to $AC$ and bisected into two segments of length $\frac{d_2}{2}$.

$\text{Area} = \text{Area}(\Delta ABC) + \text{Area}(\Delta ADC)$

$\text{Area} = \left(\frac{1}{2} \times d_1 \times \frac{d_2}{2}\right) + \left(\frac{1}{2} \times d_1 \times \frac{d_2}{2}\right) = \frac{1}{4} d_1 d_2 + \frac{1}{4} d_1 d_2 = \frac{1}{2} d_1 d_2$

4. Area of General Polygons

To find the area of an irregular polygon (such as a pentagon or hexagon), split the polygon into non-overlapping standard geometric shapes (triangles, rectangles, trapeziums). This process is known as triangulation or decomposition.

General Algorithm:

1. Identify a main diagonal passing through the polygon.
2. Drop perpendicular segments (offsets) from the remaining vertices onto this diagonal.
3. Calculate the area of each resulting sub-figure (triangles and trapeziums).
4. Sum the areas of all sub-figures to find the total polygon area.

Part 2: Surface Area of Solid Figures (3D Geometry)

Three-dimensional objects occupy space and possess three dimensions: length ($l$), breadth ($b$), and height ($h$).

• Total Surface Area (TSA): The total area covered by all outer faces of the 3D figure.
• Lateral Surface Area (LSA) or Curved Surface Area (CSA): The area of the boundary surfaces, excluding the top and bottom bases.

1. Cuboid

A cuboid is a 3D box bounded by six rectangular faces.

• Dimensions: Length = $l$, Breadth = $b$, Height = $h$
• It consists of three pairs of identical rectangular faces: $(l \times b)$, $(b \times h)$, and $(h \times l)$.

$\text{Total Surface Area (TSA)} = 2(lb + bh + hl)$

$\text{Lateral Surface Area (LSA)} = \text{Area of 4 walls} = 2h(l + b)$

2. Cube

A cube is a special case of a cuboid where all three dimensions are equal ($l = b = h = a$).

• Edge length = $a$
• It consists of 6 identical square faces, each of area $a^2$.

$\text{Total Surface Area (TSA)} = 6a^2$

$\text{Lateral Surface Area (LSA)} = 4a^2$

3. Right Circular Cylinder

A right circular cylinder consists of two parallel, congruent circular bases joined by a curved lateral surface.

• Base radius = $r$
• Height = $h$

If a cylindrical shell is unrolled, it forms a rectangle of length equal to the circumference of the base ($2\pi r$) and breadth equal to the height ($h$).

$\text{Curved Surface Area (CSA)} = 2\pi rh$

$\text{Total Surface Area (TSA)} = \text{CSA} + 2 \times (\text{Area of circular base})$

$\text{TSA} = 2\pi rh + 2\pi r^2 = 2\pi r(r + h)$

Part 3: Volume and Capacity

• Volume: The total amount of three-dimensional space enclosed by a closed surface. Measured in cubic units ($\text{cm}^3$, $\text{m}^3$).
• Capacity: The maximum volume of fluid (liquid or gas) that a container can hold. Measured in liters ($\text{L}$), milliliters ($\text{mL}$), or kiloliters ($\text{kL}$).

Fundamental Unit Conversions:

• $1 \text{ cm}^3 = 1 \text{ mL}$
• $1000 \text{ cm}^3 = 1 \text{ Liter}$
• $1 \text{ m}^3 = 1,000,000 \text{ cm}^3 = 1000 \text{ Liters} = 1 \text{ kL}$

General Formula for Volume of Uniform Solids:

$\text{Volume} = \text{Base Area} \times \text{Height}$

Volume Formulas for Cuboid, Cube, and Cylinder:

1. Cuboid: $\text{Base Area} = l \times b$ $\text{Volume} = l \times b \times h$

2. Cube: $\text{Base Area} = a \times a = a^2$ $\text{Volume} = a^3$

3. Right Circular Cylinder: $\text{Base Area} = \pi r^2$ $\text{Volume} = \pi r^2 h$

Formula Summary Tables

2D Geometric Shapes

3D Geometric Solids

Real-World Applications

1. Construction and Flooring Calculations: When laying tiles on an irregularly shaped field or room (often composite shapes made of trapeziums and rectangles), builders compute the area to estimate the quantity of tiles, concrete, or carpet required, minimizing material waste and cost.

2. Packaging and Material Cost Optimization: Manufacturing companies design tin cans (cylindrical) or cardboard boxes (cuboidal) using surface area formulas. By calculating the minimal total surface area needed for a given target volume, companies reduce packaging material expenditures.

3. Water Storage and Liquid Logistics: Civil engineers calculate the volume of cylindrical overhead water tanks or cuboidal reservoirs to determine their total storage capacity in liters ($1 \text{ m}^3 = 1000 \text{ L}$), ensuring adequate daily water supply for residential areas.

Step-by-Step Solved Textbook Examples

Example 1: Area of a Trapezium-Shaped Table Top

Question: The top surface of a dining table is in the shape of a trapezium. Find its area if its parallel sides are $1.6\text{ m}$ and $1.2\text{ m}$, and the perpendicular distance between them is $0.8\text{ m}$.

Solution:

• Step 1: Identify given parameters.

  • First parallel side, $a = 1.6\text{ m}$
  • Second parallel side, $b = 1.2\text{ m}$
  • Perpendicular distance (height), $h = 0.8\text{ m}$

• Step 2: State the relevant formula. $\text{Area of a Trapezium} = \frac{1}{2} (a + b) h$

• Step 3: Substitute the values. $\text{Area} = \frac{1}{2} \times (1.6 + 1.2) \times 0.8$ $\text{Area} = \frac{1}{2} \times 2.8 \times 0.8$ $\text{Area} = 1.4 \times 0.8 = 1.12\text{ m}^2$

Final Answer: The surface area of the table top is $1.12\text{ m}^2$.

Example 2: Area of a Polygonal Field

Question: Find the area of polygon $ABCDE$ shown below. $AD$ is the main diagonal of length $80\text{ m}$. Perpendicular offsets drawn onto $AD$ are:

• $BF \perp AD$ at $F$, where $BF = 30\text{ m}$, $AF = 20\text{ m}$
• $CH \perp AD$ at $H$, where $CH = 40\text{ m}$, $AH = 50\text{ m}$
• $EG \perp AD$ at $G$, where $EG = 25\text{ m}$, $AG = 30\text{ m}$

       B
      /| 
     / |  
    /  |   
   A---F---G---H---D
    \     /    |
     \   /     |
      \ /      |
       E       C

Solution:

• Step 1: Segment the main diagonal $AD$.

  • Total $AD = 80\text{ m}$
  • $AF = 20\text{ m}$
  • $AG = 30\text{ m} \implies FG = AG - AF = 30 - 20 = 10\text{ m}$
  • $AH = 50\text{ m} \implies FH = AH - AF = 50 - 20 = 30\text{ m}$
  • $GD = AD - AG = 80 - 30 = 50\text{ m}$
  • $HD = AD - AH = 80 - 50 = 30\text{ m}$

• Step 2: Divide the polygon into basic shapes and calculate individual areas.

  1. $\Delta ABF$: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AF \times BF = \frac{1}{2} \times 20 \times 30 = 300\text{ m}^2$

  2. Trapezium $BFHC$ (with parallel sides $BF$ and $CH$, height $FH$): $\text{Area} = \frac{1}{2} \times (BF + CH) \times FH = \frac{1}{2} \times (30 + 40) \times 30 = \frac{1}{2} \times 70 \times 30 = 1050\text{ m}^2$

  3. $\Delta CHD$: $\text{Area} = \frac{1}{2} \times HD \times CH = \frac{1}{2} \times 30 \times 40 = 600\text{ m}^2$

  4. $\Delta AGE$: $\text{Area} = \frac{1}{2} \times AG \times EG = \frac{1}{2} \times 30 \times 25 = 375\text{ m}^2$

  5. $\Delta EGD$: $\text{Area} = \frac{1}{2} \times GD \times EG = \frac{1}{2} \times 50 \times 25 = 625\text{ m}^2$

• Step 3: Sum all component areas. $\text{Total Area} = \text{Area}(\Delta ABF) + \text{Area}(\text{Trapezium } BFHC) + \text{Area}(\Delta CHD) + \text{Area}(\Delta AGE) + \text{Area}(\Delta EGD)$ $\text{Total Area} = 300 + 1050 + 600 + 375 + 625 = 2950\text{ m}^2$

Final Answer: The total area of polygon $ABCDE$ is $2950\text{ m}^2$.

Example 3: Surface Area and Painting Cost of a Cylinder

Question: An open cylindrical water tank (open at the top) has a base radius of $7\text{ m}$ and a height of $3\text{ m}$. Calculate:

1. The area of metal sheet required to make the tank.
2. The total cost of painting its outer curved surface at a rate of ₹$25$ per $\text{m}^2$. (Use $\pi = \frac{22}{7}$)

Solution:

• Step 1: Calculate the area of metal sheet required. Since the tank is open at the top, it consists of the Curved Surface Area plus the area of ONE circular base. $\text{Area of Sheet} = \text{CSA} + \text{Base Area} = 2\pi rh + \pi r^2 = \pi r(2h + r)$

  Substitute values ($r = 7\text{ m}, h = 3\text{ m}$): $\text{Area of Sheet} = \frac{22}{7} \times 7 \times (2(3) + 7)$ $\text{Area of Sheet} = 22 \times (6 + 7) = 22 \times 13 = 286\text{ m}^2$

• Step 2: Calculate the cost of painting the outer curved surface. $\text{CSA} = 2\pi rh = 2 \times \frac{22}{7} \times 7 \times 3 = 2 \times 22 \times 3 = 132\text{ m}^2$ $\text{Cost of painting} = \text{CSA} \times \text{Rate per m}^2$ $\text{Cost} = 132 \times 25 = \text{₹ } 3,300$

Final Answer:

1. The metal sheet required is $286\text{ m}^2$.
2. The cost of painting the curved surface is ₹$3,300$.

Example 4: Volume and Capacity of a Cuboidal Tank

Question: A cuboidal water tank is $6\text{ m}$ long, $5\text{ m}$ wide, and $4.5\text{ m}$ deep. How many liters of water can it hold? ($1\text{ m}^3 = 1000\text{ Liters}$)

Solution:

• Step 1: Calculate the volume of the cuboidal tank.

  • Length $l = 6\text{ m}$
  • Breadth $b = 5\text{ m}$
  • Height (depth) $h = 4.5\text{ m}$

  $\text{Volume} = l \times b \times h$ $\text{Volume} = 6 \times 5 \times 4.5 = 30 \times 4.5 = 135\text{ m}^3$

• Step 2: Convert volume into liquid capacity in Liters. Since $1\text{ m}^3 = 1000\text{ Liters}$: $\text{Capacity in Liters} = 135 \times 1000 = 135,000\text{ Liters}$

Final Answer: The tank can hold $135,000\text{ Liters}$ of water.

Common Student Mistakes to Avoid

1. Inconsistent Measurement Units

• The Error: Using mixed units directly in formulas (e.g., length in meters, breadth in centimeters).
• Correction: Always convert all dimensions to the same unit system (preferably SI units like meters or centimeters) before applying any formula.
• Example: If $l = 2\text{ m}$ and $b = 40\text{ cm}$, convert $b$ to $0.4\text{ m}$ first.

2. Misidentifying Open vs. Closed 3D Objects

• The Error: Automatically applying the Total Surface Area (TSA) formula to hollow containers, rooms without ceilings, or tanks open at the top.
• Correction:
  • For a room (4 walls): Use $\text{LSA} = 2h(l + b)$.
  • For a box/tank open at the top: Use $\text{LSA} + \text{Base Area} = 2h(l + b) + lb$ (for cuboid) or $2\pi rh + \pi r^2$ (for cylinder).

3. Confusing Radius and Diameter

• The Error: Substituting the given diameter directly as $r$ in cylinder formulas $\pi r^2 h$ or $2\pi rh$.
• Correction: Always verify whether the question gives diameter ($d$) or radius ($r$). If diameter is given, compute $r = \frac{d}{2}$ before substituting.

4. Incorrect Conversion of Cubic Units to Capacity

• The Error: Assuming $1\text{ cm}^3 = 1\text{ Liter}$ or $1\text{ m}^3 = 1\text{ Liter}$.
• Correction: Memorize the precise conversions:
  • $1000\text{ cm}^3 = 1\text{ Liter}$
  • $1\text{ m}^3 = 1000\text{ Liters}$

Practice Questions for Self-Assessment

Question 1

The diagonals of a rhombus are $16\text{ cm}$ and $12\text{ cm}$. Calculate:

1. Its area.
2. The length of its side.
3. Its perimeter.

Solution:

• Given: $d_1 = 16\text{ cm}$, $d_2 = 12\text{ cm}$.

• Part 1 (Area): $\text{Area} = \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 16 \times 12 = 96\text{ cm}^2$

• Part 2 (Side): The diagonals of a rhombus bisect each other at right angles ($90^\circ$). The half-diagonals form a right triangle with the side of the rhombus ($a$) as the hypotenuse. $\text{Half-diagonal } 1 = \frac{16}{2} = 8\text{ cm}$ $\text{Half-diagonal } 2 = \frac{12}{2} = 6\text{ cm}$ By Pythagoras Theorem: $a^2 = 8^2 + 6^2 = 64 + 36 = 100$ $a = \sqrt{100} = 10\text{ cm}$

• Part 3 (Perimeter): $\text{Perimeter} = 4 \times a = 4 \times 10 = 40\text{ cm}$

Question 2

The area of a trapezium is $34\text{ cm}^2$ and the length of one of the parallel sides is $10\text{ cm}$. If its height is $4\text{ cm}$, find the length of the other parallel side.

Solution:

• Given: $\text{Area} = 34\text{ cm}^2$, $a = 10\text{ cm}$, $h = 4\text{ cm}$.
• Let the unknown parallel side be $b$.
• Formula: $\text{Area} = \frac{1}{2} (a + b) h$ $34 = \frac{1}{2} \times (10 + b) \times 4$ $34 = (10 + b) \times 2$ $10 + b = \frac{34}{2} = 17$ $b = 17 - 10 = 7\text{ cm}$

Answer: The length of the other parallel side is $7\text{ cm}$.

Question 3

A closed cylindrical storage tank of radius $1.4\text{ m}$ and height $5\text{ m}$ is constructed using steel sheets.

1. How much steel sheet was used in total?
2. What is the capacity of the tank in liters? (Take $\pi = \frac{22}{7}$)

Solution:

• Given: $r = 1.4\text{ m}$, $h = 5\text{ m}$.

• Part 1 (Total Steel Sheet Required = TSA): $\text{TSA} = 2\pi r(r + h)$ $\text{TSA} = 2 \times \frac{22}{7} \times 1.4 \times (1.4 + 5)$ $\text{TSA} = 2 \times 22 \times 0.2 \times 6.4$ $\text{TSA} = 8.8 \times 6.4 = 56.32\text{ m}^2$

• Part 2 (Capacity in Liters): $\text{Volume} = \pi r^2 h$ $\text{Volume} = \frac{22}{7} \times 1.4 \times 1.4 \times 5$ $\text{Volume} = 22 \times 0.2 \times 1.4 \times 5 = 4.4 \times 7 = 30.8\text{ m}^3$ $\text{Capacity in Liters} = 30.8 \times 1000 = 30,800\text{ Liters}$

Answer:

1. Steel sheet used = $56.32\text{ m}^2$
2. Capacity = $30,800\text{ Liters}$

Question 4

If each edge of a solid cube is doubled (multiplied by 2):

1. How many times will its total surface area increase?
2. How many times will its volume increase?

Solution:

• Let original edge length of cube = $a$.

• Original Surface Area $S_1 = 6a^2$

• Original Volume $V_1 = a^3$

• New edge length $a' = 2a$.

• Part 1 (New Surface Area $S_2$): $S_2 = 6(a')^2 = 6(2a)^2 = 6 \times 4a^2 = 4 \times (6a^2) = 4 S_1$ The surface area increases $4$ times (or by $300\%$).

• Part 2 (New Volume $V_2$): $V_2 = (a')^3 = (2a)^3 = 8a^3 = 8 V_1$ The volume increases $8$ times (or by $700\%$).

Exam Revision & FAQs

Q1: What is the difference between Lateral Surface Area and Total Surface Area?

Answer:

• Lateral Surface Area (LSA) (or Curved Surface Area for curved solids) includes only the area of the vertical or surrounding faces of a 3D solid, ignoring top and bottom bases.
• Total Surface Area (TSA) includes the areas of all outer surfaces of the 3D solid, including top and bottom bases.

Q2: How do you find the area of an irregular polygon using triangulation?

Answer:

1. Choose a primary diagonal line through the polygon.
2. Draw perpendicular offsets from all remaining vertices onto this diagonal line.
3. Compute the individual areas of all resulting triangles and trapeziums formed between offsets along the main diagonal line.
4. Add all individual areas together to obtain the total area of the polygon.

Q3: A cubical box has an edge length of $10\text{ cm}$. Find the cost of painting its outer surface at ₹$0.50$ per $\text{cm}^2$.

Answer:

• Edge $a = 10\text{ cm}$.
• Total Surface Area $\text{TSA} = 6a^2 = 6 \times (10)^2 = 600\text{ cm}^2$.
• $\text{Cost} = \text{TSA} \times \text{Rate} = 600 \times 0.50 = \text{₹ } 300$.

Q4: Two cylinders have equal volumes. The ratio of their radii is $1 : 2$. What is the ratio of their heights?

Answer:

• Let radii be $r_1$ and $r_2$, such that $\frac{r_1}{r_2} = \frac{1}{2}$.
• Let heights be $h_1$ and $h_2$.
• Since $V_1 = V_2$: $\pi (r_1)^2 h_1 = \pi (r_2)^2 h_2$ $\frac{h_1}{h_2} = \left(\frac{r_2}{r_1}\right)^2$ $\frac{h_1}{h_2} = \left(\frac{2}{1}\right)^2 = \frac{4}{1}$
• The ratio of their heights is $4 : 1$.