Volume And Surface Area

Let the base radius be ‘r’ and height be ‘h’

⇒ Given that r = 5cm and h = 21cm

Volume of cylinder V= Area of circle at bottom × height

∴ Volume of cylinder is

Let the base radius be ‘r’ and height be ‘h’ and diameter be ‘d’

⇒ Given that d= 14cm

∴ r = 7cm

and h = 17cm

Volume of cylinder V= Area of circle at bottom × height

∴ Volume of cylinder is

Let the base radius be ‘r’

height be ‘h’

diameter be ‘d’

and volume be ‘V’

⇒ Given that

and h = 12.5cm

Volume of cylinder V= Area of circle at bottom × height

Radius of cylinder is 8cm.

Let the base radius be ‘r’ and height be ‘h’ and diameter be ‘d’

⇒ Given that d= 70cm

∴ r = 35cm

and h = 40cm

⇒ Volume of cylinder V= Area of circle at bottom × height

But, given that

The tank can hold 154 litres of water.

Ans. 154 l

Let the base radius be ‘r’ and height be ‘h’ and diameter be ‘d’

⇒ Given that circumference ‘c’ = 132 cm

We know that circumference

C=2 π r

⇒ r=21cm

and h = 25cm given

⇒ Volume of cylinder V= Area of circle at bottom × height

V=34650 cm³

∴ Volume of cylinder is 34650 cm³

Let the base radius be ‘r’ and length be ‘l’ and diameter be ‘d’ of the cylinder rod

Volume of the iron required is equal to volume of the rod

⇒ Given that d= 2.1cm

∴ r = 1.05cm

and l = 70cm

⇒ Volume of cylinder V= Area of circle at bottom × height

Volume of iron required

Ans. 242.55 cu cm

Let the base radius be ‘r’ and height be ‘h’ and diameter be ‘d’

⇒ Given that ∴ r = 0.4m = 40 cm

and h = 0.8m = 80 cm

⇒ Volume of cylinder V= Area of circle at bottom × height

V=401.92 litres

The tank can hold 401.92 litres

For the Larger cylinder let the

base radius be ‘R’ = 5cm

height be ‘H’ = X cm

Volume V = 1100 cm

⇒ Volume of cylinder V= Area of circle at bottom × height

H=14cm

Let Height of each smaller cylinder is h

∴ Number of discs of height h

We can cut 7 discs from this block.

Let the base radius be ‘r’ and height be ‘h’ and diameter be ‘d’

⇒ Given that ∴ r = 8 cm

and h = 35 cm

⇒ Curved surface area A= Circumference × height
A=2πr× h

Curved surface area

Ans. 1760 sq cm

Let the base radius be ‘r’ and height be ‘h’ and diameter be ‘d’

⇒ Given that circumference of base C = 176 cm

C=2 π× r=176 cm

r=28cm

and h = 1m=100cm

⇒ Curved surface area A1= Circumference* height

The Surface area at top and bottom are


Total surface area A=Curved surface area + Area at top and bottom
A=A₁ + A₂
A=17600 + 4928

Total surface area

Ans. 22528 sq cm

Let the base radius be ‘r’ and height be ‘h’ and diameter be ‘d’

⇒ Given that ∴ r = 5 cm

and h = 15 cm

⇒ Curved surface area A= Circumference× height
A=2πr× h

A=2π× 5× 15

Total curved surface area

Ans. 471 sq cm

Let the base radius be ‘r’ and height be ‘h’ and diameter be ‘d’

⇒ Given that

Radius of base = 28 cm

circumference of base

C=176cm

Curved surface area

SA=Circumference× height

SA=2× π× h

SA=176× h

Area at top and bottom circles

CA=π× 28× 28× 2

Total are of cylinder= Curved Surface Area + Area at top and bottom

∴Surface area SA =circumference× height
3520=176× h

h=20cm

∴ height of cylinder is 20cm

Radius of base r= 25cm

Height h= 3.5m = 350cm

Total Surface Area of column to be painted= curved Surface Area

Curved Surface Area

Cost of painting 1 column

=Rate of painting wall× Area per wall

=Rs 85.5

For painting 10 walls= 85.25× 10=852.5 Rs

Total cost = Rs 852.5

let Height of cylinder be h

Radius be r

Total surface area=Area at top and bottom + curved surface area

given r=h

r=14 cm

Radius of base is 14cm

We have

Given that,

Height of cone, h = 7 cm

Radius of cone, r = 9 cm

Volume of cone is given by

Volume = 1/3 πr²h

⇒

⇒

⇒

Thus, volume of the cone is 594 cm³.

We have

Given: Height, h = 18 cm

Volume = 924 cm³

We need to find the radius,

Volume of a cone is given by

Volume = 1/3 πr²h

⇒

Substituting the given values, we get

⇒

⇒

⇒ r = √49 = 7 cm

Thus, radius is 7 cm.

We have

Given: radius of cone, r = 7 cm

Slant height of cone, l = 25 cm

We need to find volume of the cone.

First, let’s find height of this cone.

In right-angled ∆BOC, using Pythagoras theorem,

BC² = BO² + OC²

⇒ OC² = BC² – BO²

⇒ OC² = 25² – 7²

⇒ OC² = 625 – 49 = 576

⇒ OC = √576 = 24

⇒ height of cone, h = 24 cm

Now, volume of cone is given by

Volume = 1/3 πr²h

Substituting r = 7 cm and h = 24 cm in above equation,

⇒

⇒

Thus, volume is 1232 cm³.

We have

Given: height of the cone, h = 9 cm

Volume of the cone = 462 cm³

To find radius of the cone, we know

Volume = 1/3 πr²h

⇒

⇒

⇒

⇒

⇒ r = √49 = 7 cm

Thus, radius is 7 cm.

We have

Given: diameter of the base of the cone = 28 cm

⇒ radius, r = 28/2 = 14 cm

Volume of the cone = 9856 cm³

We know,

Volume of cone = 1/3 πr²h

⇒

⇒

⇒

⇒

So, we have r = 14 cm and h = 48 cm. In right-angled ∆COB, using Pythagoras theorem

CB² = OB² + CO²

⇒ l² = r² + h², where l = slant height

⇒ l² = 14² + 48²

⇒ l² = 196 + 2304 = 2500

⇒ l = √2500 = 50

Thus, height is 48 cm and slant height is 50 cm.

We have

Given: radius of the cone, r = 5 cm

Height of the cone, h = 12 cm

We know, volume of cone is given by

Volume = 1/3 πr²h

Substituting given values in the above equation, we get

Volume = 1/3 × 3.14 × 5² × 12

⇒ Volume = 942/3 = 314

Thus, volume of cone is 314 cm³.

We have

Given: slant height of the cone, l = 10 cm

Radius of the base of the cone, r = 7 cm

We need to find curved surface area of this cone.

Curved surface area of the cone is given by

CSA = πrl

Substituting the given values in the above equation,

CSA = 22/7 × 7 × 10

⇒ CSA = (22 × 7 × 10)/7

⇒ CSA = 220

Thus, curved surface area of cone is 220 cm².

We have

Given: radius of the base of the cone, r = 7 cm

Slant height of the cone, l = 9 cm

(i). Curved surface area of cone is given by

CSA = πrl

⇒ CSA = 22/7 × 7 × 9

⇒ CSA = (22 × 7 × 9)/7

⇒ CSA = 198

Thus, curved surface area of cone is 198 cm².

(ii). For total surface area of cone, just add surface area of base of cone to the curved surface area of cone.

So, we have

Total surface area = curved surface area + area of base of the cone (area of solid circle)

⇒ TSA = πrl + πr²

⇒ TSA = πr (l + r)

⇒

⇒ TSA = 22 × (9 + 7)

⇒ TSA = 22 × 16

⇒ TSA = 352

Thus, total surface area of cone is 352 cm².

We have

Given: radius of the base of the cone, r = 9 cm

Height of the cone, h = 40 cm

First, we need to find slant height, l.

In right-angled ∆BOC, using Pythagoras theorem, we can write

BC² = OB² + OC²

⇒ l² = r² + h²

⇒ l² = 9² + 40²

⇒ l² = 81 + 1600 = 1681

⇒ l = √1681

⇒ l = 41

Curved surface area of the cone is given by

CSA = πrl

Using r = 9 cm and l = 41 cm in above equation,

CSA = 3.14 × 9 × 41

⇒ CSA = 1158.66

Total surface area of the cone is given by

TSA = CSA + area of the base of the cone (area of solid circle)

⇒ TSA = 1158.66 + πr²

⇒ TSA = 1158.66 + (3.14 × 9²)

⇒ TSA = 1158.66 + 254.34

⇒ TSA = 1413

Thus, curved surface area is 1158.66 cm² and total surface area is 1413 cm².

We have

Given: height of the conic tent, h = 10 m

Radius of its base, r = 24 m

(i). In right-angled ∆BOC, using Pythagoras theorem

BC² = OB² + OC²

⇒ BC² = 24² + 10²

⇒ BC² = 576 + 100

⇒ BC² = 676

⇒ BC = √676 = 26

⇒ l = 26

(where l = slant height of the cone)

Thus, slant height of conic tent is 26 m.

(ii). To find the amount of fabric required to make the tent, we need to find curved surface area of the cone as the fabric is required to cover the curved surface not the base of the conic tent.

So, curved surface area of cone is given by

CSA = πrl

⇒ CSA = 3.14 × 24 × 26

⇒ CSA = 1959.36

Thus, 1959.36 m² of fabric is required to make this cone-shaped tent.

We have

Given: radius of base of cone, r = 3 m

Height of the cone, h = 4 m

So, in right-angled ∆BOC, by using Pythagoras theorem,

BC² = OB² + OC²

⇒ BC² = r² + h²

⇒ BC² = 3² + 4²

⇒ BC² = 9 + 16 = 25

⇒ BC = √25 = 5

⇒ l = 5 m

Slant height of cone = 5 m

Metal required to make cone will cover the curved surface of the cone. So, curved surface area of the cone is given by

CSA = πrl

⇒ CSA = 3.14 × 3 × 5

⇒ CSA = 47.1

Thus, curved surface area is 47.1 m².

We have

Given: slant height of the cone, l = 12 cm

Curved surface area of the cone, CSA = 113.04 cm²

We know curved surface area of cone is given by

CSA = πrl, where r = radius of the base of the cone

⇒ r = CSA/πl

Substituting the given values in the above equation, we get

⇒

⇒ r = 3

Thus, radius is 3 cm.

We have

Given: height of the cone, h = 24 cm

Radius of the base of the cone, r = 7 cm

In right-angled ∆BOC, using Pythagoras theorem, we can write

BC² = OB² + OC²

⇒ BC² = 7² + 24²

⇒ BC² = 49 + 576 = 625

⇒ BC = √625 = 25

⇒ l = 25 cm

⇒ slant height of the cone = 25 cm

If this is a hat (cone-shaped), then the paper will cover only the curved surface of the hat, not the base. Base of the hat remains open.

So, we just need to find curved surface area of the cone, which is given by

CSA = πrl

Substituting values r = 7 cm and l = 25 cm in the above equation, we get

CSA = 22/7 × 7 × 25

⇒ CSA = (22 × 7 × 25)/7

⇒ CSA = 550

The paper required to make 1 hat = 550 cm²

Then, paper required to make 10 hats = 550 × 10 = 5500 cm²

Thus, paper required to make 10 hats is 5500 cm².

We have

It’s given that, radius of the sphere, r = 30 cm

We need to find its volume.

We know the volume of the sphere is given by

Volume = 4/3 πr³

Substituting r = 30 cm in above equation, we get

Volume = 4/3 × 3.14 × 30³

⇒ Volume = (4 × 3.14 × 30 × 30 × 30)/3

⇒ Volume = 339120/3 = 113040

Thus, volume of the sphere is 113040 cm³.

Given is, volume of the sphere = 36000π cm³

And we know volume of the sphere is given by

Volume = 4/3 πr³

⇒

Substituting given value in the above equation, we get

⇒ r³ = 27000

⇒ r = (27000)^1/3

⇒ r = 30

Thus, radius of the sphere is 30 cm.

According to the question,

There are 27 spheres, each of radius ‘r’.

Now, for 1 sphere:

Radius = r

Volume of this 1 sphere = 4/3 πr³

Then, for 27 spheres, each of radius ‘r’:

Volume = 27 × 4/3 πr³

⇒ Volume = 36 πr³ …(i)

Even if these 27 spheres were melted to make it into one new sphere, the volume remains unaltered, which means

Volume of this new sphere = Volume of 27 spheres.

By equation (i), we can say that

Volume of this new sphere = 36 πr³ …(ii)

Let radius of this new sphere formed by melting twenty-seven spheres of radius ‘r’ be ‘R’.

Now, Volume of this new sphere = 4/3 πR³ …(iii)

Comparing equations (ii) and (iii), we get

4/3 πR³ = 36 πr³

⇒ 4/3 R³ = 36 r³

⇒ 1/3 R³ = 9 r³

⇒ R³ = 3 × 9 r³

⇒ R³ = 27 r³

⇒ R³ = 3³ r³

⇒ R³ = (3r)³

⇒ R = (3r)^3×1/3

⇒ R = 3r

Thus, radius of this new sphere formed is 3r.

(1). When given is radius of the sphere, r = 7 cm.

We have

Surface area of sphere is given by

Surface area = 4 πr²

Substituting the given value in the above equation, we get

Surface area = 4 × 22/7 × 7²

⇒ Surface area = (4 × 22 × 7 × 7)/7

⇒ Surface area = 616

Thus, surface area is 616 cm².

(2). When given is radius of the sphere, r = 10.5 cm.

We have

Surface area of sphere is given by

Surface area = 4 πr²

Substituting the given value in the above equation, we get

Surface area = 4 × 22/7 × (10.5)²

⇒ Surface area = (4 × 22 × 10.5 × 10.5)/7

⇒ Surface area = 9702/7 = 1386

Thus, surface area is 1386 cm².

(3). When given is radius of the sphere, r = 10 cm.

We have

Surface area of sphere is given by

Surface area = 4 πr²

Substituting the given value in the above equation, we get

Surface area = 4 × 3.14 × (10)²

⇒ Surface area = 4 × 3.14 × 100

⇒ Surface area = 1256

Thus, surface area is 1256 cm².

(4). When given is radius of the sphere, r = 2.8 cm.

We have

Surface area of sphere is given by

Surface area = 4 πr²

Substituting the given value in the above equation, we get

Surface area = 4 × 22/7 × (2.8)²

⇒ Surface area = (4 × 22 × 2.8 × 2.8)/7

⇒ Surface area = 689.92/7 = 98.56

Thus, surface area is 98.56 cm².

(5). When given is radius of the sphere, r = 9.8 m.

We have

Surface area of sphere is given by

Surface area = 4 πr²

Substituting the given value in the above equation, we get

Surface area = 4 × 22/7 × (9.8)²

⇒ Surface area = (4 × 22 × 9.8 × 9.8)/7

⇒ Surface area = 8451.52/7 = 1207.36

Thus, surface area is 1207.36 m².

(6). When given is radius of the sphere, r = 42 m.

We have

Surface area of sphere is given by

Surface area = 4 πr²

Substituting the given value in the above equation, we get

Surface area = 4 × 22/7 × (42)²

⇒ Surface area = (4 × 22 × 42 × 42)/7

⇒ Surface area = 155232/7 = 22176

Thus, surface area is 22176 m².

Given is the surface area of a sphere, that is, 616 cm².

And we know that surface area of sphere is given by

Surface area = 4 πr²

⇒ r² = Surface area/4π

⇒

⇒ r² = (616 × 7)/(4 × 22)

⇒ r² = 4312/88

⇒ r² = 49

⇒ r = √49 = 7

Thus, radius of the sphere is 7 cm.

Given that, surface area of sphere = 314 cm²

And we know that surface area of sphere is given by

Surface area = 4 πr², r = radius of the sphere

⇒ r² = Surface area/4π

⇒

⇒ r² = (314)/(4 × 3.14)

⇒ r² = 314/12.56

⇒ r² = 25

⇒ r = √25 = 5

So, radius of the sphere = 5 cm

Volume of the sphere is given by

Volume = 4/3 πr³

⇒ Volume = 4/3 × 3.14 × 5³

⇒ Volume = (4 × 3.14 × 5 × 5 × 5)/3

⇒ Volume = 1570/3

⇒ Volume = 523.33

Thus, volume of the sphere is 523.33 cm³.

Given is that, diameter of the inflated ball (sphere) = 18 cm

⇒ Radius of the sphere = 18/2 = 9 cm

The amount of air that this ball contain is the volume of the sphere.

So, volume of sphere is given by

Volume = 4/3 πr³

Substituting value of radius, r = 9 cm in the above equation, we get

Volume = 4/3 × 3.14 × 9³

⇒ Volume = (4 × 3.14 × 9 × 9 × 9)/3

⇒ Volume = 9156.24/3

⇒ Volume = 3052.08

Surface area of the spherical ball is given by

Surface area = 4 πr²

⇒ Surface area = 4 × 3.14 × 9²

⇒ Surface area = 4 × 3.14 × 9 × 9

⇒ Surface area = 1017.36

Thus, volume of inflated ball is 3052.08 cm³ and surface area of inflated ball is 1017.36 cm².