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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 **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.

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 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.

Total Surface Area (TSA):

Cuboid = 2(l x b) + 2(b x h) + 2(h x l)

Cube = 6a

^{2}Right Circular Cylinder = 2πr(h + r)

Right Circular Cone = πr(l + r)

Sphere = 4πr

^{2}Hemisphere = 3πr

^{2}

Lateral/Curved Surface Area (CSA):

Cuboid = 2h(l + b)

Cube = 4a

^{2}Right Circular Cylinder = 2πrh

Right Circular Cone = πrl

Volume:

Cuboid = l x b x h

Cube = a

^{3}Right Circular Cylinder = πr

^{2}hRight Circular Cone = (1/3)πr

^{2}hSphere = (4/3)πr

^{3}Hemisphere = (2/3)πr

^{3}

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,**

** Class 9 maths chapter 13 question answer - exercise: 13.1**

** Answer: **

Given, the dimensions of the plastic box

Length,

Width,

Depth,

(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.

=

The required area of the sheet required for making the box is

** Answer: **

Given, dimensions of the plastic box

Length,

Width,

Depth,

We know, area of the sheet required for making the box is

(ii) Cost for of sheet = Rs 20

Cost for of sheet =

Required cost of the sheet is

** Answer: **

Given,

Dimensions of the room =

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

=

Cost of white-washing per area =

Cost of white-washing area =

Therefore, the required cost of whitewashing the walls of the room and the ceiling is

** Answer: **

Given,

The perimeter of rectangular hall =

Cost of painting the four walls at the rate of Rs 10 per = Rs 15000

Let the height of the wall be

Area to be painted =

Required cost =

Therefore, the height of the hall is

** Answer: **

Given, dimensions of the brick =

We know, Surface area of a cuboid =

The surface area of a single brick =

Number of bricks that can be painted =

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

** Q5 (i) ** A cubical box has each edge and another cuboidal box is long, wide and high.

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

** Answer: **

Given,

Edge of the cubical box =

The lateral surface area of the cubical box =

The lateral surface area of the cuboidal box =

=360cm^{2}

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

Difference between them =

** Q5 (ii) ** A cubical box has each edge and another cuboidal box is long, wide and high.

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

** Answer: **

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 =

** Answer: **

Given, dimensions of the greenhouse =

Area of the glass =

Therefore, the area of glass is

** Answer: **

Given, dimensions of the greenhouse =

(ii) Tape needed for all the 12 edges = Perimeter =

Therefore, of tape is needed for the edges.

** Answer: **

Given,

Dimensions of the bigger box = ,

Dimensions of smaller box =

We know,

Total surface area of a cuboid =

Total surface area of the bigger box =

Area of the overlap for the bigger box =

Similarly,

Total surface area of the smaller box =

Area of the overlap for the smaller box =

Since, 250 of each box is required,

Total area of carboard required =

Cost of of the cardboard = Rs 4

Cost of of the cardboard =

Therefore, the cost of the cardboard sheet required for 250 such boxes of each kind is

** Answer: **

Given, Dimensions of the tarpaulin =

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

= Required

Therefore, tarpaulin is required.

** Class 9 maths chapter 13 NCERT solutions - exercise: 13.2**

** Answer: **

Given,

The curved surface area of the cylinder =

And, the height of the cylinder,

We know, Curved surface area of a right circular cylinder =

Therefore, the diameter of the cylinder =

** Answer: **

Given,

Height of the cylindrical tank =

Base diameter =

We know,

The total surface area of a cylindrical tank =

Therefore, square metres of the sheet is

** Answer: **

Note: There are two surfaces, inner and outer.

Given,

Height of the cylinder,

Outer diameter =

Inner diameter =

Inner curved surface area =

Therefore, the inner curved surface area of the cylindrical pipe is

** Answer: **

Note: There are two surfaces, inner and outer.

Given,

Height of the cylinder,

Outer diameter =

Inner diameter =

Outer curved surface area =

Therefore, the outer curved surface area of the cylindrical pipe is

**Answer: **

Note: There are two surfaces, inner and outer.

Given,

Height of the cylinder,

Outer diameter =

Inner diameter =

Outer curved surface area =

Inner curved surface area =

Area of the circular rings on top and bottom =

The total surface area of the pipe =

Therefore, the total surface area of the cylindrical pipe is

** Answer: **

Given,

The diameter of the cylindrical roller =

Length of the cylindrical roller =

The curved surface area of the roller =

Area of the playground =

Therefore, the required area of the playground =

** Answer: **

Given,

Radius of the cylindrical pillar, r =

Height of the cylinder, h =

We know,

Curved surface area of a cylinder =

Curved surface area of the pillar =

Now,

Cost of painting of the pillar =

Cost of painting the curved surface area of the pillar =

Therefore, the cost of painting curved surface area of the pillar is

** Answer: **

Given, a right circular cylinder

Curved surface area of the cylinder =

The radius of the base =

Let the height of the cylinder be

We know,

Curved surface area of a cylinder of radius and height =

Therefore, the required height of the cylinder is

** Q7 (i) ** The inner diameter of a circular well is . It is deep. Find its inner curved surface area.

** Answer: **

Given,

The inner diameter of the circular well =

Depth of the well =

We know,

The curved surface area of a cylinder =

The curved surface area of the well =

Therefore, the inner curved surface area of the circular well is

** Answer: **

Given,

The inner diameter of the circular well =

Depth of the well =

The inner curved surface area of the circular well is

Now, the cost of plastering the curved surface per = Rs. 40

Cost of plastering the curved surface of =

Therefore, the cost of plastering the well is

** Answer: **

Given,

Length of the cylindrical pipe =

Diameter =

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

We know,

The curved surface area of a cylindrical pipe of radius and length =

CSA of this pipe =

Therefore, the total radiating surface of the system is

** Answer: **

Given, a closed cylindrical petrol tank.

The diameter of the tank =

Height of the tank =

We know,

The lateral surface area of a cylinder of radius and height =

The lateral surface area of a cylindrical tank =

Therefore, the lateral or curved surface area of a closed cylindrical petrol storage tank is

** Q9 (ii) ** Find: how much steel was actually used, if of the steel actually used was wasted in making the tank.

** Answer: **

Given, a closed cylindrical petrol tank.

The diameter of the tank =

Height of the tank =

Now, Total surface area of the tank =

Now, let of steel sheet be actually used in making the tank

Since of steel was wasted, the left of the total steel sheet was used to made the tank.

The total surface area of the tank =

Therefore, of steel was actually used in making the tank.

** Answer: **

Given, a cylindrical lampshade

The diameter of the base =

Height of the cylinder =

The total height of lampshade=

We know,

Curved surface area of a cylinder of radius and height =

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

Therefore, cloth will be required for covering the lampshade.

** Answer: **

Given, a cylinder with a base.

The radius of the cylinder =

Height of the cylinder =

We know,

The lateral surface area of a cylinder of radius and height =

Area of the cylindrical penholder = Lateral areal + Base area

Area of 35 penholders =

Therefore, the area of carboard required is

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

** Q1 ** Diameter of the base of a cone is and its slant height is . Find its curved surface area.

** Answer: **

Given,

Base diameter of the cone =

Slant height =

We know, Curved surface area of a cone

Required curved surface area of the cone=

** Q2 ** Find the total surface area of a cone, if its slant height is and diameter of its base is .

** Answer: **

Given,

Base diameter of the cone =

Slant height =

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

Required total surface area of the cone=

** Q3 (i) ** Curved surface area of a cone is and its slant height is 14 cm. Find radius of the base .

** Answer: **

Given,

The curved surface area of a cone =

Slant height

(i) Let the radius of cone be

We know, the curved surface area of a cone=

Therefore, the radius of the cone is

** Q3 (ii) ** Curved surface area of a cone is and its slant height is . Find total surface area of the cone.

** Answer: **

Given,

The curved surface area of a cone =

Slant height

The radius of the cone is

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

Therefore, the total surface area of the cone is

** 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 =

Height of the conical tent =

Slant height =

Therefore, the slant height of the conical tent is

** Answer: **

Given,

Base radius of the conical tent =

Height of the conical tent =

Slant height =

We know, Curved surface area of a cone

Curved surface area of the tent

Cost of of canvas =

Cost of of canvas =

Therefore, required cost of canvas to make tent is

** Answer: **

Given,

Base radius of the conical tent =

Height of the tent =

We know,

Curved surface area of a cone =

Area of tarpaulin required = Curved surface area of the tent

Now, let the length of the tarpaulin sheet be

Since is wasted, effective length =

Breadth of tarpaulin =

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

** Answer: **

Given, a conical tomb

The base diameter of the cone =

Slant height

We know, Curved surface area of a cone

Now, Cost of whitewashing per =

Cost of whitewashing per =

Therefore, the cost of white-washing its curved surface of the tomb is .

** Answer: **

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

Base radius of the cone =

Height

We know, Curved surface area of a right circular cone

The curved surface area of a cap =

The curved surface area of 10 caps =

Therefore, the area of the sheet required for 10 caps =

** Answer: **

Given, hollow cone.

The base diameter of the cone =

Height of the cone =

Slant height =

We know, Curved surface area of a cone =

The curved surface area of 1 cone =

The curved surface area of 50 cones

Now, the cost of painting area =

Cost of the painting area

Therefore, the cost of painting 50 such hollow cones is

**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: .

** Answer: **

We know,

The surface area of a sphere of radius =

Required surface area =

** Q1 (ii) ** Find the surface area of a sphere of radius:

** Answer: **

We know,

The surface area of a sphere of radius =

Required surface area =

** Q1 (iii) ** Find the surface area of a sphere of radius:

** Answer: **

We know,

The surface area of a sphere of radius =

Required surface area =

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

** Answer: **

Given,

The diameter of the sphere =

We know,

The surface area of a sphere of radius =

Required surface area =

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

** Answer: **

Given,

The diameter of the sphere =

We know,

The surface area of a sphere of radius =

Required surface area =

** Q2 (iii) ** Find the surface area of a sphere of diameter:

** Answer: **

Given,

The diameter of the sphere =

We know,

The surface area of a sphere of radius =

Required surface area =

** Q3 ** Find the total surface area of a hemisphere of radius 10 cm. (Use )

** Answer: **

We know,

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

The required total surface area of the hemisphere =

** Answer: **

Given,

We know,

The surface area of a sphere of radius =

The ratio of surface areas of the ball in the two cases =

Therefore, the required ratio is

** Answer: **

Given,

The inner radius of the hemispherical bowl =

We know,

The curved surface area of a hemisphere =

The surface area of the hemispherical bowl =

Now,

Cost of tin-plating = Rs 16

Cost of tin-plating =

Therefore, the cost of tin-plating it on the inside is

** Q6 ** Find the radius of a sphere whose surface area is .

** Answer: **

Given,

The surface area of the sphere =

We know,

The surface area of a sphere of radius =

Therefore, the radius of the sphere is

** Answer: **

Let diameter of Moon be and diameter of Earth be

We know,

The surface area of a sphere of radius =

The ratio of their surface areas =

Therefore, the ratio of the surface areas of the moon and earth is

** Answer: **

Given,

The inner radius of the bowl =

The thickness of the bowl =

Outer radius of the bowl = (Inner radius + thickness) =

We know, Curved surface area of a hemisphere of radius =

The outer curved surface area of the bowl =

Therefore, the outer curved surface area of the bowl is

**Answer: **

Given,

The radius of the sphere =

Surface area of the sphere =

** Answer: **

Given,

The radius of the sphere =

The surface area of the sphere =

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 =

Therefore, the curved surface area of the cylinder is

** Answer: **

The surface area of the sphere =

And, Surface area of the cylinder =

So, the ratio of the areas =

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

** Q1 ** A matchbox measures . What will be the volume of a packet containing 12 such boxes?

** Answer: **

Given,

Dimensions of a matchbox =

We know,

The volume of a cuboid =

The volume of a matchbox=

Volume of 12 such matchboxes =

Therefore, the volume of a packet containing 12 matchboxes is

** Q2 ** A cuboidal water tank is 6 m long, 5 m wide and deep. How many litres of water can it hold? ( )

** Answer: **

Given,

Dimensions of the cuboidal water tank =

We know,

Volume of a cuboid =

Volume of the water tank =

We know,

Volume of water the tank can hold =

** Answer: **

Let the height of the vessel be

Dimensions of the cuboidal water tank =

We know,

The volume of a cuboid =

The volume of the water tank =

According to question,

Therefore, the required height of the vessel is

** 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 .

** Answer: **

Given,

Dimensions of the cuboidal pit =

We know , Volume of a cuboid =

Volume of the cuboidal pit =

Now, Cost of digging = Rs. 30

Cost of digging =

** Answer: **

Given,

Length of the tank =

Depth of the tank =

Let the breadth of the tank be

We know, Volume of a cuboid =

The volume of the cuboidal tank =

We know,

Therefore, the breadth of the tank is

** Answer: **

Given,

Dimensions of the tank =

4000 people requiring 150 litres of water per head per day

The total volume of water required =

Let the number of days the water will last be

We know,

The volume of a cuboid =

The volume of the water tank =

According to question,

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

** Answer: **

Given,

Dimensions of the godown =

Dimension of each wooden crate =

We know , Volume of a cuboid =

Volume of the godown =

Volume of the each crate=

Let number of wooden crates be

Volume of wooden crates = Volume of the godown

** Answer: **

This is an important question.

Given,

Side of a solid cube =

We know, the volume of a cube of side =

The volume of the given cube =

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

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

Therefore, the side of the new cube is

Now, we know,

The surface area of a cube of side =

The ratio between their surface areas

Therefore, the ratio of their surface areas is

** Answer: **

Given,

Rate of water flow =

Depth of river,

Width of the river,

The volume of water flowing in 1 min =

Therefore, water falling into the sea in a minute =

**Surface area and volumes class 9 solutions - exercise:** 13.6

** Answer: **

Given,

Circumference of the base of a cylindrical vessel =

Height =

Let the radius of the base of the cylinder be

Now, Circumference of the circular base of the cylinder =

The volume of the cylinder =

Also,

Therefore, the cylindrical vessel can hold of water.

** Answer: **

Given, A hollow cylinder made of wood.

The inner diameter of the cylindrical wooden pipe =

Outer diameter =

Length of the pipe =

The volume of the wooden cylinder =

Mass of wood =

Mass of wood =

Therefore, the mass of the wooden hollow cylindrical pipe is

** Answer: **

Given,

(i) The dimension of the rectangular base of the tin can =

Height of the can =

The volume of the tin can =

(ii) The radius of the circular base of the plastic cylinder =

Height of the cylinder =

The volume of the plastic cylinder =

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

The difference in capacity =

** Q4 (i) ** If the lateral surface of a cylinder is and its height is 5 cm, then find radius of its base (Use )

** Answer: **

Given,

The lateral surface area of the cylinder =

Height of the cylinder =

(i) Let the radius of the base be

We know,

The lateral surface area of a cylinder =

Therefore, the radius of the base is

** Q4 (ii) ** If the lateral surface of a cylinder is and its height is 5 cm, then find it's volume. (Use )

** Answer: **

Given,

The lateral surface area of the cylinder =

Height of the cylinder =

The radius of the base is

(ii) We know,

The volume of a cylinder =

Therefore, the volume of the cylinder is .

** Answer: **

(i) Given,

Rs 20 is the cost of painting area of the inner curved surface of the cylinder.

Rs 2200 is the cost of painting area of the inner curved surface of the cylinder.

The inner curved surface area of the cylindrical vessel =

** Answer: **

The inner curved surface area of the cylindrical vessel =

Height of the cylinder =

Let the radius of the circular base be

The inner curved surface area of the cylindrical vessel =

Therefore, the radius of the base of the vessel is

** Answer: **

(iii) Height of the cylinder =

Radius of the base of the vessel =

Volume of the cylindrical vessel =

Therefore, the capacity of the cylindrical vessel is

** Answer: **

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

The capacity of the vessel = Volume of the vessel = litres

Height of the cylindrical vessel =

Let the radius of the circular base be

The volume of the cylindrical vessel =

Therefore, the total surface area of the vessel =

Therefore, square metres of metal sheet needed to make it is

** Answer: **

Given,

Length of the cylindrical pencil =

The radius of the graphite (Inner solid cylinder) =

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

=

We know, Volume of a cylinder=

The volume of graphite =

And, Volume of wood =

Therefore, the volume of wood is and the volume of graphite is

** Answer: **

Given,

Height =

The radius of the cylindrical bowl =

The volume of soup in a bowl for a single person =

The volume of soup given for 250 patients =

Therefore, the amount of soup the hospital has to prepare daily to serve 250 patients is

** 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 =

Height =

We know,

Volume of a right circular cone =

Required volume =

** Q1 (ii) ** Find the volume of the right circular cone with: radius cm, height 12 cm

** Answer: **

Given,

Radius =

Height =

We know,

Volume of a right circular cone =

Required volume =

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

** Answer: **

Given,

Radius =

Slant height =

Height =

We know,

Volume of a right circular cone =

Volume of the vessel=

Required capacity of the vessel =

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

** Answer: **

Given,

Height =

Slant height =

Radius =

We know,

Volume of a right circular cone =

Volume of the vessel=

Required capacity of the vessel =

** Q3 ** The height of a cone is 15 cm. If its volume is 1570 , find the radius of the base. (Use )

** Answer: **

Given,

Height of the cone =

Let the radius of the base of the cone be

We know,

The volume of a right circular cone =

** Q4 ** If the volume of a right circular cone of height 9 cm is , find the diameter of its base.

** Answer: **

Given,

Height of the cone =

Let the radius of the base of the cone be

We know,

The volume of a right circular cone =

Therefore the diameter of the right circular cone is

** Q5 ** A conical pit of top diameter m is 12 m deep. What is its capacity in kilolitres?

** Answer: **

Given,

Depth of the conical pit =

The top radius of the conical pit =

We know,

The volume of a right circular cone =

The volume of the conical pit =

Now,

The capacity of the pit =

** Answer: **

Given, a right circular cone.

The radius of the base of the cone =

The volume of the cone =

(i) Let the height of the cone be

We know,

The volume of a right circular cone =

Therefore, the height of the cone is

** Answer: **

Given, a right circular cone.

The volume of the cone =

The radius of the base of the cone =

And the height of the cone =

(ii) We know, Slant height,

Therefore, the slant height of the cone is .

** Answer: **

Given, a right circular cone.

The radius of the base of the cone =

And Slant height of the cone =

(iii) We know,

The curved surface area of a cone =

Required curved surface area=

** 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=

Base radius of the cone =

And, Slant height of the cone =

We know,

The volume of a cone =

The required volume of the cone formed =

Therefore, the volume of the solid cone obtained is

** 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=

Base radius of the cone =

And, Slant height of the cone =

We know,

The volume of a cone =

The required volume of the cone formed =

Now, Ratio of the volumes of the two solids =

Therefore, the required ratio is

** Answer: **

Given,

Height of the conical heap =

Base radius of the cone =

We know,

The volume of a cone =

The required volume of the cone formed =

Now,

The slant height of the cone =

We know, the curved surface area of a cone =

The required area of the canvas to cover the heap =

** 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 =

We know, Volume of a sphere =

The required volume of the sphere =

** Q1 (ii) ** Find the volume of a sphere whose radius is

** Answer: **

Given,

The radius of the sphere =

We know, Volume of a sphere =

The required volume of the sphere =

** 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 =

We know, Volume of a sphere =

The required volume of the sphere =

Therefore, the amount of water displaced will be

** Q2 (ii) ** Find the amount of water displaced by a solid spherical ball of diameter

** Answer: **

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

Given,

The radius of the sphere =

We know, Volume of a sphere =

The required volume of the sphere =

Therefore, amount of water displaced will be

** Answer: **

Given,

The radius of the metallic sphere =

We know, Volume of a sphere =

The required volume of the sphere =

Now, the density of the metal is per ,which means,

Mass of of the metallic sphere =

Mass of of the metallic sphere =

** Answer: **

Given,

Let be the diameters of Earth

The diameter of the Moon =

We know, Volume of a sphere =

The ratio of the volumes =

Therefore, the required ratio of the volume of the moon to the volume of the earth is

** Q5 ** How many litres of milk can a hemispherical bowl of diameter hold?

** Answer: **

The radius of the hemispherical bowl =

We know, Volume of a hemisphere =

The volume of the given hemispherical bowl =

The capacity of the hemispherical bowl =

** Answer: **

Given,

Inner radius of the hemispherical tank =

Thickness of the tank =

Outer radius = Internal radius + thickness =

We know, Volume of a hemisphere =

Volume of the iron used = Outer volume - Inner volume

** Q7 ** Find the volume of a sphere whose surface area is .

** Answer: **

Given,

The surface area of the sphere =

We know, Surface area of a sphere =

The volume of the sphere =

** Answer: **

Given,

is the cost of white-washing of the inside area

is the cost of white-washing of inside area

(i) Therefore, the surface area of the inside of the dome is

** Answer: **

Let the radius of the hemisphere be

Inside the surface area of the dome =

We know, Surface area of a hemisphere =

The volume of the hemisphere =

** Answer: **

Given,

The radius of a small sphere =

The radius of the bigger sphere =

The volume of each small sphere=

And, Volume of the big sphere of radius =

According to question,

** Answer: **

Given,

The radius of a small sphere =

The surface area of a small sphere =

The radius of the bigger sphere =

The surface area of the bigger sphere =

And,

We know, the surface area of a sphere =

The ratio of their surface areas =

Therefore, the required ratio is

** Answer: **

Given,

The radius of the spherical capsule =

The volume of the capsule =

Therefore, of medicine is needed to fill the capsule.

** Class 9 maths chapter 13 ncert solutions - exercise: 13.9 **

** Answer: **

External dimension od the bookshelf =

(Note: There is no front face)

The external surface area of the shelf =

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

Area of front face =

Area to be polished =

Cost of polishing area =

Cost of polishing area =

Now,

Dimension of inner part =

Area to be painted in 3 rows =

Cost of painting area =

Cost of painting area =

Total expense required for polishing and painting =

**Answer: **

Given,

The radius of the wooden spheres =

The surface area of a single sphere =

Again, the Radius of the cylinder support =

Height of the support =

The base area of the cylinder =

Now, Cost of painting silver =

Cost of painting 1 wooden sphere = Cost of painting silver

=

Now, Curved surface area of the cylindrical support =

Now, Cost of painting black =

Cost of painting 1 such stand = Cost of painting silver =

The total cost of painting 1 sphere and its support =

Therefore, total cost of painting 8 such spheres and their supports =

** Q3 ** The diameter of a sphere is decreased by . By what per cent does its curved surface area decrease?

** Answer: **

Let the radius of the sphere be

Diameter of the sphere =

According to question,

Diameter is decreased by

So, the new diameter =

So, the new radius =

New surface area =

Decrease in surface area =

Percentage decrease in the surface area =

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:

- Class 9 Maths Chapter 13 NCERT Exercise: 13.1
- Class 9 Maths Chapter 13 NCERT Exercise: 13.2
- Class 9 Maths Chapter 13 NCERT Exercise: 13.3
- Class 9 Maths Chapter 13 NCERT Exercise: 13.4
- Class 9 Maths Chapter 13 NCERT Exercise: 13.5
- Class 9 Maths Chapter 13 NCERT Exercise: 13.6
- Class 9 Maths Chapter 13 NCERT Exercise: 13.7
- Class 9 Maths Chapter 13 NCERT Exercise: 13.8
- Class 9 Maths Chapter 13 NCERT Exercise: 13.9

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 |

**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 **

** 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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