NCERT Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes

Edited By Ramraj Saini | Updated on May 08, 2023 01:12 PM IST

NCERT Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes

NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes are provided here. These NCERT solutions are prepared by expert team at careers360 keeping in mind the latest CBSE syllabus 2023. These solutions are simple, comprehensive and provide indepth understanding of concepts. Practicing these solutions give confidence which ultimately lead to score well in the exam. In the NCERT syllabus of this chapter, you will study the shapes- cube, cuboid, cylinder, cone, sphere, hemisphere, etc. You have to face the questions related to the curved surface area, total surface area, volumes, and many others.

Scholarship Test: Vidyamandir Intellect Quest (VIQ)

Don't Miss: JEE Main 2027: Narayana Scholarship Test Preparation Kit for Class 9

This Story also Contains

NCERT Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes
Surface Area and Volumes Class 9 Questions And Answers PDF Free Download
Surface Area and Volumes Class 9 Solutions - Important Formulae
Surface Area and Volumes Class 9 NCERT Solutions (Intext Questions and Exercise)
Summary Of NCERT Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes
NCERT solutions for class 9 maths - Chapter Wise
Key Features of NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes

NCERT Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes

NCERT solutions for chapter 13 maths class 9 Surface Area and Volumes are designed to provide you assistance while solving the practice exercises. The language of the questions sometimes is not direct, you have to identify what the question is specifically about. There are a total of 9 exercises consisting of a total of 102 questions including the optional exercise. Surface Area and Volumes Class 9 Questions And Answers have exam-oriented solutions to all the questions of practice exercises. Here you will get NCERT solutions for class 9 Maths also.

Surface Area and Volumes Class 9 Questions And Answers PDF Free Download

Download PDF

Surface Area and Volumes Class 9 Solutions - Important Formulae

Total Surface Area (TSA):

Cuboid = 2(l x b) + 2(b x h) + 2(h x l)
Cube = 6a²
Right Circular Cylinder = 2πr(h + r)
Right Circular Cone = πr(l + r)
Sphere = 4πr²
Hemisphere = 3πr²

Lateral/Curved Surface Area (CSA):

Cuboid = 2h(l + b)
Cube = 4a²
Right Circular Cylinder = 2πrh
Right Circular Cone = πrl

Volume:

Cuboid = l x b x h
Cube = a³
Right Circular Cylinder = πr²h
Right Circular Cone = (1/3)πr²h
Sphere = (4/3)πr³
Hemisphere = (2/3)πr³

In these formulas,

l = length

b = breadth

h = height

r = radius

a = side length of the respective geometric figure

Free download NCERT Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes for CBSE Exam.

Also Read,

Surface Area and Volumes Class 9 NCERT Solutions (Intext Questions and Exercise)

Class 9 maths chapter 13 question answer - exercise: 13.1

Q1 (i) A plastic box $\small 1.5\hspace {1mm}m$ long, $\small 1.25\hspace {1mm}m$ wide and $\small 65\hspace {1mm}cm$ deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine: The area of the sheet required for making the box.

Answer:

Given, the dimensions of the plastic box

Length, $l = 1.5\ m$

Width, $b = 1.25\ m$

Depth, $h = 65\ cm = 0.65\ m$

(i) The area of the sheet required for making the box (open at the top)= Lateral surface area of the box. + Area of the base.

= $2(bh+hl)+lb$

$\\ = 2(1.25\times0.65+0.65\times1.5)+1.5\times1.25 \\ = 2(0.8125+ 0.975)+1.875= 5.45$

The required area of the sheet required for making the box is $5.45\ m^2$

Q1 (ii) A plastic box $\small 1.5\hspace {1mm} m$ long, $\small 1.25\hspace {1mm} m$ wide and $\small 65\hspace {1mm} cm$ deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine: The cost of sheet for it, if a sheet measuring $\small 1m^2$ costs Rs 20.

Answer:

Given, dimensions of the plastic box

Length, $l = 1.5\ m$

Width, $b = 1.25\ m$

Depth, $h = 65\ cm = 0.65\ m$

We know, area of the sheet required for making the box is $5.45\ m^2$

(ii) Cost for $\small 1m^2$ of sheet = Rs 20

$\therefore$ Cost for $5.45\ m^2$ of sheet = $5.45\times20 = 109$

Required cost of the sheet is $Rs.\ 109$

Q2 The length, breadth and height of a room are $\small 5\hspace {1mm}m$ , $\small 4\hspace {1mm}m$ and $\small 3\hspace {1mm}m$ respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of $\small Rs \hspace {1mm} 7.50$ per $\small m^2$ .

Answer:

Given,

Dimensions of the room = $5\ m\times4\ m\times3\ m$

Required area to be whitewashed = Area of the walls + Area of the ceiling

= $2(lh+bh) + lb$

$\\ = 2(5\times4+4\times3)+ 5\times4 \\ = 2(32)+20 = 74\ m^2$
Cost of white-washing per $\small m^2$ area = $\small Rs \hspace {1mm} 7.50$
Cost of white-washing $74\ m^2$ area = $Rs (74 \times 7.50)$
$= Rs.\ 555$

Therefore, the required cost of whitewashing the walls of the room and the ceiling is $Rs.\ 555$

Q3 The floor of a rectangular hall has a perimeter $\small 250\hspace {1mm} m$ . If the cost of painting the four walls at the rate of Rs 10 per $\small m^2$ is Rs 15000, find the height of the hall. [ Hint : Area of the four walls $\small =$ Lateral surface area.]

Answer:

Given,

The perimeter of rectangular hall = $\small 250\hspace {1mm} m$

Cost of painting the four walls at the rate of Rs 10 per $\small m^2$ = Rs 15000

Let the height of the wall be $h\ m$

$\therefore$ Area to be painted = $Perimeter\times height$

$= 250h\ m^2$

$\therefore$ Required cost = $250h\times10\ m^2 = 15000\ m^2$

$\\ \implies 2500h = 15000 \\ \implies h = \frac{150}{25} = 6\ m$

Therefore, the height of the hall is $6\ m$

Q4 The paint in a certain container is sufficient to paint an area equal to $\small 9.375\hspace{1mm} m^2$ . How many bricks of dimensions $\small 22.5\hspace {1mm}cm \times 10\hspace {1mm}cm \times 7.5\hspace {1mm}cm$ can be painted out of this container?

Answer:

Given, dimensions of the brick = $\small 22.5\hspace {1mm}cm \times 10\hspace {1mm}cm \times 7.5\hspace {1mm}cm$

We know, Surface area of a cuboid = $2(lb+bh+hl)$

$\therefore$ The surface area of a single brick = $2(22.5\times10+10\times7.5+7.5\times22.5)$

$= 2(225+75+166.75) = 937.5\ cm^2 = 0.09375\ m^2$

$\therefore$ Number of bricks that can be painted = $\frac{Total\ area\ the\ container\ can\ paint}{Surface\ area\ of\ a\ single\ brick}$

$= \frac{9.375}{0.09375} = 100$

Therefore, the required number of bricks that can be painted = 100

Q5 (i) A cubical box has each edge $\small 10 \hspace{1mm}cm$ and another cuboidal box is $\small 12.5 \hspace{1mm}cm$ long, $\small 10 \hspace{1mm}cm$ wide and $\small 8 \hspace{1mm}cm$ high.

Which box has the greater lateral surface area and by how much?

Answer:

Given,

Edge of the cubical box = $\small 10 \hspace{1mm}cm$

The lateral surface area of the cubical box = $4\times(10\times10)\ cm^2 = 400\ cm^2$

The lateral surface area of the cuboidal box = $2[lh + bh]$

=360cm²

Clearly, Lateral surface area of the cubical box is greater than the cuboidal box.

Difference between them = $400-360 = 40\ cm^2$

Q5 (ii) A cubical box has each edge $\small 10 \hspace {1mm}cm$ and another cuboidal box is $\small 12.5 \hspace {1mm}cm$ long, $\small 10 \hspace {1mm}cm$ wide and $\small 8 \hspace {1mm}cm$ high.

Which box has the smaller total surface area and by how much?

Answer:

Given,

Edge of the cubical box =

Dimensions of the cuboid = $12.5\cm \times 10\ cm \times 8\ cm$

(ii) The total surface area of the cubical box = $6\times(10\times10)\ cm^2 = 600\ cm^2$

The total surface area of the cuboidal box = 2[lh + bh+lb]

$\\ = [2(12.5 × 8 + 10 × 8 + 12.5\times10)]\ cm^2 \\ = 610\ cm^2$

Clearly, the total surface area of a cuboidal box is greater than the cubical box.

Difference between them = $610-600 = 10\ cm^2$

Q6 (i) A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is $\small 30 \hspace{1mm}cm$ long, $\small 25 \hspace{1mm}cm$ wide and $\small 25 \hspace{1mm}cm$ high.What is the area of the glass?

Answer:

Given, dimensions of the greenhouse = $30\ cm \times25\ cm \times25\ cm$

Area of the glass = $2[lb + lh + bh]$
$\\ = [2(30 \times 25 + 30 \times 25 + 25 \times 25)] \\ = [2(750 + 750 + 625)] \\ = (2 \times 2125) \\ = 4250\ cm^2$
Therefore, the area of glass is $4250\ cm^2$

Q6 (ii) A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high. How much of tape is needed for all the 12 edges?

Answer:

Given, dimensions of the greenhouse = $30\ cm \times25\ cm \times25\ cm$

(ii) Tape needed for all the 12 edges = Perimeter = $4(l+b+h)$

$4(30+25+25) = 320\ cm$

Therefore, $320\ cm$ of tape is needed for the edges.

Q7 Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions $\small 25\hspace {1mm}cm \times 20\hspace {1mm}cm \times 5 \hspace {1mm}cm$ , and the smaller of dimensions $\small 15\hspace {1mm}cm \times 12\hspace {1mm}cm \times 5 \hspace {1mm}cm$ . For all the overlaps, $\small 5\%$ of the total surface area is required extra. If the cost of the cardboard is Rs 4 for $\small 1000\hspace {1mm}cm^2$ , find the cost of cardboard required for supplying 250 boxes of each kind.

Answer:

Given,

Dimensions of the bigger box = $\small 25\hspace {1mm}cm \times 20\hspace {1mm}cm \times 5 \hspace {1mm}cm$ ,

Dimensions of smaller box = $\small 15\hspace {1mm}cm \times 12\hspace {1mm}cm \times 5 \hspace {1mm}cm$

We know,

Total surface area of a cuboid = $2(lb+bh+hl)$

$\therefore$ Total surface area of the bigger box = $2(25\times20+20\times5+5\times25)$

$= 2(500+100+125) = 1450\ cm^2$

$\therefore$ Area of the overlap for the bigger box = $5\%\ of\ 1450\ cm^2 = \frac{5}{100}\times1450 = 72.5\ cm^2$

Similarly,

Total surface area of the smaller box = $2(15\times12+12\times5+5\times15)$

$= 2(180+60+75) = 630\ cm^2$

$\therefore$ Area of the overlap for the smaller box = $5\%\ of\ 630\ cm^2 = \frac{5}{100}\times630 = 31.5\ cm^2$

Since, 250 of each box is required,

$\therefore$ Total area of carboard required = $250[(1450+72.5)+(630 + 31.5)] = 546000\ cm^2$

Cost of $\small 1000\hspace {1mm}cm^2$ of the cardboard = Rs 4

$\therefore$ Cost of $\small 546000\hspace {1mm}cm^2$ of the cardboard = $\small Rs. (\frac{4}{1000}\times546000) = Rs. 2184$

Therefore, the cost of the cardboard sheet required for 250 such boxes of each kind is $\small Rs.\ 2184$

Q8 Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height $\small 2.5\hspace {1mm}m$ , with base dimensions $\small 4\hspace {1mm}m\times 3\hspace {1mm}m$ ?

Answer:

Given, Dimensions of the tarpaulin = $4\ m\times 3\ m \times2.5\ m$

The required amount of tarpaulin = Lateral surface area of the shelter + Area of top

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

$\\ = 2(4\times2.5 + 3\times2.5 )+ 4\times3 \\ = 2(10+7.5) + 12 = 47\ m^2$

Therefore, $47\ m^2$ tarpaulin is required.

Class 9 maths chapter 13 NCERT solutions - exercise: 13.2

Q1 The curved surface area of a right circular cylinder of height $\small 14\hspace{1mm}cm$ is $\small 88\hspace{1mm}cm^2$ . Find the diameter of the base of the cylinder.

Answer:

Given,

The curved surface area of the cylinder = $\small 88\hspace{1mm}cm^2$

And, the height of the cylinder, $h= 14\ cm$

We know, Curved surface area of a right circular cylinder = $2\pi rh$

$\\ \therefore 2\pi rh = 88 \\ \implies 2.\frac{22}{7}. r. (14) = 88 \\ \\ \implies r = 1$

Therefore, the diameter of the cylinder = $1\ cm$

Q2 It is required to make a closed cylindrical tank of height $\small 1\hspace{1mm}m$ and base diameter $\small 140\hspace{1mm}cm$ from a metal sheet. How many square metres of the sheet are required for the same?

Answer:

Given,

Height of the cylindrical tank = $h = 1\ m$

Base diameter = $\small d = 140\ cm = 1.4\ m$

We know,

The total surface area of a cylindrical tank = $\small 2\pi r h+2\pi r^2 = 2\pi r(r+h)$

$\small = 2.\frac{22}{7}. \frac{1.4}{2}.(0.7+1) = 2.\frac{22}{7}. (0.7).(1.7)$

$\small = 44\times0.17$

$\small = 7.48\ m^2$

Therefore, square metres of the sheet is $\small 7.48\ m^2$

Q3 (i) A metal pipe is $\small 77\hspace{1mm}cm$ long. The inner diameter of a cross section is $\small 4\hspace{1mm}cm$ , the outer diameter being $\small 4.4\hspace{1mm}cm$ (see Fig. $\small 13.11$ ). Find its inner curved surface area,

1640782638324

Answer:

Note: There are two surfaces, inner and outer.

Given,

Height of the cylinder, $h = 77\ cm$

Outer diameter = $r_1 = 4.4\ cm$

Inner diameter = $r_2 = 4\ cm$

Inner curved surface area = $2\pi r_2h$

$\\ = 2\times\frac{22}{7}\times2\times77 \\ = 968\ cm^2$

Therefore, the inner curved surface area of the cylindrical pipe is $968\ cm^2$

Q3 (ii) A metal pipe is $\small 77\hspace{1mm}cm$ long. The inner diameter of a cross section is $\small 4\hspace{1mm}cm$ , the outerdiameter being $\small 4.4\hspace{1mm}cm$ (see Fig. $\small 13.11$ ). Find its outer surface area.

1640782678422

Answer:

Note: There are two surfaces, inner and outer.

Given,

Height of the cylinder, $h = 77\ cm$

Outer diameter = $r_1 = 4.4\ cm$

Inner diameter = $r_2 = 4\ cm$

Outer curved surface area = $2\pi r_1h$

$\\ = 2\times\frac{22}{7}\times2.2\times77 \\ = 1064.8\ cm^2$

Therefore, the outer curved surface area of the cylindrical pipe is $1064.8\ cm^2$

Q3 (iii) A metal pipe is $\small 77\hspace{1mm}cm$ long. The inner diameter of a cross section is 4 cm, the outer diameter being $\small 4.4 \hspace{1mm}cm$ (see Fig. $\small 13.11$ ). Find its total surface area.

1640782694386

Answer:

Note: There are two surfaces, inner and outer.

Given,

Height of the cylinder, $h = 77\ cm$

Outer diameter = $r_1 = 4.4\ cm$

Inner diameter = $r_2 = 4\ cm$

Outer curved surface area = $2\pi r_1h$

Inner curved surface area = $2\pi r_2h$

Area of the circular rings on top and bottom = $2\pi(r_2^2-r_1^2)$

$\therefore$ The total surface area of the pipe = $2\pi r_1h +2\pi r_2h+ 2\pi(r_2^2-r_1^2)$

$\\ = [968 + 1064.8 + 2\pi {(2.2)^2 - (2)^2}]\\ \\ = (2032.8 + 2\times \frac{22}{7}\times 0.84) \\ \\ = (2032.8 + 5.28) \\ = 2038.08\ cm^2$

Therefore, the total surface area of the cylindrical pipe is $2038.08\ cm^2$

Q4 The diameter of a roller is $\small 84\hspace{1mm}cm$ and its length is $\small 120\hspace{1mm}cm$ . It takes $\small 500$ complete revolutions to move once over to level a playground. Find the area of the playground in $\small m^2$ .

Answer:

Given,

The diameter of the cylindrical roller = $\small d = 84\hspace{1mm}cm$

Length of the cylindrical roller = $\small h = 120\hspace{1mm}cm$

The curved surface area of the roller = $2\pi r h = \pi dh$

$\\ = \frac{22}{7}\times84\times120 \\ \\ = 31680\ cm^2$

$\therefore$ Area of the playground = $Area\ covered\ in\ 1\ rotation \times500$

$\\ = 31680 \times500 \\ = 15840000 cm^2 \\ = 1584\ m^2$

Therefore, the required area of the playground = $1584\ m^2$

Q5 A cylindrical pillar is $\small 50\hspace{1mm}cm$ in diameter and $3.5\ m$ in height. Find the cost of painting the curved surface of the pillar at the rate of $\small Rs \hspace{1mm} 12.50$ per $\small m^2$ .

Answer:

Given,

Radius of the cylindrical pillar, r = $\frac{50}{2}\ cm= 25\ cm= 0.25\ m$

Height of the cylinder, h = $3.5\hspace{1mm}m$

We know,

Curved surface area of a cylinder = $2\pi r h$

$\therefore$ Curved surface area of the pillar = $2\times\frac{22}{7}\times 0.25 \times3.5$

$= 1\times22\times 0.25\ m^2$

$= 5.5\ m^2$

Now,

Cost of painting $\small 1\ m^2$ of the pillar = $\small Rs \hspace{1mm} 12.50$

$\therefore$ Cost of painting the curved surface area of the pillar = $\small Rs.\ (12.50\times5.5)$

$\small = Rs.\ 68.75$

Therefore, the cost of painting curved surface area of the pillar is $\small Rs.\ 68.75$

Q6 Curved surface area of a right circular cylinder is $\small 4.4\hspace{1mm}m^2$ . If the radius of the base of the cylinder is $\small 0.7\hspace{1mm}m$ , find its height.

Answer:

Given, a right circular cylinder

Curved surface area of the cylinder = $\small 4.4\hspace{1mm}m^2$

The radius of the base = $r = 0.7\ m$

Let the height of the cylinder be $h$

We know,

Curved surface area of a cylinder of radius $r$ and height $h$ = $2\pi r h$

$\therefore$ $2\pi r h = 4.4$

$\\ \Rightarrow 2\times\frac{22}{7}\times0.7 \times h = 4.4 \\ \\ \Rightarrow 44\times0.1 \times h = 4.4 \\ \Rightarrow h = 1\ m$

Therefore, the required height of the cylinder is $1\ m$

Q7 (i) The inner diameter of a circular well is $\small 3.5\hspace{1mm}m$ . It is $\small 10\hspace{1mm}m$ deep. Find its inner curved surface area.

Answer:

Given,

The inner diameter of the circular well = $d = \small 3.5\hspace{1mm}m$

Depth of the well = $h = 10\ m$

We know,

The curved surface area of a cylinder = $2\pi rh$

$\therefore$ The curved surface area of the well = $2\times\frac{22}{7}\times\frac{3.5}{2}\times10$

$\\ = 44 \times 0.25 \times 10 \\ = 110\ m^2$

Therefore, the inner curved surface area of the circular well is $110\ m^2$

Q7 (ii) The inner diameter of a circular well is $\small 3.5\hspace{1mm}m$ . It is $\small 10\hspace{1mm}m$ deep. Find the cost of plastering this curved surface at the rate of Rs 40 per $\small m^2$ .

Answer:

Given,

The inner diameter of the circular well = $d = \small 3.5\hspace{1mm}m$

Depth of the well = $h = 10\ m$

$\therefore$ The inner curved surface area of the circular well is $110\ m^2$

Now, the cost of plastering the curved surface per $\small m^2$ = Rs. 40

$\therefore$ Cost of plastering the curved surface of $110\ m^2$ = $Rs.\ (110 \times 40) = Rs.\ 4400$

Therefore, the cost of plastering the well is $Rs.\ 4400$

Q8 In a hot water heating system, there is a cylindrical pipe of length $\small 28\hspace{1mm}m$ and diameter $\small 5\hspace{1mm}cm$ . Find the total radiating surface in the system.

Answer:

Given,

Length of the cylindrical pipe = $l = \small 28\hspace{1mm}m$

Diameter = $\small d = 5\hspace{1mm}cm = 0.05\ m$

The total radiating surface will be the curved surface of this pipe.

We know,

The curved surface area of a cylindrical pipe of radius $r$ and length $l$ = $2\pi r l$

$\therefore$ CSA of this pipe =

$= 2\times\frac{22}{7}\times\frac{0.05}{2}\times28$

$\\ = 22\times0.05\times4 \\ = 4.4\ m^2$

Therefore, the total radiating surface of the system is $4.4\ m^2$

Q9 (i) Find the lateral or curved surface area of a closed cylindrical petrol storage tank that is $\small 4.2\hspace{1mm}m$ in diameter and $\small 4.5\hspace{1mm}m$ high.

Answer:

Given, a closed cylindrical petrol tank.

The diameter of the tank = $\small d = 4.2\hspace{1mm}m$

Height of the tank = $\small h = 4.5\hspace{1mm}m$

We know,

The lateral surface area of a cylinder of radius $\small r$ and height $\small h$ = $\small 2\pi r h$

$\small \therefore$ The lateral surface area of a cylindrical tank = $\small 2\times\frac{22}{7}\times\frac{4.2}{2}\times4.5$

$\small \\ = (44 \times 0.3 \times 4.5) \\ = 59.4\ m^2$

Therefore, the lateral or curved surface area of a closed cylindrical petrol storage tank is $\small 59.4\ m^2$

Q9 (ii) Find: how much steel was actually used, if $\small \frac{1}{12}$ of the steel actually used was wasted in making the tank.

Answer:

Given, a closed cylindrical petrol tank.

The diameter of the tank = $\small d = 4.2\hspace{1mm}m$

Height of the tank = $\small h = 4.5\hspace{1mm}m$

Now, Total surface area of the tank = $2\pi r (r + h)$

$\\ = 2 \times \frac{22}{7} \times 2.1 \times (2.1 + 4.5) \\ = (44 \times 0.3 \times 6.6) \\ = 87.12\ m^2$

Now, let $x\ m^2$ of steel sheet be actually used in making the tank

Since $\small \frac{1}{12}$ of steel was wasted, the left $\small \frac{11}{12}$ of the total steel sheet was used to made the tank.
$\small \therefore$ The total surface area of the tank = $\small \frac{11}{12}\times (Total\ area\ of\ steel\ sheet )$
$\small \\ \Rightarrow 87.12= \frac{11}{12}x \\ \Rightarrow x = \frac{87.12\times12}{11} \\ \Rightarrow x = 95.04\ m^2$

Therefore, $\small 95.04\ m^2$ of steel was actually used in making the tank.

Q10 In Fig. $\small 13.12$ , you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of $\small 20\hspace{1mm}cm$ and height of $\small 30\hspace{1mm}cm$ . A margin of $\small 2.5\hspace{1mm}cm$ is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

1640782739530 Answer:

Given, a cylindrical lampshade

The diameter of the base = $d = 20\ cm$

Height of the cylinder = $30\ cm$

The total height of lampshade= $(30+ 2.5 + 2.5) = 35\ cm$

We know,

Curved surface area of a cylinder of radius $r$ and height $h$ = $2\pi r h$

Now, Cloth required for covering the lampshade = Curved surface area of the cylinder

$\\ = 2\times\frac{22}{7} \times10 \times 35 \\ \\ = 22\times10\times10 = 2200\ cm^2$

Therefore, $2200\ cm^2$ cloth will be required for covering the lampshade.

Q11 The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius $\small 3\hspace{1mm}cm$ and height $\small 10.5\hspace{1mm}cm$ . The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?

Answer:

Given, a cylinder with a base.

The radius of the cylinder = $r = 3\ cm$

Height of the cylinder = $h = 10.5\ cm$

We know,

The lateral surface area of a cylinder of radius $r$ and height $h$ = $2\pi r h$

$\therefore$ Area of the cylindrical penholder = Lateral areal + Base area

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

$= \frac{22}{7}\times3\times[2(10.5)+3]$

$\\ = \frac{22}{7}\times3\times[24] \\ \\ = \frac{1584}{7}\ cm^2$
Area of 35 penholders = $\frac{1584}{7}\times35\ cm^2$

$= 7920\ cm^2$

Therefore, the area of carboard required is $7920\ cm^2$

Class 9 surface area and volumes NCERT solutions - exercise: 13.3

Q1 Diameter of the base of a cone is $\small 10.5 \hspace{1mm}cm$ and its slant height is $\small 10 \hspace{1mm}cm$ . Find its curved surface area.

Answer:

Given,

Base diameter of the cone = $d=10.5\ cm$

Slant height = $l=10\ cm$

We know, Curved surface area of a cone $= \pi r l$

$\therefore$ Required curved surface area of the cone=

$\\ = \frac{22}{7}\times \frac{10.5}{2}\times10 \\ \\ = 165\ cm^2$

Q2 Find the total surface area of a cone, if its slant height is $\small 21\hspace{1mm}m$ and diameter of its base is $\small 24\hspace{1mm}m$ .

Answer:

Given,

Base diameter of the cone = $d=24\ m$

Slant height = $l=21\ cm$

We know, Total surface area of a cone = Curved surface area + Base area

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

$\therefore$ Required total surface area of the cone=

$\\ = \frac{22}{7}\times\frac{24}{2}\times(21+12) \\ \\ =\frac{22}{7}\times\frac{24}{2}\times33 \\ = 1244.57 \ m^2$

Q3 (i) Curved surface area of a cone is $\small 308\hspace{1mm}cm^2$ and its slant height is 14 cm. Find radius of the base .

Answer:

Given,

The curved surface area of a cone = $\small 308\hspace{1mm}cm^2$

Slant height $= l = 14\ cm$

(i) Let the radius of cone be $r\ cm$

We know, the curved surface area of a cone= $\pi rl$

$\therefore$ $\\ \pi rl = 308 \\ \\ \Rightarrow \frac{22}{7}\times r\times14 = 308 \\ \Rightarrow r = \frac{308}{44} = 7$

Therefore, the radius of the cone is $7\ cm$

Q3 (ii) Curved surface area of a cone is $\small 308\hspace{1mm}cm^2$ and its slant height is $\small 14\hspace{1mm}cm$ . Find total surface area of the cone.

Answer:

Given,

The curved surface area of a cone = $\small 308\hspace{1mm}cm^2$

Slant height $= l = 14\ cm$

The radius of the cone is $r =$ $7\ cm$

(ii) We know, Total surface area of a cone = Curved surface area + Base area

$= \pi r l + \pi r^2$

$\\ = 308+\frac{22}{7}\times 7^2 \\ = 308+154 = 462\ cm^2$

Therefore, the total surface area of the cone is $462\ cm^2$

Q4 (i) A conical tent is 10 m high and the radius of its base is 24 m. Find slant height of the tent.

Answer:

Given,

Base radius of the conical tent = $r=24\ m$

Height of the conical tent = $h=10\ m$

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

$\\ =\sqrt{10^2+24^2} \\ = \sqrt{676} \\ = 26\ m$

Therefore, the slant height of the conical tent is $26\ m$

Q4 (ii) A conical tent is 10 m high and the radius of its base is 24 m. Find cost of the canvas required to make the tent, if the cost of $\small 1\hspace{1mm}m^2$ canvas is Rs 70.

Answer:

Given,

Base radius of the conical tent = $r=24\ m$

Height of the conical tent = $h=10\ m$

$\therefore$ Slant height = $l=\sqrt{h^2+r^2} = 26\ m$

We know, Curved surface area of a cone $= \pi r l$

$\therefore$ Curved surface area of the tent

$\\ = \frac{22}{7}\times24\times26 \\ \\ =\frac{13728}{7}\ m^2$

Cost of $1\ m^2$ of canvas = $Rs.\ 70$

$\therefore$ Cost of $\frac{13728}{7}\ m^2$ of canvas =

$Rs.\ (\frac{13728}{7}\times70) = Rs.\ 137280$

Therefore, required cost of canvas to make tent is $Rs.\ 137280$

Q5 What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use $\small \pi =3.14$ ).

Answer:

Given,

Base radius of the conical tent = $r =6\ m$

Height of the tent = $h =8\ m$

We know,

Curved surface area of a cone = $\pi rl = \pi r\sqrt{h^2 + r^2}$

$\therefore$ Area of tarpaulin required = Curved surface area of the tent

$\\ =3.14\times6\times \sqrt{8^2+ 6^2} \\ = 3.14\times6\times 10 \\ = 188.4\ m^2$

Now, let the length of the tarpaulin sheet be $x\ m$

Since $20\ cm$ is wasted, effective length = $x - 20 cm = (x - 0.2)\ m$

Breadth of tarpaulin = $3\ m$

$\\ \therefore [(x - 0.2) \times 3] = 188.4 \\ \Rightarrow x - 0.2 = 62.8 \\ \Rightarrow x = 63\ m$

Therefore, the length of the required tarpaulin sheet will be 63 m.

Q6 The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs 210 per $\small 100\hspace{1mm}m^2$ .

Answer:

Given, a conical tomb

The base diameter of the cone = $d =14\ m$

Slant height $= l = 25\ m$

We know, Curved surface area of a cone $= \pi r l$

$\\ = \frac{22}{7}\times\frac{14}{2}\times25 \\ \\ = 22\times25 \\ = 550\ m^2$

Now, Cost of whitewashing per $\small 100\hspace{1mm}m^2$ = $\small Rs.\ 210$

$\therefore$ Cost of whitewashing per $\small 550\hspace{1mm}m^2$ = $\small \\ Rs. (\frac{210}{100}\times550 )$

$\small \\ = Rs.\ (21\times55 ) = Rs.\ 1155$

Therefore, the cost of white-washing its curved surface of the tomb is $\small Rs.\ 1155$ .

Q7 A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Answer:

Given, a right circular cone cap (which means no base)

Base radius of the cone = $r=7\ cm$

Height $= h = 24\ cm$

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

We know, Curved surface area of a right circular cone $= \pi r l$

$\therefore$ The curved surface area of a cap =

$\\ = \frac{22}{7}\times7\times\sqrt{24^2+7^2} \\ \\ = 22\times\sqrt{625} \\ = 22\times25\ \\ = 550\ cm^2$

$\therefore$ The curved surface area of 10 caps = $550\times10 = 5500\ cm^2$

Therefore, the area of the sheet required for 10 caps = $5500\ cm^2$

Q8 A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs 12 per $\small m^2$ , what will be the cost of painting all these cones? (Use $\small \pi =3.14$ and take $\small \sqrt{1.04}=1.02$ )

Answer:

Given, hollow cone.

The base diameter of the cone = $d = 40\ cm = 0.4\ m$

Height of the cone = $h = 1\ m$

$\therefore$ Slant height = $l = \sqrt{h^2+r^2}$ $= \sqrt{1^2+0.2^2}$

We know, Curved surface area of a cone = $\pi r l = \pi r\sqrt{h^2+r^2}$

$\therefore$ The curved surface area of 1 cone = $3.14\times0.2\times\sqrt{1.04} = 3.14\times0.2\times1.02$

$= 0.64056\ m^2$

$\therefore$ The curved surface area of 50 cones $= (50\times0.64056)\ m^2$

$= 32.028\ m^2$

Now, the cost of painting $\small 1\ m^2$ area = $\small Rs.\ 12$

$\therefore$ Cost of the painting $32.028\ m^2$ area $= Rs.\ (32.028\times12)$

$= Rs.\ 384.336$

Therefore, the cost of painting 50 such hollow cones is $Rs.\ 384.34\ (approx)$

NCERT Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes - exercise: 13.4

Q1 (i) Find the surface area of a sphere of radius: $\small 10.5\hspace{1mm}cm$ .

Answer:

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ Required surface area = $4\times\frac{22}{7} \times(10.5)^2$

$\\ =88\times1.5\times10.5 \\ = 1386\ cm^2$

Q1 (ii) Find the surface area of a sphere of radius: $\small 5.6\hspace{1mm}cm$

Answer:

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ Required surface area = $4\times\frac{22}{7} \times(5.6)^2$

$\\ =88\times0.8\times5.6 \\ = 394.24\ cm^2$

Q1 (iii) Find the surface area of a sphere of radius: $\small 14\hspace{1mm}cm$

Answer:

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ Required surface area = $4\times\frac{22}{7} \times(14)^2$

$\\ = 88\times2\times14 \\ = 2464 \ cm^2$

Q2 (i) Find the surface area of a sphere of diameter: 14 cm

Answer:

Given,

The diameter of the sphere = $14\ cm$

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ Required surface area = $4\times\frac{22}{7} \times\left (\frac{14}{2} \right )^2$

$= 4\times\frac{22}{7} \times\frac{14}{2}\times\frac{14}{2}$

$\\ = 22\times2\times14 \\ = 616\ cm^2$

Q2 (ii) Find the surface area of a sphere of diameter: 21 cm

Answer:

Given,

The diameter of the sphere = $21\ cm$

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ Required surface area = $4\times\frac{22}{7} \times\left (\frac{21}{2} \right )^2$

$= 4\times\frac{22}{7} \times\frac{21}{2}\times\frac{21}{2}$

$\\ = 22\times3\times21 \\ = 1386\ cm^2$

Q2 (iii) Find the surface area of a sphere of diameter: $\small 3.5\hspace{1mm}m$

Answer:

Given,

The diameter of the sphere = $3.5\ m$

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ Required surface area = $4\times\frac{22}{7} \times\left (\frac{3.5}{2} \right )^2$

$= 4\times\frac{22}{7} \times\frac{3.5}{2}\times\frac{3.5}{2}$

$\\ = 22\times0.5\times3.5 \\ = 38.5\ m^2$

Q3 Find the total surface area of a hemisphere of radius 10 cm. (Use $\small \pi =3.14$ )

Answer:

We know,

The total surface area of a hemisphere = Curved surface area of hemisphere + Area of the circular end

$= 2\pi r^2 + \pi r^2 = 3\pi r^2$

$\therefore$ The required total surface area of the hemisphere = $3\times3.14\times(10)^2$

$\\ = 942\ cm^2$

Q4 The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Answer:

Given,

$r_1 = 7\ cm$

$r_2 = 14\ cm$

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ The ratio of surface areas of the ball in the two cases = $\frac{Initial}{Final} = \frac{4\pi r_1^2}{4\pi r_2^2}$

$= \frac{r_1^2}{r_2^2}$

$\\ = \left (\frac{7}{14} \right )^2 \\ \\ = \left (\frac{1}{2} \right )^2 \\ \\ = \frac{1}{4}$

Therefore, the required ratio is $1:4$

Q5 A hemispherical bowl made of brass has inner diameter $\small 10.5\hspace{1mm}cm$ . Find the cost of tin-plating it on the inside at the rate of Rs 16 per $\small 100\hspace{1mm}cm^2$ .

Answer:

Given,

The inner radius of the hemispherical bowl = $r = \frac{10.5}{2}\ cm$

We know,

The curved surface area of a hemisphere = $2\pi r^2$

$\therefore$ The surface area of the hemispherical bowl = $2\times\frac{22}{7}\times\left (\frac{10.5}{2} \right )^2$

$=11\times1.5\times10.5$

$= 173.25 \ cm^2$

Now,

Cost of tin-plating $\small 100\hspace{1mm}cm^2$ = Rs 16

$\therefore$ Cost of tin-plating $\small 33\hspace{1mm}cm^2$ = $\small \\ Rs. \left (\frac{16}{100}\times173.25 \right )$

$\small = Rs. 27.72$

Therefore, the cost of tin-plating it on the inside is $\small Rs. 27.72$

Q6 Find the radius of a sphere whose surface area is $\small 154\hspace{1mm}cm^2$ .

Answer:

Given,

The surface area of the sphere = $\small 154\hspace{1mm}cm^2$

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore 4\pi r^2 = 154$

$\\ \Rightarrow 4\times\frac{22}{7}\times r^2 = 154$

$\\ \Rightarrow r^2 = \frac{154\times7}{4\times22} = \frac{7\times7}{4}$

$\\ \Rightarrow r = \frac{7}{2}$

$\\ \Rightarrow r = 3.5\ cm$

Therefore, the radius of the sphere is $3.5\ cm$

Q7 The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Answer:

Let diameter of Moon be $d_m$ and diameter of Earth be $d_e$

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ The ratio of their surface areas = $\frac{Surface\ area\ of\ moon}{Surface\ area\ of\ Earth}$

$= \frac{4\pi \left (\frac{d_m}{2} \right )^2}{4\pi \left (\frac{d_e}{2} \right )^2}$

$= \frac{d_m^2}{d_e^2}$

$=\left ( \frac{\frac{1}{4}d_e}{d_e} \right )^2$

$= \frac{1}{16}$

Therefore, the ratio of the surface areas of the moon and earth is $= 1:16$

Q8 A hemispherical bowl is made of steel, $\small 0.25\hspace{1mm}cm$ thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Answer:

Given,

The inner radius of the bowl = $r_1 = 5\ cm$

The thickness of the bowl = $\small 0.25\hspace{1mm}cm$

$\therefore$ Outer radius of the bowl = (Inner radius + thickness) =

$r_2 = 5+0.25 = 5.25\ cm$

We know, Curved surface area of a hemisphere of radius $r$ = $2\pi r^2$

$\therefore$ The outer curved surface area of the bowl = $2\pi r_2^2$

$= 2\times\frac{22}{7}\times (5.25)^2$

$= 2\times\frac{22}{7}\times5.25\times5.25 = 173.25\ cm^2$

Therefore, the outer curved surface area of the bowl is $173.25\ cm^2$

Q9 (i) A right circular cylinder just encloses a sphere of radius $\small r$ (see Fig. $\small 13.22$ ). Find surface area of the sphere,

1640782797156

Answer:

Given,

The radius of the sphere = $r$

$\therefore$ Surface area of the sphere = $4\pi r^2$

Q9 (ii) A right circular cylinder just encloses a sphere of radius $\small r$ (see Fig. $\small 13.22$ ). Find curved surface area of the cylinder,

1640782822853

Answer:

Given,

The radius of the sphere = $r$

$\therefore$ The surface area of the sphere = $4\pi r^2$

According to the question, the cylinder encloses the sphere.

Hence, the diameter of the sphere is the diameter of the cylinder.

Also, the height of the cylinder is equal to the diameter of the sphere.

We know, the curved surface area of a cylinder = $2\pi rh$

$= 2\pi r(2r) = 4\pi r^2$

Therefore, the curved surface area of the cylinder is $4\pi r^2$

Q9 (iii) A right circular cylinder just encloses a sphere of radius $\small r$ (see Fig. $\small 13.22$ ). Find ratio of the areas obtained in (i) and (ii).

1640782835952

Answer:

The surface area of the sphere = $4\pi r^2$

And, Surface area of the cylinder = $4\pi r^2$

So, the ratio of the areas = $\frac{4\pi r^2}{4\pi r^2} = 1$

NCERT Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes - exercise: 13.5

Q1 A matchbox measures $\small 4\hspace{1mm}cm\times 2.5\hspace{1mm}cm\times 1.5\hspace{1mm}cm$ . What will be the volume of a packet containing 12 such boxes?

Answer:

Given,

Dimensions of a matchbox = $\small 4\hspace{1mm}cm\times 2.5\hspace{1mm}cm\times 1.5\hspace{1mm}cm$

We know,

The volume of a cuboid = $l\times b \times h$

$\small \therefore$ The volume of a matchbox= $\small (4 \times2.5\times 1.5)\ cm^3 = 15\ cm^3$

$\small \therefore$ Volume of 12 such matchboxes = $\small (15\times12)\ cm^3 = 180\ cm^3$

Therefore, the volume of a packet containing 12 matchboxes is $\small 180\ cm^3$

Q2 A cuboidal water tank is 6 m long, 5 m wide and $\small 4.5\hspace{1mm}m$ deep. How many litres of water can it hold? ( $\small 1\hspace{1mm}m^3=1000\hspace{1mm}l$ )

Answer:

Given,

Dimensions of the cuboidal water tank = $6\ m \times5\ m\times 4.5\ m$

We know,

Volume of a cuboid = $l\times b \times h$

$\small \therefore$ Volume of the water tank = $\small (6 \times5\times 4.5)\ m^3 = 135\ m^3$

We know,

$\small 1\hspace{1mm}m^3=1000\hspace{1mm}l$

$\small \therefore$ Volume of water the tank can hold = $\small (135\times1000) = 135000\ litres$

Q3 A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?

Answer:

Let the height of the vessel be $h\ m$

Dimensions of the cuboidal water tank = $10\ m \times8\ m\times h\ m$

We know,

The volume of a cuboid = $l\times b \times h$

$\small \therefore$ The volume of the water tank = $\small (10 \times8\times h)\ m^3 = 80h\ m^3$

According to question,

$\small \\ \Rightarrow h= \frac{380}{80} = 4.75$

Therefore, the required height of the vessel is $\small 4.75\ m$

Q4 Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of Rs 30 per $\small m^3$ .

Answer:

Given,

Dimensions of the cuboidal pit = $8\ m\times 6\ m\times 3\ m$

We know , Volume of a cuboid = $l\times b\times h$

$\therefore$ Volume of the cuboidal pit = $(8\times 6\times 3\) = 144\ m^3$

Now, Cost of digging $\small 1\ m^3$ = Rs. 30

$\therefore$ Cost of digging $144\ m^3$ = $Rs.(144\times30)$

$= Rs.4320$

Q5 The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively $\small 2.5\hspace{1mm}m$ and 10 m.

Answer:

Given,

Length of the tank = $\small l = 2.5\hspace{1mm}m$

Depth of the tank = $h = 10\ m$

Let the breadth of the tank be $\small b\ m$

We know, Volume of a cuboid = $l\times b\times h$

$\therefore$ The volume of the cuboidal tank = $(2.5\times b\times 10) = 25b\ m^3$

We know, $1\ m^3 = 1000\ litre$

$\therefore$ $25b\ m^3 = 50000\ litre = 50\ m^3$

$\\ \Rightarrow 25b = 50 \\ \Rightarrow b = 2\ m$

Therefore, the breadth of the tank is $2\ m$

Q6 A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring $\small 20\hspace{1mm}m\times 15\hspace{1mm}m\times 6\hspace{1mm}m$ . For how many days will the water of this tank last?

Answer:

Given,

Dimensions of the tank = $\small 20\hspace{1mm}m\times 15\hspace{1mm}m\times 6\hspace{1mm}m$

4000 people requiring 150 litres of water per head per day

The total volume of water required = $4000\times150 = 600000\ litres$

Let the number of days the water will last be $n$

We know,

The volume of a cuboid = $l\times b \times h$

$\small \therefore$ The volume of the water tank = $\small (20 \times15\times 6)\ m^3 = 1800\ m^3 = 1800000\ litres$

According to question,

$\small n\times600000 = 1800000$

$\small \\ \Rightarrow n= 3$

Therefore, the water in the tank will last for 3 days.

Q7 A godown measures $\small 40\hspace{1mm}m\times 25\hspace{1mm}m\times 15\hspace{1mm}m$ . Find the maximum number of wooden crates each measuring $1.5\hspace{1mm}m\times 1.25\hspace{1mm}m\times 0.5\hspace{1mm}m$ that can be stored in the godown.

Answer:

Given,

Dimensions of the godown = $\small 40\hspace{1mm}m\times 25\hspace{1mm}m\times 15\hspace{1mm}m$

Dimension of each wooden crate = $\small 1.5\hspace{1mm}m\times 1.25\hspace{1mm}m\times 0.5\hspace{1mm}m$

We know , Volume of a cuboid = $l\times b\times h$

$\therefore$ Volume of the godown = $(40\times 25\times 15)\ m^3$

$\therefore$ Volume of the each crate= $(1.5\times 1.25\times 0.5)\ m^3$

Let number of wooden crates be $n$

$\therefore$ Volume of $n$ wooden crates = Volume of the godown

$n\times(1.5\times 2.5\times 0.5)\ m^3 = (40\times 25\times 15)\ m^3$

$\\ \Rightarrow n = \frac{40\times 25\times 15}{1.5\times 1.25\times 0.5} \\ \\ \Rightarrow n = 80\times 20\times 10 = 16000$

Q8 A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.

Answer:

This is an important question.

Given,

Side of a solid cube = $l = 12\ cm$

We know, the volume of a cube of side $l$ = $l^3$

$\therefore$ The volume of the given cube = $12^3\ cm^3$

Now, the cube is cut into 8 equal cubes of side $a$ (let)

$\therefore$ The total volume of these 8 cubes = Volume of the bigger cube

$\Rightarrow (8\times a^3)\ cm^3 = 12\ cm^3$

$\\ \Rightarrow a^3 = \frac{12}{8} \\ \Rightarrow a^3 = \left (\frac{12}{2} \right ) \\ \Rightarrow a = 6\ cm$

Therefore, the side of the new cube is $6\ cm$

Now, we know,

The surface area of a cube of side $l$ = $6l^2$

$\therefore$ The ratio between their surface areas $= \frac{Surface\ area\ of\ bigger\ cube}{Surface\ area\ of\ a\ smaller\ cube}$

$\\ = \frac{6l^2}{6a^2} \\ = \left (\frac{l}{a} \right )^2 \\ = \left (\frac{12}{6} \right )^2 \\ = 4$

Therefore, the ratio of their surface areas is $4:1$

Q9 A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?

Answer:

Given,
Rate of water flow = $2\ km\ per\ hour$
$= \frac{2\times1000}{60}\ m/min$
$= \frac{100}{3}\ m/min$
Depth of river, $h = 3\ m$
Width of the river, $b= 40\ m$
The volume of water flowing in 1 min = $Rate\ of\ flow \times Cross\ sectional\ area$

$= \frac{100}{3}\times40\times3$

$= 4000\ m^3$

Therefore, water falling into the sea in a minute = $4000\ m^3$

Surface area and volumes class 9 solutions - exercise: 13.6

Q1 The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? ( $\small 1000\hspace{1mm}cm^3=1l$ )

Answer:

Given,

Circumference of the base of a cylindrical vessel = $132\ cm$

Height = $h = 25\ cm$

Let the radius of the base of the cylinder be $r\ cm$

Now, Circumference of the circular base of the cylinder = $2\pi r = 132\ cm$

$\\ \Rightarrow 2\times\frac{22}{7}\times r = 12\times11 \\ \\ \Rightarrow r = \frac{12\times7}{4} = 21\ cm$

$\therefore$ The volume of the cylinder = $\pi r^2 h$

$\\ = \frac{22}{7}\times (21)^2 \times25 \\ = 22\times3\times21\times25 \\ = 34650\ cm^3$

Also, $\small 1000\hspace{1mm}cm^3=1l$

$\therefore$ $34650\ cm^3 = \frac{34650}{1000} = 34.65\ litres$

Therefore, the cylindrical vessel can hold $34.65\ litres$ of water.

Q2 The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if $\small 1\hspace{1mm}cm^3$ of wood has a mass of $\small 0.6\hspace{1mm}g$ .

Answer:

Given, A hollow cylinder made of wood.

The inner diameter of the cylindrical wooden pipe = $r_1 = 24\ cm$
Outer diameter = $r_2 = 28\ cm$

Length of the pipe = $h= 35\ cm$

$\therefore$ The volume of the wooden cylinder = $\pi r^2 h = \pi (r_2^2 - r_1^2)h$

$\\ = \frac{22}{7}\times (14^2-12^2) \times35 \\ = 22\times (14-12)(14+12) \times5 \\ = 22\times (2)(26) \times5 \\ = 5720\ cm^3$

Mass of $\small 1\hspace{1mm}cm^3$ wood = $\small 0.6\hspace{1mm}g$
Mass of $5720\ cm^3$ wood = $5720\times0.6\ g$
$\\ = 3432\ g \\ = 3.432\ kg$

Therefore, the mass of the wooden hollow cylindrical pipe is $3.432\ kg$

Q3 A soft drink is available in two packs – (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

Answer:

Given,

(i) The dimension of the rectangular base of the tin can = $5\ cm\times4\ cm$

Height of the can = $h = 15\ cm$

$\therefore$ The volume of the tin can = $Rectangular\ area\times height$

$(5\times4\times15)\ cm^3 = 300\ cm^3$

(ii) The radius of the circular base of the plastic cylinder = $r = \frac{7}{2}= 3.5\ cm$

Height of the cylinder = $h = 10\ cm$

$\therefore$ The volume of the plastic cylinder = $\pi r^2 h$

$\\ = \frac{22}{7}\times (3.5)^2 \times10 \\ = 22\times0.5\times3.5\times10 \\ = 11\times35 \\ = 385\ cm^3$

Clearly, the plastic cylinder has more capacity than the rectangular tin can.

The difference in capacity = $(385-300)\ cm^3 = 85\ cm^3$

Q4 (i) If the lateral surface of a cylinder is $\small 94.2\hspace{1mm}cm^2$ and its height is 5 cm, then find radius of its base (Use $\small \pi =3.14$ )

Answer:

Given,

The lateral surface area of the cylinder = $\small 94.2\hspace{1mm}cm^2$

Height of the cylinder = $h = 5\ cm$

(i) Let the radius of the base be $r\ cm$

We know,

The lateral surface area of a cylinder = $2\pi r h$

$\\ \therefore 2\pi r h = 94.2 \\ \Rightarrow 2\times3.14\times r\times5 = 94.2 \\ \\ \Rightarrow r = \frac{94.2}{31.4} = 3\ cm$

Therefore, the radius of the base is $3\ cm$

Q4 (ii) If the lateral surface of a cylinder is $\small 94.2\hspace{1mm}cm^2$ and its height is 5 cm, then find it's volume. (Use $\small \pi =3.14$ )

Answer:

Given,

The lateral surface area of the cylinder = $\small 94.2\hspace{1mm}cm^2$

Height of the cylinder = $h = 5\ cm$

The radius of the base is $3\ cm$

(ii) We know,

The volume of a cylinder = $\pi r^2 h$

$\\ = 3.14\times3^2\times5 \\ = 3.14 \times 9 \times 5= 3.14 \times 45 \\ = 141.3\ cm^3$

Therefore, the volume of the cylinder is $141.3\ cm^3$ .

Q5 (i) It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs 20 per $\small m^2$ , find inner curved surface area of the vessel.

Answer:

(i) Given,

Rs 20 is the cost of painting $1\small m^2$ area of the inner curved surface of the cylinder.

$\therefore$ Rs 2200 is the cost of painting $\frac{1}{20}\times2200 = 110\ m^2$ area of the inner curved surface of the cylinder.

$\therefore$ The inner curved surface area of the cylindrical vessel = $110\ m^2$

Q5 (ii) It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs 20 per $\small m^2$ , find radius of the base.

Answer:

The inner curved surface area of the cylindrical vessel = $110\ m^2$

Height of the cylinder = $h = 10\ m$

Let the radius of the circular base be $r \ m$

$\therefore$ The inner curved surface area of the cylindrical vessel = $2\pi r h$

$\\ \Rightarrow 2\times\frac{22}{7}\times r \times10 = 110 \\ \Rightarrow 44\times r = 11\times7 \\ \Rightarrow 4\times r = 7 \\ \Rightarrow r = \frac{7}{4} = 1.75\ m$

Therefore, the radius of the base of the vessel is $1.75 \ m$

Q5 (iii) It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs 20 per $\small m^2$ , find capacity of the vessel.

Answer:

(iii) Height of the cylinder = $h = 10\ m$

Radius of the base of the vessel = $r =1.75 \ m$

$\therefore$ Volume of the cylindrical vessel = $\pi r^2 h$

$\\ = \frac{22}{7}\times (1.75)^2 \times10 \\ = 22\times2.5\times1.75\times10 \\ = 55\times17.5 \\ = 96.25 \ m^3$

Therefore, the capacity of the cylindrical vessel is $96.25 \ m^3$

Q6 The capacity of a closed cylindrical vessel of height 1 m is $\small 15.4$ litres. How many square metres of metal sheet would be needed to make it?

Answer:

(Using capacity(volume), we will find the radius and then find the surface area)

The capacity of the vessel = Volume of the vessel = $\small 15.4$ litres

Height of the cylindrical vessel = $h = 1\ m$

Let the radius of the circular base be $r \ m$

$\therefore$ The volume of the cylindrical vessel = $\pi r^2 h$

$\\ \Rightarrow \frac{22}{7}\times r^2 \times1 = 15.4\ litres= 0.0154\ m^3 \\ \Rightarrow r^2 = \frac{0.0014\times11\times7}{22} \\ \Rightarrow r^2 = \frac{7\times7}{10000} \\ \Rightarrow r = \frac{7}{100}= 0.07\ m$

Therefore, the total surface area of the vessel = $2\pi r h+ 2\pi r^2 = 2\pi r(r+h)$

$\\ = 2\times\frac{22}{7}\times0.07\times(0.07+1) \\ = 0.44 \times 1.07 \\ = 0.4708\ m^2$

Therefore, square metres of metal sheet needed to make it is $0.4708\ m^2$

Q7 A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

Answer:

Given,

Length of the cylindrical pencil = $h = 14\ cm$

The radius of the graphite (Inner solid cylinder) = $r_1 = \frac{1}{2}\ mm = 0.05\ cm$

Radius of the pencil (Inner solid graphite cylinder + Hollow wooden cylinder) =

= $r_2 = \frac{7}{2}\ mm = 0.35\ cm$

We know, Volume of a cylinder= $\pi r^2 h$

$\therefore$ The volume of graphite = $\pi r_1^2 h$

$\\ = \frac{22}{7}\times 0.05^2 \times14 \\ = 44\times0.0025 \\ = 0.11\ cm^3$

And, Volume of wood = $\pi (r_2^2- r_1^2) h$

$\\ = \frac{22}{7}\times (0.35^2-0.05^2) \times14 \\ = 44\times(0.35-0.05)(0.35+0.05) \\ = 44\times(0.30)(0.40) \\ = 5.28 \ cm^3$

Therefore, the volume of wood is $5.28 \ cm^3$ and the volume of graphite is $0.11 \ cm^3$

Q8 A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?

Answer:

Given,

Height = $h = 4\ cm$

The radius of the cylindrical bowl = $r_1 = \frac{7}{2}\ cm = 3.5\ cm$

$\therefore$ The volume of soup in a bowl for a single person = $\pi r^2 h$

$\\ = \frac{22}{7}\times (3.5)^2 \times4 \\ = 88\times0.5\times3.5 \\ = 154\ cm^3$
$\therefore$ The volume of soup given for 250 patients = $(250 \times 154)\ cm^3$
$\\ = 38500\ cm^3 \\ = 38.5\ litres$

Therefore, the amount of soup the hospital has to prepare daily to serve 250 patients is $38.5\ litres$

Surface area and volumes class 9 ncert solutions - exercise: 13.7

Q1 (i) Find the volume of the right circular cone with radius 6 cm, height 7 cm

Answer:

Given,

Radius = $r =6\ cm$

Height = $h =7\ cm$

We know,

Volume of a right circular cone = $\frac{1}{3}\pi r^2 h$

$\therefore$ Required volume = $\frac{1}{3}\times\frac{22}{7}\times6^2\times7$

$\\ = 22\times2\times6 \\ = 264\ cm^3$

Q1 (ii) Find the volume of the right circular cone with: radius $\small 3.5$ cm, height 12 cm

Answer:

Given,

Radius = $r =3.5\ cm$

Height = $h =12\ cm$

We know,

Volume of a right circular cone = $\frac{1}{3}\pi r^2 h$

$\therefore$ Required volume = $\frac{1}{3}\times\frac{22}{7}\times3.5^2\times12$

$\\ = 22\times0.5\times3.5\times4 \\ = 11\times14 \\ = 154\ cm^3$

Q2 (i) Find the capacity in litres of a conical vessel with radius 7 cm, slant height 25 cm

Answer:

Given,

Radius = $r =7\ cm$

Slant height = $l = \sqrt{r^2 + h^2} = 25\ cm$

Height = $h =\sqrt{l^2-r^2} = \sqrt{25^2-7^2}$

$= \sqrt{(25-7)(25+7)} = \sqrt{(18)(32)}$

$= 24\ cm$

We know,
Volume of a right circular cone = $\frac{1}{3}\pi r^2 h$

$\therefore$ Volume of the vessel= $\frac{1}{3}\times\frac{22}{7}\times7^2\times24$

$\\ = 22\times7\times8\\ = 154\times8 \\ = 1232\ cm^3$

$\therefore$ Required capacity of the vessel =

$= \frac{1232}{1000} = 1.232\ litres$

Q2 (ii) Find the capacity in litres of a conical vessel with height 12 cm, slant height 13 cm

Answer:

Given,

Height = $h =12\ cm$

Slant height = $l = \sqrt{r^2 + h^2} = 13\ cm$

Radius = $r =\sqrt{l^2-h^2} = \sqrt{13^2-12^2}$

$= \sqrt{(13-12)(13+12)} = \sqrt{(1)(25)}$

$= 5\ cm$

We know,
Volume of a right circular cone = $\frac{1}{3}\pi r^2 h$

$\therefore$ Volume of the vessel= $\frac{1}{3}\times\frac{22}{7}\times5^2\times12$

$\\ = \frac{22}{7}\times25\times4\\ = \frac{2200}{7}\ cm^3$

$\therefore$ Required capacity of the vessel =

$= \frac{2200}{7\times1000} = \frac{11}{35}\ litres$

Q3 The height of a cone is 15 cm. If its volume is 1570 $\small cm^3$ , find the radius of the base. (Use $\small \pi =3.14$ )

Answer:

Given,

Height of the cone = $h =15\ cm$

Let the radius of the base of the cone be $r\ cm$

We know,
The volume of a right circular cone = $\frac{1}{3}\pi r^2 h$

$\therefore$ $\frac{1}{3}\times3.14\times r^2\times15 = 1570$

$\\ \Rightarrow 3.14\times r^2\times5 = 1570 \\ \Rightarrow r^2 = \frac{1570}{15.7} \\ \Rightarrow r^2 = 100 \\ \Rightarrow r = 10\ cm$

Q4 If the volume of a right circular cone of height 9 cm is $\small 48\pi \hspace{1mm}cm^3$ , find the diameter of its base.

Answer:

Given,

Height of the cone = $h =9\ cm$

Let the radius of the base of the cone be $r\ cm$

We know,
The volume of a right circular cone = $\frac{1}{3}\pi r^2 h$

$\therefore$ $\frac{1}{3}\times\pi\times r^2\times9 = 48\pi$

$\\ \Rightarrow 3r^2 = 48 \\ \Rightarrow r^2 = 16\\ \Rightarrow r = 4\ cm$

Therefore the diameter of the right circular cone is $8\ cm$

Q5 A conical pit of top diameter $\small 3.5$ m is 12 m deep. What is its capacity in kilolitres?

Answer:

Given,

Depth of the conical pit = $h =12\ m$

The top radius of the conical pit = $r = \frac{3.5}{2}\ m$

We know,
The volume of a right circular cone = $\frac{1}{3}\pi r^2 h$

$\therefore$ The volume of the conical pit =

$= \frac{1}{3}\times\frac{22}{7}\times \left (\frac{3.5}{2} \right )^2\times12$

$\\ = \frac{1}{3}\times\frac{22}{7}\times \frac{3.5\times 3.5}{4}\times12 \\ \\ = 22\times 0.5\times 3.5 \\ = 38.5\ m^3$

Now, $1\ m^3 = 1\ kilolitre$

$\therefore$ The capacity of the pit = $38.5\ kilolitre$

Q6 (i) The volume of a right circular cone is $\small 9856\hspace{1mm}cm^3$ . If the diameter of the base is 28 cm, find height of the cone

Answer:

Given, a right circular cone.

The radius of the base of the cone = $r = \frac{28}{2} = 14\ cm$

The volume of the cone = $\small 9856\hspace{1mm}cm^3$

(i) Let the height of the cone be $h\ m$

We know,
The volume of a right circular cone = $\frac{1}{3}\pi r^2 h$

$\therefore$ $\frac{1}{3}\times\frac{22}{7}\times(14)^2\times h = 9856$

$\\ \Rightarrow \frac{1}{3}\times\frac{22}{7}\times14\times14\times h = 9856 \\ \Rightarrow \frac{1}{3}\times22\times2\times14\times h = 9856 \\ \Rightarrow h = \frac{9856\times3}{22\times2\times14} \\ \\ \Rightarrow h =48\ cm$

Therefore, the height of the cone is $48\ cm$

Q6 (ii) The volume of a right circular cone is $\small 9856\hspace{1mm}cm^3$ . If the diameter of the base is 28 cm, find slant height of the cone

Answer:

Given, a right circular cone.

The volume of the cone = $\small 9856\hspace{1mm}cm^3$

The radius of the base of the cone = $r = \frac{28}{2} = 14\ cm$

And the height of the cone = $h = 48\ cm$

(ii) We know, Slant height, $l = \sqrt{r^2+h^2}$

$\\ \Rightarrow l = \sqrt{14^2+48^2} \\ \Rightarrow l = \sqrt{196+2304} = \sqrt{2500} \\ \Rightarrow l = 50\ cm$

Therefore, the slant height of the cone is $50\ cm$ .

Q7 (iii) The volume of a right circular cone is $\small 9856\hspace{1mm}cm^3$ . If the diameter of the base is 28 cm, find curved surface area of the cone

Answer:

Given, a right circular cone.

The radius of the base of the cone = $r = \frac{28}{2} = 14\ cm$

And Slant height of the cone = $l = 50\ cm$

(iii) We know,

The curved surface area of a cone = $\pi r l$

$\therefore$ Required curved surface area= $\frac{22}{7}\times14\times50$

$\\ = 22\times2\times50 \\ = 2200\ cm^2$

Q7 A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.

Answer:

When a right-angled triangle is revolved about the perpendicular side, a cone is formed whose,

Height of the cone = Length of the axis= $h = 12\ cm$

Base radius of the cone = $r = 5\ cm$

And, Slant height of the cone = $l = 13\ cm$

We know,

The volume of a cone = $\frac{1}{3}\pi r^2 h$

The required volume of the cone formed = $\frac{1}{3}\times\pi\times5^2\times12$

$\\ = \pi\times25\times4 \\ = 100\pi\ cm^3$

Therefore, the volume of the solid cone obtained is $100\pi\ cm^3$

Q8 If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.

Answer:

When a right-angled triangle is revolved about the perpendicular side, a cone is formed whose,

Height of the cone = Length of the axis= $h = 5\ cm$

Base radius of the cone = $r = 12\ cm$

And, Slant height of the cone = $l = 13\ cm$

We know,

The volume of a cone = $\frac{1}{3}\pi r^2 h$

The required volume of the cone formed = $\frac{1}{3}\times\frac{22}{7}\times12^2\times5$

$\\ = \pi\times4\times60 \\ = 240\pi\ cm^3$

Now, Ratio of the volumes of the two solids = $\\ = \frac{100\pi}{240\pi}$

$\\ = \frac{5}{12}$

Therefore, the required ratio is $5:12$

Q9 A heap of wheat is in the form of a cone whose diameter is $\small 10.5$ m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

Answer:

Given,

Height of the conical heap = $h = 3\ m$

Base radius of the cone = $r = \frac{10.5}{2}\ m$

We know,

The volume of a cone = $\frac{1}{3}\pi r^2 h$

The required volume of the cone formed = $\frac{1}{3}\times\frac{22}{7}\times\left (\frac{10.5}{2} \right )^2\times3$

$\\ = 22\times\frac{1.5\times10.5}{4} \\ = 86.625\ m^3$

Now,

The slant height of the cone = $l = \sqrt{r^2+h^2}$

$\\ \Rightarrow l = \sqrt{3^2+5.25^2} = \sqrt{9+27.5625} \approx 6.05$

We know, the curved surface area of a cone = $\pi r l$

The required area of the canvas to cover the heap = $\frac{22}{7}\times\frac{10.5}{2}\times6.05$

$= 99.825\ m^2$

Class 9 maths chapter 13 question answer - exercise: 13.8

Q1 (i) Find the volume of a sphere whose radius is 7 cm

Answer:

Given,

The radius of the sphere = $r = 7\ cm$

We know, Volume of a sphere = $\frac{4}{3}\pi r^3$

The required volume of the sphere = $\frac{4}{3}\times\frac{22}{7}\times (7)^3$

$\\ = \frac{4}{3}\times22\times 7\times 7$

$\\ = \frac{4312}{3}$

$\\ = 1437\frac{1}{3}\ cm^3$

Q1 (ii) Find the volume of a sphere whose radius is $\small 0.63\hspace{1mm}m$

Answer:

Given,

The radius of the sphere = $r = 0.63\ m$

We know, Volume of a sphere = $\frac{4}{3}\pi r^3$

The required volume of the sphere = $\frac{4}{3}\times\frac{22}{7}\times (0.63)^3$

$\\ = 4\times22\times 0.03\times 0.63\times 0.63$

$\\ = 1.048\ m^3$

$\\ = 1.05\ m^3\ \ \ (approx.)$

Q2 (i) Find the amount of water displaced by a solid spherical ball of diameter of 28 cm

Answer:

The solid spherical ball will displace water equal to its volume.

Given,

The radius of the sphere = $r = \frac{28}{2}\ cm = 14\ cm$

We know, Volume of a sphere = $\frac{4}{3}\pi r^3$

$\therefore$ The required volume of the sphere = $\frac{4}{3}\times\frac{22}{7}\times (14)^3$

$\\ = \frac{4}{3}\times22\times 2\times 14\times 14$

$\\ = \frac{34469}{3} \\ = 11489\frac{2}{3}\ cm^3$

Therefore, the amount of water displaced will be $11489\frac{2}{3}\ cm^3$

Q2 (ii) Find the amount of water displaced by a solid spherical ball of diameter $\small 0.21\hspace{1mm}m$

Answer:

The solid spherical ball will displace water equal to its volume.

Given,

The radius of the sphere = $r = \frac{0.21}{2}\ m$

We know, Volume of a sphere = $\frac{4}{3}\pi r^3$

$\therefore$ The required volume of the sphere = $\frac{4}{3}\times\frac{22}{7}\times \left(\frac{0.21}{2} \right )^3$

$\\ = 4\times22\times \frac{0.01\times 0.21\times 0.21}{8}$

$\\ = 11\times 0.01\times 0.21\times 0.21$

$\\ =0.004851\ m^3$

Therefore, amount of water displaced will be $0.004851\ m^3$

Q3 The diameter of a metallic ball is $\small 4.2\hspace{1mm}cm$ . What is the mass of the ball, if the density of the metal is $\small 8.9\hspace{1mm}g$ per $\small cm^3$ ?

Answer:

Given,

The radius of the metallic sphere = $r = \frac{4.2}{2}\ cm = 2.1\ cm$

We know, Volume of a sphere = $\frac{4}{3}\pi r^3$

$\therefore$ The required volume of the sphere = $\frac{4}{3}\times\frac{22}{7}\times 2.1^3$

$\\ = 4\times22\times 0.1\times 2.1\times 2.1$

$\\ =38.808\ cm^3$

Now, the density of the metal is $\small 8.9\hspace{1mm}g$ per $\small cm^3$ ,which means,

Mass of $\small 1\ cm^3$ of the metallic sphere = $\small 8.9\hspace{1mm}g$

Mass of $38.808\ cm^3$ of the metallic sphere = $\small (8.9\times38.808)\ g$

$\small \approx 345.39\ g$

Q4 The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?

Answer:

Given,

Let $d_e$ be the diameters of Earth

$\therefore$ The diameter of the Moon = $d_m = \frac{1}{4}d_e$

We know, Volume of a sphere =

$\frac{4}{3}\pi r^3 =\frac{4}{3}\pi \left (\frac{d}{2} \right )^3 = \frac{1}{6}\pi d^3$

$\therefore$ The ratio of the volumes = $\frac{Volume\ of\ the\ Earth}{Volume\ of\ the\ Moon}$

$\\ = \frac{\frac{1}{6}\pi d_e^3}{\frac{1}{6}\pi d_m^3} \\ = \frac{ d_e^3}{(\frac{d_e}{4})^3} \\ = 64: 1$

Therefore, the required ratio of the volume of the moon to the volume of the earth is $1: 64$

Q5 How many litres of milk can a hemispherical bowl of diameter $\small 10.5\hspace{1mm}cm$ hold?

Answer:

The radius of the hemispherical bowl = $r = \frac{10.5}{2}\ cm$

We know, Volume of a hemisphere = $\frac{2}{3}\pi r^3$

The volume of the given hemispherical bowl = $\frac{2}{3}\times\frac{22}{7}\times \left (\frac{10.5}{2} \right )^3$

$= \frac{2}{3\times8}\times22\times1.5\times10.5\times10.5$

$= 303.1875\ cm^3$

The capacity of the hemispherical bowl = $= \frac{303.1875}{1000} \approx 0.303\ litres\ \ \ (approx.)$

Q6 A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

Answer:

Given,

Inner radius of the hemispherical tank = $r_1 = 1\ m$

Thickness of the tank = $1\ cm = 0.01\ m$

$\therefore$ Outer radius = Internal radius + thickness = $r_2 = (1+0.01)\ m = 1.01\ m$

We know, Volume of a hemisphere = $\frac{2}{3}\pi r^3$

$\therefore$ Volume of the iron used = Outer volume - Inner volume

$= \frac{2}{3}\pi r_2^3 - \frac{2}{3}\pi r_1^3$

$= \frac{2}{3}\times\frac{22}{7}\times (1.01^3 - 1^3)$

$= \frac{44}{21}\times0.030301$

$= 0.06348\ m^3\ \ (approx)$

Q7 Find the volume of a sphere whose surface area is $\small 154\hspace{1mm}cm^2$ .

Answer:

Given,

The surface area of the sphere = $\small 154\hspace{1mm}cm^2$

We know, Surface area of a sphere = $4\pi r^2$

$\therefore 4\pi r^2 = 154$

$\\ \Rightarrow 4\times\frac{22}{7}\times r^2 = 14\times11 \\ \Rightarrow r^2 = \frac{7\times7}{4} \\ \Rightarrow r = \frac{7}{2} \\ \Rightarrow r = 3.5\ cm$

$\therefore$ The volume of the sphere = $\frac{4}{3}\pi r^3$

$= \frac{4}{3}\times\frac{22}{7}\times (3.5)^3$

$= 179\frac{2}{3}\ cm^3$

Q8 (i) A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of $\small Rs\hspace{1mm}4989.60$ . If the cost of white-washing is Rs 20 per square metre, find the inside surface area of the dome

Answer:

Given,

$\small Rs\hspace{1mm}20$ is the cost of white-washing $1\ m^2$ of the inside area

$\small Rs\hspace{1mm}4989.60$ is the cost of white-washing $\frac{1}{20}\times4989.60\ m^2 = 249.48\ m^2$ of inside area

(i) Therefore, the surface area of the inside of the dome is $249.48\ m^2$

Q8 (ii) A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of $\small Rs\hspace{1mm}4989.60$ . If the cost of white-washing is Rs 20 per square metre, find the volume of the air inside the dome.

Answer:

Let the radius of the hemisphere be $r\ m$

Inside the surface area of the dome = $249.48\ m^2$

We know, Surface area of a hemisphere = $2\pi r^2$

$\\ \therefore 2\pi r^2 = 249.48 \\ \Rightarrow r^2 = \frac{249.48\times7}{2\times22} \\ \Rightarrow r = 6.3\ m$

$\therefore$ The volume of the hemisphere = $\frac{2}{3}\pi r^3$

$= \frac{2}{3}\times\frac{22}{7}\times (6.3)^3$

$= 523.908\ m^3$

Q9 (i) Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area $\small S'$ . Find the radius $\small r'$ of the new sphere

Answer:

Given,

The radius of a small sphere = $r$

The radius of the bigger sphere = $r'$

$\therefore$ The volume of each small sphere= $\frac{4}{3}\pi r^3$

And, Volume of the big sphere of radius $r'$ = $\frac{4}{3}\pi r'^3$

According to question,

$27\times\frac{4}{3}\pi r^3=\frac{4}{3}\pi r'^3$

$\\ \Rightarrow r'^3 = 27\times r^3 \\ \Rightarrow r' = 3\times r$

$\therefore r' = 3r$

Q9 (ii) Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area $\small S'$ . Find the ratio of S and $\small S'$ .

Answer:

Given,

The radius of a small sphere = $r$

The surface area of a small sphere = $S$

The radius of the bigger sphere = $r'$

The surface area of the bigger sphere = $S'$

And, $r' = 3r$

We know, the surface area of a sphere = $4\pi r^2$

$\therefore$ The ratio of their surface areas = $\frac{4\pi r'^2}{4\pi r^2}$

$\\ = \frac{ (3r)^2}{ r^2} \\ = 9$

Therefore, the required ratio is $1:9$

Q10 A capsule of medicine is in the shape of a sphere of diameter $\small 3.5\hspace{1mm}mm$ . How much medicine (in $\small mm^3$ ) is needed to fill this capsule?

Answer:

Given,

The radius of the spherical capsule = $r =\frac{3.5}{2}$

$\therefore$ The volume of the capsule = $\frac{4}{3}\pi r^3$

$= \frac{4}{3}\times\frac{22}{7}\times(\frac{3.5}{2})^3$

$= \frac{4}{3}\times22\times\frac{0.5\times3.5\times3.5}{8}$

$= 22.458\ mm^3 \approx 22.46\ mm^3\ \ (approx)$

Therefore, $22.46\ mm^3\ \ (approx)$ of medicine is needed to fill the capsule.

Class 9 maths chapter 13 ncert solutions - exercise: 13.9

Q1 A wooden bookshelf has external dimensions as follows: Height $\small =110\hspace{1mm}cm$ , Depth $\small =25\hspace{1mm}cm$ , Breadth $\small =85\hspace{1mm}cm$ (see Fig. $\small 13.31$ ). The thickness of the plank is $\small 5\hspace{1mm}cm$ everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is $\small 20\hspace{1mm}$ paise per $\small cm^2$ and the rate of painting is $\small 10\hspace{1mm}$ paise per $\small cm^2$ , find the total expenses required for polishing and painting the surface of the bookshelf.

1640782890665

Answer:

External dimension od the bookshelf = $85\ cm\times25\ cm\times110\ cm$

(Note: There is no front face)

The external surface area of the shelf = $lh + 2 (lb + bh)$
$\\ = [85 \times 110 + 2 (85 \times 25 + 25 \times 110)] \\ = (9350 + 9750) \\ = 19100\ cm^2$

We know, each stripe on the front surface is also to be polished. which is 5 cm stretch.

Area of front face = $[85 \times 110 - 75 \times 100 + 2 (75 \times5)]$

$\\ = 1850 + 750 \\ = 2600\ cm^2$

Area to be polished = $(19100 + 2600) = 21700\ cm^2$

Cost of polishing $1\ cm^2$ area = $Rs\ 0.20$

Cost of polishing $21700\ cm^2$ area = $Rs.\ (21700 \times 0.20) = Rs.\ 4340$

Now,

Dimension of inner part = $75\ cm\times15\ cm\times100\ cm$

Area to be painted in 3 rows = $3\times[2 (l + h) b + lh]$

$\\ =3\times [2 (75 + 30) \times 20 + 75 \times 30] \\ = 3\times[(4200 + 2250)] \\ = 3\times6450 \\ = 19350\ cm^2$

Cost of painting $1\ cm^2$ area = $Rs\ 0.10$

Cost of painting $19350\ cm^2$ area = $Rs.\ (19350 \times 0.10)= Rs.\ 1935$

Total expense required for polishing and painting = $Rs.\ (4340 + 1935)$

$= Rs.\ 6275$

Q2 The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in Fig $\small 13.32$ . Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius $\small 1.5\hspace{1mm}cm$ and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per $\small cm^2$ and black paint costs 5 paise per $\small cm^2$ .

1640782916472

Answer:

Given,

The radius of the wooden spheres = $r_1 = \frac{21}{2}\ cm$

$\therefore$ The surface area of a single sphere = $4\pi r_1^2$

$\\ = 4\times\frac{22}{7}\times\frac{21}{2}\times\frac{21}{2} \\ = 22\times3\times21$

$= 1386\ cm^2$

Again, the Radius of the cylinder support = $\small r_2 = 1.5\hspace{1mm}cm$

Height of the support = $h = 7\ cm$

$\therefore$ The base area of the cylinder = $\pi r_2^2 = \frac{22}{7}\times1.5\times1.5 = 7.07\ cm^2$

Now, Cost of painting $1\ cm^2$ silver = $25\ paise = Rs.\ 0.25$

$\therefore$ Cost of painting 1 wooden sphere = Cost of painting $(1386-7.07)\ cm^2$ silver

= $Rs. (0.25\times1378.93) = Rs.\ 344.7325$

Now, Curved surface area of the cylindrical support = $2\pi r_2 h$

$\\ = 2\times\frac{22}{7}\times1.5\times7 \\ = 22\times3 \\ = 66\ cm^2$

Now, Cost of painting $1\ cm^2$ black = $5\ paise = Rs.\ 0.05$

$\therefore$ Cost of painting 1 such stand = Cost of painting $66\ cm^2$ silver = $Rs. (0.05\times66) = Rs.\ 3.3$

$\therefore$ The total cost of painting 1 sphere and its support = $Rs.\ (344.7325+3.3) = Rs.\ 348.0325$

Therefore, total cost of painting 8 such spheres and their supports = $Rs.\ (8\times348.0325) = Rs.\ 2784.26$

Q3 The diameter of a sphere is decreased by $\small 25\%$ . By what per cent does its curved surface area decrease?

Answer:

Let the radius of the sphere be $r$

Diameter of the sphere = $2r$

According to question,

Diameter is decreased by $\small 25\%$

So, the new diameter = $\frac{3}{4}\times2r = \frac{3r}{2}$

So, the new radius = $r' = \frac{3r}{4}$

$\therefore$ New surface area = $4\pi r'^2 = 4\pi (\frac{3r}{4})^2$

$\therefore$ Decrease in surface area = $4\pi r^2 - 4\pi (\frac{3r}{4})^2$

$= 4\pi r^2[1 -\frac{9}{16} ]$

$= 4\pi r^2[\frac{7}{16} ]$

$\therefore$ Percentage decrease in the surface area = $\frac{Difference\ in\ areas}{Original\ surface\ area}$

$= \frac{4\pi r^2[\frac{7}{16}]}{4\pi r^2}\times100 \%$

$= \frac{7}{16}\times100 \%$

$= 43.75 \%$

Summary Of NCERT Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes

Class 9 surface area and volumes NCERT solutions is designed to aid students in determining the surface areas and volumes of various objects such as cuboids, cylinders, cones, and spheres. This chapter provides a detailed explanation of how to calculate the area of these objects by multiplying their length and breadth. Additionally, NCERT Solutions for Class 9 Maths Chapter 13 offer a wide range of exercise problems, including basic and advanced level questions, to prepare students for competitive exams. The solutions also provide clear explanations of the various activities to aid students in understanding the underlying concepts before attempting the questions. The topics covered in NCERT Solutions for Class 9 Maths Chapter 13 include:

Surface area of a cuboid and cube
Surface area of a right circular cylinder
Surface area of a right circular cone
Surface area of a sphere
Volume of a cuboid
Volume of a cylinder
Volume of a right circular cone
Volume of a sphere

In addition, the class 9 surface area and volume solutions delve into the details of the cuboid, cube, right circular cone, cylinder, hemisphere, and sphere, offering a thorough understanding of these objects.

If any student is looking for class 9 maths ch 13 question answer that are listed below in one place:

NCERT solutions for class 9 maths - Chapter Wise

Chapter No.	Chapter Name
Chapter 1	Number Systems
Chapter 2	Polynomials
Chapter 3	Coordinate Geometry
Chapter 4	Linear Equations In Two Variables
Chapter 5	Introduction to Euclid's Geometry
Chapter 6	Lines And Angles
Chapter 7	Triangles
Chapter 8	Quadrilaterals
Chapter 9	Areas of Parallelograms and Triangles
Chapter 10	Circles
Chapter 11	Constructions
Chapter 12	Heron’s Formula
Chapter 13	Surface Area and Volumes
Chapter 14	Statistics
Chapter 15	Probability

Key Features of NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes

Comprehensive coverage: NCERT Solutions for maths chapter 13 class 9 cover all the important topics related to the surface areas and volumes of different geometrical objects such as cubes, cuboids, cylinders, cones, and spheres.

Easy to understand: The ch 13 maths class 9 solutions are presented in a clear and concise manner, making them easy for students to understand. The step-by-step explanations help students to grasp the concepts better.

Helpful tips and tricks: The class 9 chapter 13 maths solutions provide helpful tips and tricks to solve the problems more efficiently, saving time and effort.

NCERT solutions for class 9 subject wise

NCERT Solutions for Class 9 Maths

NCERT Solutions for Class 9 Science

How to use NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes ?

Revise the formulas and concepts regarding the figures circle, rectangle, square, etc.
Learn about the other figures introduced in this chapter.
Memorize the formulas for every shape.
Learn the formula application by going through some examples.
Jump on to the practice exercises to get command over the topic.
You can use NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes during the practice.

Keep working hard and happy learning!

Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Questions (FAQs)

1. What are the important topics in maths chapter 13 class 9 Surface Area and Volumes ?

The surface areas and volumes of a cuboid, cube, cylinder, circular cone, and sphere are covered in this chapter. These basic concepts will remain with you in your upcoming study and help you to score well in exams, so try to command these. you can practice these NCERT solutions to get indepth understanding of concepts.

2. How many chapters are there in CBSE class 9 maths ?

There are 15 chapters starting from numbers systems to probability in the CBSE class 9 maths. NCERT syllabus list all the chapters, and students can go through them.

3. Where can I find the complete solutions of NCERT for class 9 maths ?

Here you will get the detailed NCERT solutions for class 9 maths by clicking on the link. students are advised to practice these problems and solutions to get command in the concepts which is essential for exam.

4. Why should we follow NCERT Solutions for class 9 chapter 13?

NCERT Solutions for class 9th surface area and volume provide students with comprehensive and high-quality reference material that covers various mathematical concepts. The class 9th chapter 13 solutions present the questions in a simple, easy-to-remember format, making it easier for students to understand and retain the answers. By practicing these solutions, students can greatly enhance their chances of scoring well in their exams.

Also Read

NCERT Solutions for Class 9 - Subject-wise PDFs Updated for 2024-25

NCERT Solutions for Class 9 Science

NCERT Solutions for Class 9 Maths

NCERT Syllabus for Class 9 Maths 2023 - Download Syllabus Pdf

NCERT Solutions For Class 9 Science Chapter 1 Matter in Our Surroundings

NCERT Solutions for Exercise 2.5 Class 9 Maths Chapter 2 Polynomials

NCERT Class 9 Maths Notes - Download Chapter Wise PDFs

NCERT Class 9 Maths Chapter 8 Notes Quadrilaterals - Download PDF

NCERT Class 9 Maths Chapter 2 Notes Polynomials - Download PDF

NCERT Class 9 Maths Chapter 1 Notes Number System - Download Free PDF

NCERT Class 9 Maths Chapter 5 Notes Introduction To Euclid’s Geometry - Download PDF

NCERT Class 9 Maths Chapter 6 Notes Lines and Angles - Download PDF

NCERT Exemplar Class 9 Maths Solutions - Chapter Wise Problems with Solutions

NCERT Exemplar Class 9 Solutions - Subject Wise Problems with Solutions

NCERT Exemplar Class 9 Science Solutions Chapter 15 Improvement in Food Resources

NCERT Exemplar Class 9 Science Solutions Chapter 14 Natural Resources

NCERT Exemplar Class 9 Science Solutions Chapter 13 Why do we fall ill?

NCERT Exemplar Class 9 Science Solutions Chapter 12 Sound

Articles

Silverzone ABHO Result 2024-25 - Check Akhil Bhartiya Hindi Olympiad Results Here

Nov 20, 2024

Silverzone SKGKO Result 2024-25 – Check Your Scores and Performance

Nov 20, 2024

Education Loan for School Students in India – Eligibility, Benefits & Application Process

Nov 20, 2024

Silverzone ABHO Answer Key 2024 | Download Akhil Bhartiya Hindi Olympiad Solutions

Nov 20, 2024

Silverzone SKGKO Answer Key 2024-25: Download Smart Kid GK Olympiad Answer Key @ silverzone.org

Nov 20, 2024

Sainik School Entrance Exam 2025: Application, Exam Date, How to Apply, Syllabus

Nov 20, 2024

ICICI Education Loan: Eligibility, Interest Rates & Application Process

Nov 19, 2024

MPSOS 10th Result 2024 - Check MPSOS Class 10 December 2024 Exam Result @mpsos.nic.in

Nov 16, 2024

Upcoming School Exams

National Institute of Open Schooling 12th Examination

Admit Card Date:04 October,2024 - 29 November,2024

Application Process

Exam Pattern

Admit Card

National Institute of Open Schooling 10th examination

Admit Card Date:04 October,2024 - 29 November,2024

Application Process

Exam Pattern

Mock Test

Mizoram Board of School Education 12th Examination

Application Date:07 October,2024 - 22 November,2024

Exam Pattern

Result

Mock Test

Mizoram Board 10th Examination

Application Date:07 October,2024 - 22 November,2024

Exam Pattern

Admit Card

Result

Punjab Board of Secondary Education 10th Examination

Application Correction Date:08 October,2024 - 27 November,2024

Exam Pattern

Result

Admit Card

View All School Exams

Get answers from students and experts

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes

NCERT Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes

Surface Area and Volumes Class 9 Questions And Answers PDF Free Download

Surface Area and Volumes Class 9 Solutions - Important Formulae

Surface Area and Volumes Class 9 NCERT Solutions (Intext Questions and Exercise)

Summary Of NCERT Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes

NCERT solutions for class 9 maths - Chapter Wise

Key Features of NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes

Frequently Asked Questions (FAQs)

Also Read

Articles

Upcoming School Exams

National Institute of Open Schooling 12th Examination

National Institute of Open Schooling 10th examination

Mizoram Board of School Education 12th Examination

Mizoram Board 10th Examination

Punjab Board of Secondary Education 10th Examination

Popular Questions

Colleges After 12th

Popular Course After 12th

Applications for Admissions are open.

VMC VIQ Scholarship Test

JEE Main Important Physics formulas

JEE Main Important Chemistry formulas

TOEFL ® Registrations 2024

Pearson | PTE

JEE Main high scoring chapters and topics

Explore on Careers360

NCERT Solutions

NCERT Exemplars

NCERT Books

NCERT Notes

CBSE

CBSE Class 10

CBSE Class 12th

CISCE Board 10th

IMO

IEO

NSO

NMMS

Schools By Medium

By Ownership

By City

By State

NVS

JNVST

Download Careers360 App

All this at the convenience of your phone

Scan and download the app