Classwise Concept with Examples
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Class 8th Chapters
1. Rational Numbers	2. Linear Equations in One Variable	3. Understanding Quadrilaterals
4. Practical Geometry	5. Data Handling	6. Squares and Square Roots
7. Cubes and Cube Roots	8. Comparing Quantities	9. Algebraic Expressions and Identities
10. Visualising Solid Shapes	11. Mensuration	12. Exponents and Powers
13. Direct and Inverse Proportions	14. Factorisation	15. Introduction to Graphs
16. Playing with Numbers

Content On This Page
Area and Perimeter of Some Plane Figures	Area of Quadrilaterals and Polygons	Volume of Solids
Surface Area of Solids

Chapter 11 Mensuration (Concepts)

Welcome to this significantly expanded exploration of Mensuration, the branch of geometry dedicated to the measurement of geometric figures. Building upon the foundations laid in Class 7, where we primarily dealt with the perimeter (distance around) and area (space enclosed) of basic two-dimensional (2D) shapes, Class 8 takes a substantial leap forward. We will not only delve into calculating the area of more complex 2D figures but, more importantly, venture into the three-dimensional (3D) world to understand and quantify the surface area and volume of common solid shapes. This chapter equips you with essential tools for measuring the world around us, from flat surfaces to the space occupied by objects.

We begin by briefly revisiting and reinforcing the fundamental formulas for calculating the perimeter and area of familiar plane figures like squares ($P=4s, A=s^2$), rectangles ($P=2(l+b), A=l \times b$), triangles ($A=\frac{1}{2} \times base \times height$), parallelograms ($A=base \times height$), and circles (Circumference $C=2\pi r$, Area $A=\pi r^2$). Building on this, we introduce methods for tackling more intricate 2D shapes. You will learn how to calculate the area of a general quadrilateral by strategically dividing it into two triangles using a diagonal ($d$) and utilizing the lengths of the perpendiculars ($h_1, h_2$) dropped onto this diagonal from the opposite vertices: $Area = \frac{1}{2} \times d \times (h_1 + h_2)$. We also derive and apply the specific formula for the area of a trapezium (a quadrilateral with one pair of parallel sides, say $a$ and $b$, and height $h$): $Area = \frac{1}{2} \times (a+b) \times h$. Techniques for finding the areas of other polygons might also be explored by decomposing them into triangles and quadrilaterals.

The major advancement in this chapter lies in our exploration of three-dimensional solids. We introduce two crucial concepts related to the 'skin' or outer surface of these objects:

Lateral Surface Area (LSA): This refers to the area of only the side faces of the solid, excluding the area of its top and bottom bases (if they exist).
Total Surface Area (TSA): This encompasses the area of all surfaces of the solid, including the bases.

We will derive and apply the formulas for these surface areas for key solids, where $l, b, h, a, r$ represent length, breadth, height, side/edge, and radius, respectively:

Cuboid: $LSA = 2h(l+b)$; $TSA = 2(lb + bh + hl)$
Cube: $LSA = 4a^2$; $TSA = 6a^2$
Right Circular Cylinder: $LSA = 2\pi rh$; $TSA = 2\pi r(r + h)$

Following surface area, we delve into the concept of Volume, which quantifies the amount of 3D space enclosed or occupied by a solid object. Understanding volume is crucial for determining capacity or the amount of material an object contains. We will learn the formulas for calculating the volume of the same set of common solids:

Volume of Cuboid: $V = l \times b \times h$
Volume of Cube: $V = a^3$
Volume of Cylinder: $V = \pi r^2 h$

Throughout this chapter, meticulous attention must be paid to units of measurement. Perimeter is measured in linear units (e.g., cm, m), surface area in square units (e.g., $cm^2, m^2$), and volume in cubic units (e.g., $cm^3, m^3$). We will also frequently encounter practical volume units like litres (L) and need to be proficient in conversions, such as $1 \, L = 1000 \, cm^3$ and $1 \, m^3 = 1000 \, L = 1,000,000 \, cm^3$. The practical relevance of these concepts is highlighted through numerous application problems. These might involve calculating the cost of painting walls (LSA) or entire objects (TSA) often expressed in $\textsf{₹}$ per square unit, determining the capacity of tanks or containers (Volume), estimating the amount of material needed to construct objects, and solving various other real-world measurement challenges.

Area and Perimeter of Some Plane Figures

Mensuration is a branch of geometry that deals with the measurement of lengths, areas, and volumes of different geometrical shapes. In previous classes, you have learned about calculating the boundary (perimeter) and the space enclosed within the boundary (area) of various 2-dimensional (2-D) shapes or plane figures. This section serves as a review of these concepts and their formulas for common shapes.

Perimeter

The perimeter of a simple closed plane figure is the total length of its boundary. Imagine walking along the outline of a shape; the total distance you cover is its perimeter. Perimeter is a one-dimensional measurement.

The units of perimeter are units of length, such as millimetres (mm), centimetres (cm), metres (m), kilometres (km), inches, feet, etc.

Area

The area of a simple closed plane figure is the measure of the region or surface enclosed by its boundary. It quantifies the two-dimensional space that the shape occupies.

The units of area are square units, such as square millimetres (mm$^2$), square centimetres (cm$^2$), square metres (m$^2$), square kilometres (km$^2$), square inches (in$^2$), etc. A square unit represents the area of a square with side length equal to the unit of length (e.g., 1 cm$^2$ is the area of a square with side 1 cm).

Area = $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$

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

These formulas are essential for calculating the area and perimeter of various plane figures.

Example 1. Find the area and perimeter of a rectangle whose length is 8 cm and width is 5 cm.

Answer:

Given:

Length ($l$) = 8 cm.
Width ($w$) = 5 cm.

To Find:

Area and Perimeter of the rectangle.

Solution:

Using the formula for the perimeter of a rectangle:

Perimeter $= 2(l+w)$

Substitute the given values:

Perimeter $= 2(8 \text{ cm} + 5 \text{ cm}) = 2(13 \text{ cm}) = 26 \text{ cm}$

Using the formula for the area of a rectangle:

Area $= l \times w$

Substitute the given values:

Area $= 8 \text{ cm} \times 5 \text{ cm} = (8 \times 5) \text{ cm}^2 = 40 \text{ cm}^2$

The area of the rectangle is $40 \text{ cm}^2$ and the perimeter is $26 \text{ cm}$.

Example 2. The base of a triangle is 10 cm and its height is 6 cm. Find its area.

Answer:

Given:

Base ($b$) = 10 cm.
Height ($h$) = 6 cm.

To Find:

Area of the triangle.

Solution:

Using the formula for the area of a triangle:

Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times b \times h$

Substitute the given values:

Area $= \frac{1}{2} \times 10 \text{ cm} \times 6 \text{ cm}$

Calculate the value:

$= \frac{1}{2} \times (10 \times 6) \text{ cm}^2 = \frac{1}{2} \times 60 \text{ cm}^2$

$= 30 \text{ cm}^2$

The area of the triangle is $30 \text{ cm}^2$.

Area of Quadrilaterals and Polygons

In the previous section, we reviewed the formulas for the area of basic plane figures like triangles, squares, rectangles, parallelograms, and circles. Now, we will look at how to find the area of more general quadrilaterals and polygons, especially when direct formulas are not readily available or applicable.

The key strategy for finding the area of complex polygons is to decompose or divide them into simpler shapes whose areas we know how to calculate (primarily triangles and sometimes trapeziums or rectangles). We then calculate the area of each simpler part and add them together to get the total area of the polygon.

Area of a General Quadrilateral

A general quadrilateral is a four-sided polygon that does not necessarily fit into specific categories like parallelograms, trapeziums, etc. To find its area, we can divide it into two triangles by drawing one of its diagonals.

Consider a quadrilateral ABCD. Draw a diagonal, say AC. This divides the quadrilateral into two triangles: $\triangle$ABC and $\triangle$ADC.

Let the length of the diagonal AC be $d$. From the other two vertices (B and D), draw perpendiculars to the diagonal AC. Let the length of the perpendicular from B to AC be $h_1$, and the length of the perpendicular from D to AC be $h_2$. These perpendiculars are the heights of $\triangle$ABC and $\triangle$ADC respectively, with respect to the common base AC (or parts of it, but the formula generalises).

The area of the quadrilateral ABCD is the sum of the areas of $\triangle$ABC and $\triangle$ADC.

Area of Quadrilateral ABCD = Area of $\triangle$ABC + Area of $\triangle$ADC

Using the formula for the area of a triangle ($\frac{1}{2} \times \text{base} \times \text{height}$):

Area $= \frac{1}{2} \times \text{AC} \times h_1 + \frac{1}{2} \times \text{AC} \times h_2$

Substitute AC = $d$ and factor out $\frac{1}{2}d$:

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

... (i)

So, the area of a quadrilateral is half the product of its diagonal and the sum of the perpendiculars drawn from the opposite vertices to the diagonal.

Example 1. The diagonal of a quadrilateral is 15 cm. The lengths of the perpendiculars drawn from the opposite vertices to this diagonal are 8 cm and 7 cm. Find the area of the quadrilateral.

Answer:

Given:

Length of the diagonal ($d$) = 15 cm.
Lengths of the perpendiculars from opposite vertices to the diagonal ($h_1$ and $h_2$) = 8 cm and 7 cm.

To Find:

Area of the quadrilateral.

Solution:

Using the formula for the area of a general quadrilateral:

Area $= \frac{1}{2} d (h_1 + h_2)$

[Using Formula (i)]

Substitute the given values $d=15$, $h_1=8$, and $h_2=7$:

Area $= \frac{1}{2} \times 15 \text{ cm} \times (8 \text{ cm} + 7 \text{ cm})$

$= \frac{1}{2} \times 15 \text{ cm} \times 15 \text{ cm}$

Calculate the product:

$= \frac{15 \times 15}{2} \text{ cm}^2 = \frac{225}{2} \text{ cm}^2$

Convert the fraction to a decimal:

$= 112.5 \text{ cm}^2$

The area of the quadrilateral is $112.5 \text{ cm}^2$.

Area of a Polygon

Finding the area of a simple polygon like a triangle or a square is straightforward using direct formulas. However, for more complex polygons (like pentagons, hexagons, or irregular shapes), we use the strategy of decomposition. This means we break down the complex polygon into simpler shapes whose areas we know how to calculate, such as triangles, rectangles, and trapeziums.

Method 1: Triangulation from a Single Vertex

A simple way to decompose any convex polygon is to pick one vertex and draw diagonals to all other non-adjacent vertices. This process, known as triangulation, divides the polygon into a series of triangles.

For an $n$-sided polygon, this method will always create $(n-2)$ triangles. The total area of the polygon is then the sum of the areas of all these triangles.

A hexagon is shown. From one vertex, three diagonals are drawn, dividing the hexagon into four triangles.

To calculate the area of each triangle, you would need information like its base and height, or the lengths of its three sides (to use Heron's formula).

Method 2: Decomposition using Perpendiculars (Surveyor's Method)

This is a practical method often used in land surveying. It involves drawing one long diagonal or a baseline and then dropping perpendiculars (heights) from the remaining vertices onto this line. This divides the entire polygon into a combination of triangles and trapeziums.

Example 1. Find the area of the polygon ABCDE shown below, where perpendiculars are drawn from vertices B, C, and E to the diagonal AD.

Answer:

Given:

From the figure, we have the following lengths:

Lengths along the diagonal AD: AF = 3 m, FG = 4 m, GH = 2 m, HD = 3 m.
Lengths of the perpendiculars: BF = 4 m, CG = 6 m, EH = 3 m.

From the segment lengths, we can calculate the length of the base for each shape:

Base of $\triangle$AFB is AF = 3 m.
Height of Trapezium FBCG is FG = 4 m.
Base of $\triangle$CGD is GD = GH + HD = 2 + 3 = 5 m.
Base of $\triangle$EHD is HD = 3 m.

Solution:

To find the total area of the polygon ABCDE, we decompose it into simpler shapes using the given perpendiculars. The polygon is divided into four shapes: one triangle ($\triangle$AFB), one trapezium (FBCG), and two more triangles ($\triangle$CGD and $\triangle$EHD). The total area is the sum of the areas of these four figures.

1. Area of Triangle AFB:

Area($\triangle$AFB) = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \text{AF} \times \text{BF}$

$ = \frac{1}{2} \times 3 \times 4 = 6 \text{ m}^2$.

2. Area of Trapezium FBCG:

Area(FBCG) = $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} = \frac{1}{2} \times (\text{BF} + \text{CG}) \times \text{FG}$

$ = \frac{1}{2} \times (4+6) \times 4 = \frac{1}{2} \times 10 \times 4 = 20 \text{ m}^2$.

3. Area of Triangle CGD:

Area($\triangle$CGD) = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \text{GD} \times \text{CG}$

$ = \frac{1}{2} \times 5 \times 6 = 15 \text{ m}^2$.

4. Area of Triangle EHD:

Area($\triangle$EHD) = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \text{HD} \times \text{EH}$

$ = \frac{1}{2} \times 3 \times 3 = 4.5 \text{ m}^2$.

Total Area of Polygon ABCDE:

The total area is the sum of the areas of the four parts.

Total Area = Area($\triangle$AFB) + Area(FBCG) + Area($\triangle$CGD) + Area($\triangle$EHD)

Total Area = $6 + 20 + 15 + 4.5 = 45.5 \text{ m}^2$.

The area of the polygon ABCDE is 45.5 m².

Method 3: Area of a Regular Polygon

A regular polygon has all sides equal and all interior angles equal. This symmetry allows for a simpler method of finding the area. A regular $n$-sided polygon can be divided into $n$ congruent isosceles triangles, with their common vertex at the center of the polygon.

A regular pentagon divided into 5 congruent isosceles triangles from its center.

The area of the polygon is simply the sum of the areas of these $n$ triangles.

Area of Regular Polygon = $n \times (\text{Area of one isosceles triangle})$.

Example 2. Find the area of a regular hexagon with side length 6 cm.

Answer:

Given: A regular hexagon with side length ($s$) = 6 cm.

To Find: Area of the hexagon.

Solution:

A regular hexagon is a special case because when it is divided into 6 triangles from its center, these triangles are not just isosceles but are equilateral. The side length of each equilateral triangle is equal to the side length of the hexagon.

Regular hexagon divided into 6 equilateral triangles

So, we need to find the area of 6 equilateral triangles, each with a side length of 6 cm.

The formula for the area of an equilateral triangle with side $s$ is $\frac{\sqrt{3}}{4} s^2$.

First, calculate the area of one such triangle:

Area of one triangle $= \frac{\sqrt{3}}{4} \times (6 \text{ cm})^2$

$ = \frac{\sqrt{3}}{4} \times 36 \text{ cm}^2 = 9\sqrt{3} \text{ cm}^2$

Now, multiply this by 6 to get the total area of the hexagon:

Area of Hexagon $= 6 \times (\text{Area of one triangle})$

$ = 6 \times 9\sqrt{3} \text{ cm}^2 = 54\sqrt{3} \text{ cm}^2$

For a numerical approximation, we can use $\sqrt{3} \approx 1.732$:

Area $\approx 54 \times 1.732 \text{ cm}^2 \approx 93.528 \text{ cm}^2$

The area of the regular hexagon is $54\sqrt{3} \text{ cm}^2$ (approximately 93.53 cm²).

Volume of Solids

In the previous sections, we discussed the measurement of 2-dimensional shapes (perimeter and area). Now, we move to the measurement of 3-dimensional (3-D) shapes or solids. A key measurement for 3-D shapes is their volume.

Understanding Volume

The volume of a three-dimensional solid is the amount of space that it occupies. It is a measure of the capacity of the solid if it is hollow, or the amount of substance it contains if it is solid throughout. Volume is a three-dimensional measurement.

The standard unit of volume in the International System of Units (SI) is the cubic metre (m$^3$). Other common units include cubic centimetres (cm$^3$), cubic millimetres (mm$^3$), cubic kilometres (km$^3$), cubic inches (in$^3$), cubic feet (ft$^3$), etc.

The volume of a cylinder is the product of the area of its circular base and its height.

Volume = (Area of base) $\times$ height

Volume = $(\pi r^2) \times h$

Volume = $\pi r^2 h$

Note: The volume of a prism (with uniform cross-section) is generally the Area of the base multiplied by its height. For a cuboid, the base is a rectangle (area $l \times b$), so volume is $(l \times b) \times h$. For a cylinder, the base is a circle (area $\pi r^2$), so volume is $(\pi r^2) \times h$.

Example 1. Find the volume of a cuboid whose length is 10 cm, breadth is 6 cm, and height is 4 cm.

Answer:

Given:

Length ($l$) = 10 cm.
Breadth ($b$) = 6 cm.
Height ($h$) = 4 cm.

To Find:

Volume of the cuboid.

Solution:

Using the formula for the volume of a cuboid:

Volume $= l \times b \times h$

Substitute the given values:

Volume $= 10 \text{ cm} \times 6 \text{ cm} \times 4 \text{ cm}$

Perform the multiplication:

Volume $= (10 \times 6 \times 4) \text{ cm}^3 = (60 \times 4) \text{ cm}^3 = 240 \text{ cm}^3$

The volume of the cuboid is $240 \text{ cm}^3$.

Example 2. Find the volume of a cylinder whose radius is 7 cm and height is 10 cm. (Use $\pi = \frac{22}{7}$)

Answer:

Given:

Radius of the base ($r$) = 7 cm.
Height of the cylinder ($h$) = 10 cm.
Value of $\pi = \frac{22}{7}$.

To Find:

Volume of the cylinder.

Solution:

Using the formula for the volume of a cylinder:

Volume $= \pi r^2 h$

Substitute the given values:

Volume $= \frac{22}{7} \times (7 \text{ cm})^2 \times 10 \text{ cm}$

$= \frac{22}{7} \times (7 \text{ cm} \times 7 \text{ cm}) \times 10 \text{ cm}$

$= \frac{22}{7} \times 49 \times 10 \text{ cm}^3$

Simplify and calculate:

$= \frac{22}{\cancel{7}_{\normalsize 1}} \times \cancel{49}^{\normalsize 7} \times 10 \text{ cm}^3$

$= 22 \times 7 \times 10 \text{ cm}^3$

$= 154 \times 10 \text{ cm}^3 = 1540 \text{ cm}^3$

The volume of the cylinder is $1540 \text{ cm}^3$.

Surface Area of Solids

In addition to volume (the space occupied by a solid), another important measurement for 3-dimensional (3-D) shapes is their surface area. The surface area refers to the total area of all the surfaces (faces) that enclose the solid.

Understanding Surface Area

Imagine wrapping a gift box with paper. The amount of paper needed to cover the entire box without overlap is equal to the surface area of the box. For solids bounded by flat faces, the surface area is simply the sum of the areas of all its polygonal faces.

Surface area is measured in square units, just like the area of a 2-D shape (cm$^2$, m$^2$, etc.).

Types of Surface Area:

For some solids, it's useful to distinguish between the total surface area and the area of specific parts:

Lateral Surface Area (LSA): This refers to the sum of the areas of the side faces of a solid. It generally excludes the area of the top and bottom faces (bases). For solids with flat sides like prisms and pyramids, it's the sum of the areas of the lateral faces.
Curved Surface Area (CSA): This term is used for solids that have a curved surface, such as a cylinder or a cone. It refers only to the area of the curved part of the surface.
Total Surface Area (TSA): This is the sum of the areas of all the surfaces (faces) that enclose the solid. It includes the lateral surface area (or curved surface area) plus the area(s) of the base(s). For a closed solid, TSA is the entire outer boundary area.

Formulas for Surface Area of Common Solids

Here are the formulas for calculating the Lateral/Curved Surface Area and Total Surface Area of some common solids:

Cube

A cube is a three-dimensional solid with six identical square faces. All its edges are equal in length.

Let the side length of the cube be $a$. The area of each square face is $a^2$.

Lateral Surface Area (LSA) of a Cube

The lateral surface area is the area of the four side faces (excluding the top and bottom faces).

LSA = Area of 4 square faces

LSA = $4 \times (\text{Area of one face})$

LSA = $4a^2$

Total Surface Area (TSA) of a Cube

The total surface area is the sum of the areas of all six faces.

TSA = Area of 6 square faces

TSA = $6 \times (\text{Area of one face})$

TSA = $6a^2$

Cuboid

A cuboid is a three-dimensional solid with six rectangular faces.

Let the length, breadth, and height of the cuboid be $l$, $b$, and $h$ respectively.

Area of Top and Bottom faces = $l \times b$ (2 faces)
Area of Front and Back faces = $l \times h$ (2 faces)
Area of Left and Right side faces = $b \times h$ (2 faces)

Lateral Surface Area (LSA) of a Cuboid

The lateral surface area is the area of the four side faces (Front, Back, Left, and Right).

LSA = (Area of Front face + Area of Back face) + (Area of Left face + Area of Right face)

LSA = $(l \times h + l \times h) + (b \times h + b \times h)$

LSA = $2(lh) + 2(bh)$

LSA = $2(l+b)h$

Total Surface Area (TSA) of a Cuboid

The total surface area is the sum of the areas of all six rectangular faces.

TSA = (Area of Top/Bottom faces) + (Area of Front/Back faces) + (Area of Side faces)

TSA = $2(lb) + 2(lh) + 2(bh)$

TSA = $2(lb + bh + hl)$

Cylinder

A right circular cylinder has two parallel circular bases and a curved surface connecting them.

Let the radius of the circular base be $r$ and the height of the cylinder be $h$.

If the curved surface is unrolled, it forms a rectangle with length equal to the circumference of the base ($2\pi r$) and width equal to the height ($h$).

Curved Surface Area (CSA) of a Cylinder

The curved surface area is the area of the rectangular sheet formed by unrolling the curved part.

CSA = Area of the rectangle

CSA = (Circumference of base) $\times$ (Height)

CSA = $2\pi rh$

Total Surface Area (TSA) of a Cylinder

The total surface area is the sum of the curved surface area and the area of the two circular bases.

TSA = (Curved Surface Area) + (Area of two bases)

TSA = $2\pi rh + 2 \times (\pi r^2)$

TSA = $2\pi rh + 2\pi r^2$

TSA = $2\pi r(h+r)$

Note: For prisms and pyramids with polygonal bases, the LSA is the sum of the areas of the lateral faces. The TSA is the LSA plus the area(s) of the base(s).

Example 1. Find the lateral surface area and total surface area of a cube with side length 5 cm.

Answer:

Given:

Side length of cube ($a$) = 5 cm.

To Find:

Lateral Surface Area (LSA) and Total Surface Area (TSA).

Solution:

Using the formula for the Lateral Surface Area of a cube:

LSA $= 4a^2$

Substitute the value of $a$:

LSA $= 4 \times (5 \text{ cm})^2 = 4 \times (5 \times 5) \text{ cm}^2 = 4 \times 25 \text{ cm}^2 = 100 \text{ cm}^2$

Using the formula for the Total Surface Area of a cube:

TSA $= 6a^2$

Substitute the value of $a$:

TSA $= 6 \times (5 \text{ cm})^2 = 6 \times 25 \text{ cm}^2 = 150 \text{ cm}^2$

The lateral surface area of the cube is $100 \text{ cm}^2$ and the total surface area is $150 \text{ cm}^2$.

Example 2. Find the total surface area of a cuboid whose length is 10 cm, breadth is 8 cm, and height is 5 cm.

Answer:

Given:

Length ($l$) = 10 cm.
Breadth ($b$) = 8 cm.
Height ($h$) = 5 cm.

To Find:

Total Surface Area (TSA) of the cuboid.

Solution:

Using the formula for the Total Surface Area of a cuboid:

TSA $= 2(lb + bh + hl)$

Substitute the given values:

TSA $= 2((10 \text{ cm} \times 8 \text{ cm}) + (8 \text{ cm} \times 5 \text{ cm}) + (5 \text{ cm} \times 10 \text{ cm}))$

Calculate the areas of the pairs of faces:

TSA $= 2(80 \text{ cm}^2 + 40 \text{ cm}^2 + 50 \text{ cm}^2)$

Add the areas inside the bracket and multiply by 2:

TSA $= 2(170 \text{ cm}^2) = 340 \text{ cm}^2$

The total surface area of the cuboid is $340 \text{ cm}^2$.

Example 3. Find the curved surface area and total surface area of a cylinder with radius 14 cm and height 20 cm. (Use $\pi = \frac{22}{7}$)

Answer:

Given:

Radius of the base ($r$) = 14 cm.
Height of the cylinder ($h$) = 20 cm.
Value of $\pi = \frac{22}{7}$.

To Find:

Curved Surface Area (CSA) and Total Surface Area (TSA) of the cylinder.

Solution:

Using the formula for the Curved Surface Area (CSA) of a cylinder:

CSA $= 2\pi rh$

Substitute the given values:

CSA $= 2 \times \frac{22}{7} \times 14 \text{ cm} \times 20 \text{ cm}$

Simplify and calculate:

CSA $= 2 \times \frac{22}{\cancel{7}_{\normalsize 1}} \times \cancel{14}^{\normalsize 2} \times 20 \text{ cm}^2$

CSA $= 2 \times 22 \times 2 \times 20 \text{ cm}^2 = 44 \times 40 \text{ cm}^2 = 1760 \text{ cm}^2$

Using the formula for the Total Surface Area (TSA) of a cylinder:

TSA $= 2\pi r (h+r)$

Substitute the given values:

TSA $= 2 \times \frac{22}{7} \times 14 \text{ cm} \times (20 \text{ cm} + 14 \text{ cm})$

$= 2 \times \frac{22}{7} \times 14 \text{ cm} \times 34 \text{ cm}$

Simplify and calculate:

TSA $= 2 \times \frac{22}{\cancel{7}_{\normalsize 1}} \times \cancel{14}^{\normalsize 2} \times 34 \text{ cm}^2$

TSA $= 2 \times 22 \times 2 \times 34 \text{ cm}^2 = 44 \times 68 \text{ cm}^2$

TSA $= 2992 \text{ cm}^2$

Alternatively, calculate TSA as CSA + Area of 2 bases:

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

Area of circular base $= \pi r^2 = \frac{22}{7} \times (14 \text{ cm})^2 \ $$ = \frac{22}{7} \times 196 \text{ cm}^2 \ $$ = \frac{22}{\cancel{7}_{\normalsize 1}} \times \cancel{196}^{\normalsize 28} \text{ cm}^2 = 22 \times 28 \text{ cm}^2$.

Perform $22 \times 28$:

$22 \times 28 = 22 \times (20 + 8) = 440 + 176 = 616$

Area of 2 bases $= 2 \times 616 \text{ cm}^2 = 1232 \text{ cm}^2$.

TSA $= 1760 \text{ cm}^2 + 1232 \text{ cm}^2 = 2992 \text{ cm}^2$

Both methods give the same result. The curved surface area is $1760 \text{ cm}^2$ and the total surface area is $2992 \text{ cm}^2$.