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Chapter 13 - Surface Areas and Volume, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9 PDF Download

Surface Areas and Volumes

INTRODUCTION

Uptill now we have been dealing with plane figures that can be drawn on the page of our notebook or on the blackboard. In this chapter, we shall study about some solid figures like cuboid, cube, cylinder and sphere. We shall also learn to find the surface areas and volumes of these figures.

PLANE EKJLKES

SOLID FICTEES

Hie geometrical figure winch haw only two rfjmpnonnfi are called the plane figures.

Two dimensions or 2D are known

Le.. length and bieadth.

A figure winch liaxe three Hjinsrearwt; as length, breadth and heiglit is not a plane figure and we can not draw such figures on black board exactly. These three dimensional figures are called sofids.

Three dimensions or 3D are known Le., length, breadth and height.

Ex. Rectangle, Square. Paralkkignam, Rhonius. Triangle, circle.

Ex. Cube, Cuboid Cylinder cone. Sphere. Prism Pyramid etc.



 CUBOID

Chapter 13 - Surface Areas and Volume, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9

A rectangular solid bounded by six rectangular plane faces is called a cuboid. A match box, a tea-packet, a brick, a book, etc.,are all examples of a cuboid.

A cuboid has 6 rectangular faces, 12 edges and 8 vertices.

The following are some definitions of terms related to a cuboid:

(i) The space enclosed by a cuboid is called its volume.

(ii) The line joining opposite corners of a cuboid is called its diagonal. A cuboid has four diagonals.

A diagonal of a cuboid is the length of the longest rod that can be placed in the cuboid.

(iii) The sum of areas of all the six faces of a cuboid is known as its total surface area.

(iv) The four faces which meet the base of a cuboid are called the lateral faces of the cuboid.

(v) The sum of areas of the four walls of a cuboid is called its lateral surface area.

For a cuboid of length = ℓ units, breadth = b units and height = h units, we have :
 

FORMULAE

Smn of lengths of all edges =4 (ℓ + b + h) units.

Diagonal of cuboid = Chapter 13 - Surface Areas and Volume, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9

Total Surface Area of cuboid = 2(ℓb + bh + ℓh) sq. miits

Lateral Surface Area of a cuboid = [2(ℓ + b) x h] sq. muts

Area of fom' walls of a loom = [2(ℓ + b) x h] sq. luiits

Vblmne of cuboid = (ℓ x b x h) cubic units

 

REMARK : For the calculation of surface area, volume etc. of a cuboid, the length, breadth and height must be expressed in the same units.

CUBE

A cuboid whose length, breadth and height are all equal is called a cube.
Ice-cubes, Sugar cubes, Dice, etc. are all examples of a cube. Each edge of a cube is called its side.
For a cube of edge = a units, we have;

CROSS SECTION 

A cut which is made through a solid perpendicular to its length is called its cross section. If the cut has the same shape and size at every point of its length, then it is called uniform cross-section.
Volume of a solid with uniform cross section = (Area of its cross section) × (length).
Lateral Surface Area of a solid with uniform cross section = (Perimeter of cross section) × (length).

SOLVED EXAMPLES

Ex 1. Find the surface area of a cube whose edge is 15 cm.

Sol. The edge of the cube = 15 cm, i.e., a = 15 cm.
Surface area of the cube = 6a2 = 6 × (15)2 = 1350 cm2.

Ex 2. A small indoor greenhouse is made entirely of glass sheets (including the base) held together with tape. It is 40 cm long, 30 cm wide and 30 cm high. Find
 (i) the area of the glass sheet required and
 (ii) the total length of the tape required for all the 12 edges.

Sol. The dimensions of the greenhouse are as under :

Chapter 13 - Surface Areas and Volume, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9
Length (ℓ) = 40 cm, Width (b) = 30 cm, Height (h) = 30 cm
The area of the glass sheet required

=  2 [ℓ x b + b x  h + h x  ℓ]
=  2 [40 x 30 + 30 x 30  + 30 x 40] cm2
=  2  [1200 + 900  + 1200] cm2
=  2 x 3300  cm2  =  6600 cm2

Hence, 6600 cm2 of glass sheet is required.

Length of the tap required = Sum of the length of the 12 edges.

= 4 (ℓ + b + h)
= 4 x [40 + 30 + 30] cm = 400 cm

Hence, 400 cm of the tape is required.

Ex 3. A wall of length 10 m was to be built across an open ground. The height of the wall is 4 m andthickness of the wall is 24 cm. If this wall is to be built up with bricks of dimensions 24 cm × 12cm × 8 cm, then find the number of bricks which are required.

Sol. We know that, the volume of the wall and the sum of the volumes of the required number of bricks is same.
Length of the wall = 10 × 100 cm = 1000 cm
Breadth or the thickness of the wall = 24 cm
Height of the wall = 4 × 100 cm = 400 cm
The wall is in the shape of a cuboid and its volume = 1000 × 24 × 400 cm3
Now, a brick is also a cuboid having length = 24 cm, breadth = 12 cm and height = 8 cm.
Volume of one brick = 24 × 12 × 8 cm3
The required number of bricks = Volume ofthe wall
Volume of onebrick = Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important = 4166.6
Hence, the required number of bricks = 4167.

 Ex 4 . Aakriti playing with plastic building blocks which are of identical cubical shapes. She makes a structure as shown in fig. If the edge of each cube is 5 cm, then find the volume of the structure built by Aakriti.

Sol.

Chapter 13 - Surface Areas and Volume, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9

In fig. the structure is made with 10 cubes.

Length of an edge of each block = 5 cm.

Volume of one block - (5) 3 cm3 - 125 cm3

Volume of the structure = Volume of the ten blocks = 10 x 125 cm3 = 1250 cm3

Hence, the volume of the structure is 1250 cm3.

RIGHT CIRCULAR CYLINDER

Solids like circular pillars, circular pipes, circular pencils, measuring jars, road rollers and gas cylinders, etc., are said to be in cylindrical shape.

Chapter 13 - Surface Areas and Volume, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9

In mathematical terms, a right circular cylinder is a solid generated by the revolution of a rectangle about its sides.

Let the rectangle ABCD revolve about its side AB, so as to describe a right circular cylinder as shown in the figure.

You must have observed that the cross-sections of a right circular cylinder are circles congruent and parallel toeach other.

Cylinders Not Right Circular

There are two cases when the cylinder is not a right circular cylinder.

Chapter 13 - Surface Areas and Volume, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9

ln the case-I tap see a cylinder. which is certainly circular but is not at right angles to the base. So we cannot say it is a right circular cylinder.

Chapter 13 - Surface Areas and Volume, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9

In the case-II we see a cylinder with a non-circular base as the base is not circular. So we cannot call it a right circular cylinder

REMARK : Unless stated otherwise, here in this chapter the word cylinder would mean a right circular cylinder.

The following are definitions of some terms related to a right circular cylinder :

(i) The radius of any circular end is called the radius of the right circular cylinder.

Thus, in the above figure, AD as well as BC is a radius of the cylinder.

(ii) The line joining the centres of circular ends of the cylinder, is called the axis of the right circular cylinder.

In the above figure, the line AB is the axis of the cylinder. Clearly, the axis is perpendicular to the circular ends.

REMARK : If the line joining the centres of circular ends of a cylinder is not perpendicular to the circular ends, then the cylinder is not a right circular cylinder.

(iii) The length of the axis of the cylinder is called the height or length of the cylinder.

(iv) The curved surface joining the two bases of a right circular cylinder is called its lateral surface.

For a right circular cylinder of radius = r units & height = h units, we have :

FORMULAE

Area of each circular end = πr2 sq. units.
Curved (Lateral) Surface Area = (2πrh) sq. imits.
Total Surface Area = Curved Surface Area
+ Area of two circular ends.
= (2πrh + 2πr2) sq. units.
= [2πrh (h + r)] sq. units.
Volume of cylinder =πr2h cubic units.

 

The above formulae are applicable to solid cylinders only.

HOLLOW RIGHT CIRCULAR CYLINDERS

Solids like iron pipes, rubber tubes, etc., are in the shape of hollow cylinders.

A solid bounded by two coaxial cylinders of the same height and different radii is called a hollow cylinder.
Chapter 13 - Surface Areas and Volume, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9

For a hollow cylinder of height h and with external and internal radii R and r respectively, we have :

FORMULAE

Thickness of cylinder = (R - r) units.
Area of a cross-section = (πR2 - πr2) sq. units.
= π(R2 - r2) sq. units.
Curved (Lateral) Surface Area = (External Curved. Surface Area)
+ (Internal Curved Suiface Area)
= (2πRh + 2πrh) sq. units = 2πh (R+ r) sq. units.
Total Surface Area = (Curved Surface Area) + 2 (Area of Base Ring) 
= [(2πRh + 2πrh) + 2(πR2 - πr2)] sq. units
= 2π(Rh +rh + R2 - r) sq. units.
Volume of Material = π(R2- r2)h cubic units
Volume of Hollow region = πr2h cubic units

 

Ex 5. A cylindrical block of wood has radius 70 cm and length 2 m is to be painted with blue coloured enamel. The cost of painting is Rs. 1.25 per 100 cm2. Find the cost of painting the block. Take π = 22/7

Sol. Here, the radius r of the cylindrical block of wood = 70 cm and the length h = 200 cm
The total surface area of the cylindrical block = 2πr(r + h)
Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important cm= 440 × 270 cm2 = 118800 cm2
Cost of painting 100 cm2 = Rs. 1.25 = Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important

⇒ Cost of painting 1 cm2 = Rs. 5/400
Then, cost of painting 118800 cm2 = Rs. 5/400 × 118800 = Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important × 1188 = Rs.1485
Hence, cost of painting the block of wood = Rs.1485.

Ex 6. A cylindrical vessel, without lid, has to be tin-coated including both of its sides. If the radius ofits base is 1/2 m and its height is 1.4 m, calculate the cost of tin-coating at the rate of Rs. 50 per 1000 cm2. (Use π = 3.14)

Sol. Radius of the base (r) = 1/2m =1/2× 100 cm = 50 cm
Height (h) = 1.4 m = 1.4 × 100 cm = 140 cm
Surface area to be tin-coated = 2 (2πrh + πr2)

= 2 [2 x 3.14 x 50 x 140 + 3.14 x (50)2]
= 2 [43960 + 7850] = 2 (51810) = 103620 cm2

= Cost of tin-coating at the rate of Rs. 50 per 1000 cm2

= Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important
Hence, the cost of tin-coating is Rs. 5181.

Ex 7. The pillars of a temple are cylindrical shaped. If each pillar has a circular base of radius 20 cm and height 7 m, then find the quantity of concrete mixture used to build 20 such pillars. Alsofind the cost of the concrete mixture at the rate of Rs. 200 per m3. Take π  = 22/7

Sol. Concrete mixture used for making each pillar = Volume of each pillar = π r2 × h

Chapter 13 - Surface Areas and Volume, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9

Radius of base of pillar,  Chapter 13 - Surface Areas and Volume, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9

Height of pillar, h = 7m

Volume of mixture used for 1 pillar

Chapter 13 - Surface Areas and Volume, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9

Volume of mixture used for 20 pillars

Chapter 13 - Surface Areas and Volume, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9

Hence, volume of mixture required to make 20 pillars = 17.6 m3
Now, cost of mixture at the rate of Rs.200 per m3
= Rs.200 × 17.6 = Rs. 3520.
Hence, the cost of concrete mixture is Rs. 3520.

RIGHT CIRCULAR CONE

Solids like an ice-cream cone, a conical tent, a conical vessel, a clown's cap etc. are said to be in conical shape.
In mathematical terms, a right circular cone is a solid generated by revolving a right-angled triangle about one of the sides containing the right angle.
Let a triangle AOC revolve about it's side OC, so as to describe a right circular cone, as shown in the figure.

Chapter 13 - Surface Areas and Volume, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9

Cones Not Right Circular

Chapter 13 - Surface Areas and Volume, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9 

Chapter 13 - Surface Areas and Volume, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9

REMARK : Unless stated otherwise, by 'cone' in this chapter, we shall mean 'a right circular cone'.

The following are definitions of some terms related to right circular cone :

(i) The fixed point O is called the vertex of the cone.

(ii) The fixed line OC is called the axis of the cone.

(iii) A right circular cone has a plane end, which is in circular shape. This is called the base of the cone. The vertex of a right circular cone is farthest from its base.

(iv) The length of the line segment joining the vertex to the centre of the base is called the height of the cone. Length OC is the height of the cone.

(v) The length of the line segment joining the vertex to any point on the circular edge of the base, is called the slant height of the cone.
Length OA is slant height of the cone.

(vi) The radius AC of the base circle is called the radius of the cone. There are two cases when we cannot call a cone a right circular cone.

The following are definitions of some terms related to right circular cone :
(i) The fixed point O is called the vertex of the cone.
(ii) The fixed line OC is called the axis of the cone.
(iii) A right circular cone has a plane end, which is in circular shape. This is called the base of the cone. The vertex of a right circular cone is farthest from its base.
(iv) The length of the line segment joining the vertex to the centre of the base is called the height of the cone. Length OC is the height of the cone.
(v) The length of the line segment joining the vertex to any point on the circular edge of the base, is called the slant height of the cone. Length OA is slant height of the cone.
(vi) The radius AC of the base circle is called the radius of the cone.

Relation Between Slant Height, Radius and Vertical Height.

Chapter 13 - Surface Areas and Volume, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9
Let us take a right circular cone with vertex at O, vertical height h, slant height ℓ and radius r. A is any point on the rim of the base of the cone and C is the centre of the base. Here, OC = h, AC = r and OA = ℓ.

The cone is right circular and therefore, OC is at right angle to the base of the cone. So, we have OC Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important CA,
i.e., ΔOCA is right angled at C.
Then by Pythagoras theorem, we have :
Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important
For a right circular cone of Radius = r, Height = h & Slant Height = ℓ, we have :

 

FORMULAE
Area of the curved (lateral) surface = πrℓ) sq. units. Chapter 13 - Surface Areas and Volume, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9
Total Surface Area of cone = (Curved surface Aiea + Area of Base)
= (π rℓ + πr2) sq. units = π r (ℓ + r) sq. units.
"Solume of cone Chapter 13 - Surface Areas and Volume, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9


Hollow Right Circular Cone
Chapter 13 - Surface Areas and Volume, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9


(i) Centre of the circle is vertex of the cone.
 (ii) Radius of the circle is slant height of the cone.
 (iii) Length of arc AB is the circumference of the base of the cone.
 (iv) Area of the sector is the curved surface area of the cone.

Ex 8. The radius of the base of a conical tent is 12 m. The tent is 9 m high. Find the cost of the canvas required to make the tent, if one square metre of canvas costs Rs. 120. (Take π = 3.14)
 Sol. Here, r = 12 m and h = 9 m. Let Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important be slant height of the conical tent.

Now, Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important2 = r2 + h2 = (12)2 + (9)2 = 144 + 81 = 225
Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important  = 15 m

The curved surface area of the tent  Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important
Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important
Hence, the canvas required = Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important
Cost of the canvas at the rate of Rs. 120 per m2 = Rs. 120 × 565.2 = Rs. 67824

Ex 9. How many metres of cloth of 1.1 m width will be required to make a conical tent whose vertical height is 12 m and base radius is 16 m? Find also the cost of the cloth used at the rate of Rs.14 per metre.

Sol. Here, h = 12 m, r = 16 m
Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important = 20m
Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important Curved surface area = Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important

Width of cloth = 1.1 m
Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important
Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important Cost of the cloth used @ Rs.14 per metre = Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important × 14 = Rs. 12800.

Ex 10. The curved surface of a right circular cone is 198 cm2 and the radius of its base is 7 cm. Find the volume of the cone. Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important

Sol. Radius of the base of the cone = 7 cm.
Let h cm be the vertical height and cm be the slant height. Here, r = 7 cm.
Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important
Now, Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important  81 – 49 = 32 cm

Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important × 1.41 cm = 5.64 cm
Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important

Hence, the volume of the cone = 289.52 cm3.

The document Chapter 13 - Surface Areas and Volume, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9 is a part of the Class 9 Course Extra Documents & Tests for Class 9.
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FAQs on Chapter 13 - Surface Areas and Volume, Solved Examples, Class 9, Maths - Extra Documents & Tests for Class 9

1. What is the formula for finding the surface area of a rectangular prism?
Ans. The formula for finding the surface area of a rectangular prism is 2(lw + lh + wh), where l represents the length, w represents the width, and h represents the height of the prism.
2. How do you calculate the volume of a cylinder?
Ans. The volume of a cylinder can be calculated using the formula V = πr^2h, where V represents the volume, r represents the radius of the base, and h represents the height of the cylinder.
3. How can I find the surface area of a cone?
Ans. To find the surface area of a cone, you can use the formula A = πr(r + l), where A represents the surface area, r represents the radius of the base, and l represents the slant height of the cone.
4. What is the formula for calculating the volume of a sphere?
Ans. The formula for calculating the volume of a sphere is V = (4/3)πr^3, where V represents the volume and r represents the radius of the sphere.
5. How do I find the surface area of a triangular prism?
Ans. To find the surface area of a triangular prism, you can use the formula A = bh + (pl), where A represents the surface area, b represents the base of the triangle, h represents the height of the triangle, p represents the perimeter of the base, and l represents the slant height of the prism.
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