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Class 9th Chapters
1. Number Systems 2. Polynomials 3. Coordinate Geometry
4. Linear Equations In Two Variables 5. Introduction To Euclid’s Geometry 6. Lines And Angles
7. Triangles 8. Quadrilaterals 9. Areas Of Parallelograms And Triangles
10. Circles 11. Constructions 12. Heron’s Formula
13. Surface Areas And Volumes 14. Statistics 15. Probability

Content On This Page
Basis Terms Realted to Solid Figures Surface Area of Solid Figures Volume of Solid Figures


Chapter 13 Surface Areas And Volumes (Concepts)

Welcome to this significantly expanded chapter on Mensuration, where we take a major leap forward from the concepts explored in Class 8. While previously we focused on the perimeter and area of 2D shapes and the basics of surface area and volume for cuboids, cubes, and cylinders, this chapter delves deeper into the three-dimensional world. We will introduce methods for calculating the Surface Areas and Volumes of more complex and commonly encountered solid shapes: the cone and the sphere, alongside a thorough review and application of the formulas for the solids already studied. Our goal is to develop a comprehensive understanding of how to quantify both the exterior surface and the space occupied by these fundamental 3D forms.

Before venturing into new territory, let's consolidate our understanding of the solids introduced earlier. Recall the key measurements: Lateral Surface Area (LSA) or Curved Surface Area (CSA), which measures the area of the side surfaces excluding bases; Total Surface Area (TSA), encompassing the area of all surfaces including bases; and Volume (V), representing the total space enclosed by the solid. We revisit the formulas for:

Now, we introduce new shapes and their corresponding formulas. For the Right Circular Cone, we introduce the concept of slant height ($l$), which is the distance from the apex to any point on the circumference of the base. It relates to the radius ($r$) and perpendicular height ($h$) via the Pythagorean theorem: $l^2 = r^2 + h^2$. The formulas are:

Next, we explore the perfectly symmetrical Sphere:

We also consider the Hemisphere (exactly half a sphere):

Beyond calculating these quantities for individual, isolated shapes using their dimensions ($r, h, l, a, \dots$), a significant focus of this chapter involves tackling problems concerning combinations of solids. This requires strong spatial visualization skills. Examples include finding the surface area of a toy shaped like a cone mounted on a hemisphere, or the volume of a capsule formed by a cylinder with hemispherical ends. Solving these problems involves carefully identifying which surfaces are exposed (for TSA) or which volumes contribute to the total, often requiring adding or subtracting the respective calculated parts. Furthermore, numerous practical application problems are explored, such as determining the amount of canvas needed for a tent (surface area), finding the capacity of tanks or vessels (volume), calculating the costs (perhaps in $\textsf{₹}$) associated with painting surfaces or filling containers, and analyzing scenarios involving the conversion of a solid from one shape into another (by equating their volumes). Throughout all calculations, maintaining accuracy with units (e.g., $cm^2, m^2$ for area; $cm^3, m^3$, litres for volume) and performing necessary conversions remains critically important.



Basic Terms Related to Solid Figures

In earlier classes, our focus was primarily on two-dimensional (2D) geometric figures, such as squares, rectangles, triangles, and circles. These figures lie flat on a plane and are described by two dimensions, typically length and breadth. We learned how to calculate their perimeters (the length of their boundary) and areas (the measure of the surface enclosed within their boundary).


In this chapter, we move into the realm of three-dimensional (3D) geometry. We will study shapes that occupy space, known as solid figures or simply solids. These figures have three dimensions: length, breadth (often called width), and height (or depth). Common examples of solid figures encountered in everyday life include cuboids (like bricks or boxes), cubes (like dice), cylinders (like pipes or soda cans), cones (like ice cream cones), and spheres (like balls).

Examples of solid figures: cube, cuboid, cylinder, cone, sphere

For solid figures, the concepts of perimeter and area of 2D shapes are extended and replaced by notions of surface area (the total area of all the exterior surfaces of the solid) and volume (the amount of space enclosed or occupied by the solid).

Key Concepts and Terminology

Solid Figure (or Solid):

A solid figure, or solid, is a geometric figure that exists in three dimensions and occupies a definite amount of space. It has length, breadth, and height, and it possesses volume and surface area.

Diagram showing a 3D object with length, breadth, and height indicated

Dimensions:

Three-dimensional figures are characterized by three fundamental measurements that define their size and extent in space: length, breadth (or width), and height (or depth). For example, a cuboid is defined by its length, breadth, and height.

Surface:

The surface of a solid is the exterior boundary that separates the solid from the surrounding space. A solid's surface can consist of flat parts (called plane faces) or curved parts (called curved surfaces).

Edge:

An edge is the line segment where two faces of a solid meet. Edges are typically straight in solids with flat faces.

Vertex:

A vertex (plural: vertices) is a point where three or more edges of a solid meet. Vertices are the "corners" of a solid.


Common Solid Figures and Their Components

Let's look at some common solid figures and identify their components:

Cuboid:

A cuboid is a solid figure bounded by six rectangular faces. It is a type of rectangular prism. In a cuboid, opposite faces are congruent (identical in shape and size). A cuboid has 12 edges and 8 vertices.

Example: A matchbox, a brick, a rectangular room.

Diagram of a cuboid showing its rectangular faces, edges, and vertices

Components: 6 rectangular faces, 12 edges, 8 vertices.

Cube:

A cube is a special type of cuboid in which all edges are of equal length. Consequently, all six faces of a cube are congruent squares. A cube also has 12 edges and 8 vertices.

Example: A dice, a sugar cube, a Rubik's cube.

Diagram of a cube showing its square faces, edges, and vertices, with side length 'a'

Components: 6 square faces, 12 edges, 8 vertices.

Cylinder:

A cylinder is a solid figure with two congruent circular bases that are parallel to each other, connected by a curved lateral surface. The axis of the cylinder is the line segment joining the centres of the two bases.

Example: A can of soft drink, a circular pillar, a section of pipe.

Diagram of a cylinder showing its circular bases and curved lateral surface, with radius r and height h

Components: Two circular plane bases, one curved lateral surface.

Cone:

A cone is a solid figure with a single circular base and a curved lateral surface that tapers smoothly from the base to a point called the apex or vertex. The height of the cone is the perpendicular distance from the apex to the centre of the base. The slant height is the distance from the apex to any point on the circumference of the base.

Example: An ice-cream cone, a birthday party hat, a traffic cone.

Diagram of a cone showing its circular base, curved surface, and apex, with radius r, height h, and slant height l

Components: One circular plane base, one curved lateral surface, an apex (vertex).

Sphere:

A sphere is a perfectly round solid figure in three-dimensional space. It is defined as the set of all points in space that are at a fixed distance from a fixed central point. It has no edges or vertices.

Example: A ball, a globe, a marble.

Diagram of a sphere showing its curved surface and radius r

Components: One continuous curved surface.

In the following sections, we will learn how to calculate the surface areas and volumes of these common solid figures, along with applying these calculations to solve real-world problems.



Surface Area of Solid Figures

The surface area of a three-dimensional solid figure is the total area of all the surfaces (or faces) that enclose the solid. Imagine unfolding the solid and measuring the area of each piece; the sum of these areas is the surface area. Surface area is a measure of the extent of the boundary of a solid and is always expressed in square units (e.g., $\text{cm}^2$, $\text{m}^2$).


Types of Surface Area

For many solid figures, particularly those with distinct bases, we distinguish between two types of surface area:

Surface Area Formulas for Common Solids

The surface area formulas depend on the shape of the solid and its dimensions. These formulas are derived by calculating the area of each individual face or curved surface and summing them up.

Cuboid:

A cuboid has 6 rectangular faces. Let the dimensions be length ($l$), breadth ($b$), and height ($h$).

Diagram of a cuboid showing pairs of identical faces: top/bottom (lb), front/back (lh), left/right (bh)

Total Surface Area (TSA) of a Cuboid: Sum of the areas of all 6 faces.

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

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

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

Lateral Surface Area (LSA) of a Cuboid: Sum of the areas of the 4 side faces (excluding top and bottom bases).

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

LSA $= 2lh + 2bh$

LSA of Cuboid $= 2h(l+b)$

Note: LSA of a cuboid is also the perimeter of the base ($2(l+b)$) multiplied by the height ($h$).

Cube:

A cube is a special cuboid where all edges are equal. Let the side length be $a$. So, $l=b=h=a$. It has 6 square faces, each with area $a \times a = a^2$.

Diagram of a cube showing side length a

Total Surface Area (TSA) of a Cube: Sum of the areas of all 6 square faces.

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

TSA of Cube $= 6a^2$

Lateral Surface Area (LSA) of a Cube: Sum of the areas of 4 square side faces.

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

LSA of Cube $= 4a^2$

Cylinder:

A cylinder has two circular bases and one curved lateral surface. Let the radius of the base be $r$ and the height be $h$.

Diagram showing the net of a cylinder, with two circles and a rectangle (unrolled curved surface)

Curved Surface Area (CSA) of a Cylinder: Area of the unrolled rectangle.

CSA $= \text{Length of rectangle} \times \text{Width of rectangle}$

CSA $= (2\pi r) \times h$

CSA of Cylinder $= 2\pi rh$

Total Surface Area (TSA) of a Cylinder: CSA + Area of Top Base + Area of Bottom Base.

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

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

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

Cone:

A cone has a circular base and a curved lateral surface. Let the radius of the base be $r$, the height (perpendicular) be $h$, and the slant height be $l$. The slant height is the distance from the apex to any point on the circumference of the base. In a right circular cone, the height, radius, and slant height form a right-angled triangle, so $l^2 = r^2 + h^2$.

Slant height $l = \sqrt{r^2 + h^2}$

Diagram showing the net of a cone, with a circle (base) and a sector of a circle (unrolled curved surface)

Curved Surface Area (CSA) of a Cone: Area of the unrolled sector.

CSA $= \frac{1}{2} \times l \times (2\pi r)$

CSA $= \pi r l$

CSA of Cone $= \pi rl$

Total Surface Area (TSA) of a Cone: CSA + Area of Base.

TSA $= (\pi rl) + (\pi r^2)$

TSA of Cone $= \pi r(l+r)$

Sphere:

A sphere has only one continuous curved surface. Let the radius be $r$. The formula for the surface area of a sphere is derived using calculus, but it is given by $4\pi r^2$.

Diagram of a sphere showing radius r

Surface Area of a Sphere:

Surface Area $= 4\pi r^2$

Hemisphere:

A hemisphere is half of a sphere. It consists of a curved surface and a flat circular base. Let the radius be $r$.

Diagram of a hemisphere showing radius r and flat circular base

Curved Surface Area (CSA) of a Hemisphere:

CSA $= \frac{1}{2} \times (\text{Surface Area of Sphere})$

CSA $= \frac{1}{2} \times (4\pi r^2)$

CSA of Hemisphere $= 2\pi r^2$

Total Surface Area (TSA) of a Hemisphere: CSA + Area of the flat circular base.

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

TSA of Hemisphere $= 3\pi r^2$


Example 1. Find the TSA and LSA of a cuboid with length 8 cm, breadth 5 cm, and height 3 cm.

Answer:

Given:

Length $l = 8$ cm, breadth $b = 5$ cm, height $h = 3$ cm.

To Find:

Lateral Surface Area (LSA) and Total Surface Area (TSA) of the cuboid.

Solution:

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

LSA $= 2h(l+b)$

Substitute the given dimensions:

LSA $= 2 \times 3 \text{ cm} (8 \text{ cm} + 5 \text{ cm})$

LSA $= 6 \text{ cm} (13 \text{ cm})$

LSA $= 78 \text{ cm}^2$

... (1)

The Lateral Surface Area of the cuboid is 78 cm$^2$.

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

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

Substitute the given dimensions:

TSA $= 2((8 \text{ cm} \times 5 \text{ cm}) + (8 \text{ cm} \times 3 \text{ cm}) + (5 \text{ cm} \times 3 \text{ cm}))$

TSA $= 2(40 \text{ cm}^2 + 24 \text{ cm}^2 + 15 \text{ cm}^2)$

TSA $= 2(79 \text{ cm}^2)$

TSA $= 158 \text{ cm}^2$

... (2)

The Total Surface Area of the cuboid is 158 cm$^2$.


Example 2. The radius of a cylinder is 7 cm and its height is 10 cm. Find its CSA and TSA. (Use $\pi = \frac{22}{7}$)

Answer:

Given:

Radius of cylinder base $r = 7$ cm.

Height of cylinder $h = 10$ cm.

Use $\pi = \frac{22}{7}$.

To Find:

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

Solution:

Using the formula for the Curved Surface Area of a cylinder:

CSA $= 2\pi rh$

Substitute the given values:

CSA $= 2 \times \frac{22}{7} \times 7 \text{ cm} \times 10 \text{ cm}$

Cancel out the common factor 7 in the numerator and denominator:

CSA $= 2 \times 22 \times \cancel{7} \times \frac{10}{\cancel{7}} \text{ cm}^2$

CSA $= 44 \times 10 \text{ cm}^2$

CSA $= 440 \text{ cm}^2$

... (1)

The Curved Surface Area of the cylinder is 440 cm$^2$.

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

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

Substitute the given values:

TSA $= 2 \times \frac{22}{7} \times 7 \text{ cm} (10 \text{ cm} + 7 \text{ cm})$

TSA $= 2 \times \frac{22}{\cancel{7}} \times \cancel{7} \text{ cm} (17 \text{ cm})$

TSA $= 44 \times 17 \text{ cm}^2$

Perform the multiplication:

TSA $= 748 \text{ cm}^2$

... (2)

The Total Surface Area of the cylinder is 748 cm$^2$.


Example 3. The diameter of a sphere is 14 cm. Find its surface area. (Use $\pi = \frac{22}{7}$)

Answer:

Given:

Diameter of sphere $d = 14$ cm.

Use $\pi = \frac{22}{7}$.

To Find:

The surface area of the sphere.

Solution:

The formula for the surface area of a sphere requires the radius ($r$). The radius is half of the diameter.

Radius $r = \frac{d}{2} = \frac{14}{2} \text{ cm}$

$r = 7 \text{ cm}$

... (1)

Using the formula for the Surface Area of a Sphere:

Area $= 4\pi r^2$

Substitute the given value of $\pi$ and the calculated radius from equation (1):

Area $= 4 \times \frac{22}{7} \times (7 \text{ cm})^2$

Area $= 4 \times \frac{22}{7} \times 49 \text{ cm}^2$

Cancel out the common factor 7:

Area $= 4 \times 22 \times \frac{\cancel{49}^{7}}{\cancel{7}} \text{ cm}^2$

Area $= 4 \times 22 \times 7 \text{ cm}^2$

Perform the multiplication:

Area $= 88 \times 7 \text{ cm}^2$

Area $= 616 \text{ cm}^2$

... (2)

The surface area of the sphere is 616 cm$^2$.



Volume of Solid Figures

While surface area measures the total area of the outer boundaries of a solid, volume measures the amount of three-dimensional space enclosed or occupied by the solid. Think of volume as the capacity of the solid – how much it can hold if it were hollow, or how much "stuff" the solid itself is made of. Volume is a measure of space and is always expressed in cubic units (e.g., $\text{cm}^3$, $\text{m}^3$).


Volume Formulas for Common Solids

The volume formulas for various solid figures are derived based on their geometric properties. For many regular solids, the volume can be conceptually understood as the area of the base multiplied by the height (or an appropriate fraction thereof).

Cuboid:

Let the dimensions of a cuboid be length ($l$), breadth ($b$), and height ($h$). The base of a cuboid is a rectangle with area $l \times b$. The volume of a cuboid is found by multiplying the area of its base by its height.

Diagram of a cuboid showing length l, breadth b, and height h

Volume $= (\text{Area of base}) \times \text{height}$

Volume of Cuboid $= (l \times b) \times h$

Volume of Cuboid $= lbh$

If the area of the base is given as $A_{\text{base}}$, then Volume $= A_{\text{base}} \times h$.

Cube:

A cube is a special cuboid where all edges have the same length. Let the side length be $a$. So, $l=b=h=a$.

Diagram of a cube showing side length a

Using the volume formula for a cuboid with $l=b=h=a$:

Volume of Cube $= a \times a \times a$

Volume of Cube $= a^3$

Cylinder:

A cylinder has a circular base. Let the radius of the base be $r$ and the height be $h$. The area of the circular base is $\pi r^2$. Similar to the cuboid, the volume of a cylinder is the area of its base multiplied by its height.

Diagram of a cylinder showing radius r and height h

Volume $= (\text{Area of base}) \times \text{height}$

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

Volume of Cylinder $= \pi r^2h$

Cone:

A cone has a circular base with radius $r$ and height $h$. The volume of a cone has a specific relationship with the volume of a cylinder that has the same base radius and height. It is found that the volume of a cone is exactly one-third of the volume of the corresponding cylinder. This can be demonstrated through experiments (filling a cylinder with water using a cone of the same base and height) or proven mathematically using calculus.

Diagram of a cone showing radius r and height h

Volume $= \frac{1}{3} \times (\text{Area of base}) \times \text{height}$

Volume of Cone $= \frac{1}{3} \times (\pi r^2) \times h$

Volume of Cone $= \frac{1}{3} \pi r^2h$

Sphere:

A sphere is a perfectly round solid figure defined by its radius $r$. The formula for the volume of a sphere is derived using integral calculus or by advanced geometric arguments involving Cavalieri's principle. The formula is:

Diagram of a sphere showing radius r

Volume of Sphere $= \frac{4}{3} \pi r^3$

Hemisphere:

A hemisphere is half of a sphere. Its volume is simply half the volume of the corresponding sphere. Let the radius be $r$.

Diagram of a hemisphere showing radius r

Volume of Hemisphere $= \frac{1}{2} \times (\text{Volume of Sphere})$

Volume of Hemisphere $= \frac{1}{2} \times \left(\frac{4}{3} \pi r^3\right)$

Volume of Hemisphere $= \frac{2}{3} \pi r^3$


Example 1. Find the volume of a cube whose edge is 4 cm.

Answer:

Given:

Edge length of the cube $a = 4$ cm.

To Find:

The volume of the cube.

Solution:

Using the formula for the Volume of a Cube with edge length $a$:

Volume $= a^3$

Substitute the given edge length $a = 4$ cm:

Volume $= (4 \text{ cm})^3$

Calculate the cube of the number and apply the cubic unit:

Volume $= 4 \times 4 \times 4 \text{ cm}^3$

Volume $= 64 \text{ cm}^3$

... (1)

The volume of the cube is 64 cm$^3$.


Example 2. A cylindrical container has a base radius of 5 cm and a height of 14 cm. Find its volume. (Use $\pi = \frac{22}{7}$)

Answer:

Given:

Radius of cylinder base $r = 5$ cm.

Height of cylinder $h = 14$ cm.

Use $\pi = \frac{22}{7}$.

To Find:

The volume of the cylindrical container.

Solution:

Using the formula for the Volume of a Cylinder with radius $r$ and height $h$:

Volume $= \pi r^2h$

Substitute the given values:

Volume $= \frac{22}{7} \times (5 \text{ cm})^2 \times 14 \text{ cm}$

Volume $= \frac{22}{7} \times 25 \text{ cm}^2 \times 14 \text{ cm}$

Cancel out the common factor 7:

Volume $= \frac{22}{\cancel{7}} \times 25 \times \cancel{14}^{2} \text{ cm}^3$

Volume $= 22 \times 25 \times 2 \text{ cm}^3$

Perform the multiplication:

Volume $= 22 \times (25 \times 2) \text{ cm}^3$

Volume $= 22 \times 50 \text{ cm}^3$

Volume $= 1100 \text{ cm}^3$

... (1)

The volume of the cylindrical container is 1100 cm$^3$.


Example 3. Find the volume of a cone with height 21 cm and base diameter 12 cm. (Use $\pi = \frac{22}{7}$)

Answer:

Given:

Height of the cone, $h = 21$ cm.

Diameter of the base, $d = 12$ cm.

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To Find:

The volume of the cone.

Solution:

First, we need to find the radius of the cone's base. The radius ($r$) is half of the diameter ($d$).

Radius, $r = \frac{d}{2} = \frac{12}{2} \text{ cm}$

$r = 6 \text{ cm}$

The formula for the volume of a cone is:

Volume $(V) = \frac{1}{3} \pi r^2h$

Now, we substitute the values of $r$, $h$, and $\pi$ into the formula:

$V = \frac{1}{3} \times \frac{22}{7} \times (6)^2 \times 21$

$V = \frac{1}{3} \times \frac{22}{7} \times 36 \times 21$

We can simplify the expression by cancelling common factors:

$V = \frac{1}{\cancel{3}} \times 22 \times \cancel{36}^{12} \times \frac{21}{7}$

$V = 22 \times 12 \times \frac{\cancel{21}^{3}}{\cancel{7}}$

Now, we multiply the remaining numbers:

$V = 22 \times 12 \times 3$

$V = 22 \times 36$

$V = 792$

The volume is measured in cubic units, so the volume of the cone is 792 cm$^3$.

Therefore, the volume of the cone is 792 cm3.