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

# 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.

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 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

h = height

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.

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

Class 9 maths chapter 13 question answer - exercise: 13.1

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 $5.45\ m^2$

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 $Rs.\ 109$

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 $Rs.\ 555$

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 $6\ m$

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

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

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 = $400-360 = 40\ cm^2$

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

Given,

Edge of the cubical box =

Dimensions of the cuboid =

(ii) The total surface area of the cubical box =

The total surface area of the cuboidal box =

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

Difference between them =

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 $4250\ cm^2$

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.

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 $\small Rs.\ 2184$

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

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 = $1\ cm$

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 $\small 7.48\ m^2$

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 $968\ cm^2$

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 $1064.8\ cm^2$

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 $2038.08\ cm^2$

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 = $1584\ m^2$

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 $\small Rs.\ 68.75$

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 $1\ m$

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 $110\ m^2$

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 $Rs.\ 4400$

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 $4.4\ m^2$

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 $\small 59.4\ m^2$

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.

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.

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

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=

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

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=

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

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 $7\ cm$

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 $462\ cm^2$

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 $26\ m$

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 $Rs.\ 137280$

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.

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 $\small Rs.\ 1155$ .

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 = $5500\ cm^2$

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 $\dpi{100} Rs.\ 384.34\ (approx)$

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

We know,

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

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

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

We know,

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

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

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

We know,

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

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

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

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

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

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

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

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

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

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

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

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

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 = $3\times3.14\times(10)^2$

$\\ = 942\ cm^2$

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 $1:4$

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 $\small Rs. 27.72$

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 $3.5\ cm$

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 $= 1:16$

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 $173.25\ cm^2$

Given,

The radius of the sphere = $r$

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

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 $4\pi r^2$

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

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 $\small 180\ cm^3$

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 = $\small (135\times1000) = 135000\ litres$

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 $\small 4.75\ m$

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 $144\ m^3$ = $Rs.(144\times30)$

$= Rs.4320$

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 $2\ m$

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.

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

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

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

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 $4:1$

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

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.

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 $3.432\ kg$

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 = $(385-300)\ cm^3 = 85\ cm^3$

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 $3\ cm$

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 $141.3\ cm^3$ .

(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 = $110\ m^2$

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 $1.75 \ m$

(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 $96.25 \ m^3$

(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 $0.4708\ m^2$

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 $0.11 \ cm^3$

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

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

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

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

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

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 =

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

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 =

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

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

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

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

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 $8\ cm$

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 = $38.5\ kilolitre$

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 $48\ cm$

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 $50\ cm$ .

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= $\frac{22}{7}\times14\times50$

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

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 $100\pi\ cm^3$

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 $5:12$

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

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

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

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

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

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 = $\frac{4}{3}\times\frac{22}{7}\times (0.63)^3$

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

$\\ = 1.048\ m^3$

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

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

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 $0.004851\ m^3$

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 = $\small (8.9\times38.808)\ g$

$\small \approx 345.39\ g$

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 $1: 64$

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 = $= \frac{303.1875}{1000} \approx 0.303\ litres\ \ \ (approx.)$

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

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

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

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

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

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 = $\frac{4}{3}\pi r^3$

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

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

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 $249.48\ m^2$

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 = $\frac{2}{3}\pi r^3$

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

$= 523.908\ m^3$

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,

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

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

$\therefore r' = 3r$

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 $1:9$

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

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 = $Rs.\ (4340 + 1935)$

$= Rs.\ 6275$

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 = $Rs.\ (8\times348.0325) = Rs.\ 2784.26$

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:

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.

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