Surface Area And Volume

Length = l = 12 cm

Breadth = b = 9 cm

Height = h = 5 cm

Total surface area of cuboid = 2 × (lb + bh + lh)

= 2 × (12×9 + 9×5 + 12×5)

= 2 × (108 + 45 + 60)

= 2 × 213

= 426 cm²

Therefore, total surface area of the cuboid is 426 cm²

To find the total surface area of the new cube formed we first need to find its edge

Volume of cube = (edge)³

Volume of cube with edge 8 cm = 8³ = 512 cm³

Volume of cube with edge 6 cm = 6³ = 216 cm³

Volume of cube with edge 1 cm = 1³ = 1 cm³

Let the edge of new cube formed be ‘a’

Volume of new cube = a³

As the three cubes are melted and the new cube is formed

Volume of new cube = volume of all the three cubes

⇒ a³ = 512 + 216 + 1

⇒ a³ = 729

⇒ a = 9

Surface area of cube = 6 × (edge)²

⇒ surface area of new cube = 6 × a²

= 6 × 9²

= 6 × 81

= 486 cm²

Therefore, surface area of new cube is 486 cm²

The box is like the shape of a cuboid

Length = l = 50 cm

Breadth = b = 36 cm

Height = h = 25 cm

Total surface area of cuboid = 2 × (lb + bh + lh)

⇒ total surface area of box = 2 × (50×36 + 36×25 + 50×25)

= 2 × (1800 + 900 + 1250)

= 2 × 3950

= 7900 cm²

Amount of cloth required to make cover = total surface area of box

Therefore, 7900 cm² cloth will be required to make cover for the box

A cube has 6 surfaces

We will get two cuboids after cutting

The cutting plane will pass through 4 surfaces halving their areas

So, each part will be having 4 surfaces with area 50 cm²

each part will have remaining 2 surfaces with area 100 cm²

total surface area of each part = 100 + 100 + 50 + 50 + 50 + 50

= 400 cm²

Therefore, total surface area of each part is 400 cm²

the box is as shown in figure

The wood is 3 cm thick

And given length, breadth and height are external which means including the wood thickness

So, to find the inner surface area we need to subtract the wooden thickness from external length, breadth and height

Thus length, breadth and height of inner surface area are as follows

In the length and breadth, we need to subtract 3 from both the sides which means we have to subtract 6 but in the height as the box is open from one end so we need to subtract only 3cm base thickness as seen in the figure

Length = 146 – 6 = 140 cm

Breadth = 116 – 6 = 110 cm

Height = 83 – 3 = 80 cm

The shape of the box is a cuboid and it is open box, therefore, we should subtract the top rectangular surface area from the total inner surface area

Total surface area of cuboid = 2 × (lb + bh + lh) – lb

⇒ total surface area of box = 2 × (140×110 + 110×80 + 140×80) – 140×110

= 2 × (15400 + 8800 + 11200) – 15400

= 2 × 35400 – 15400

= 70800 – 15400

= 55400 cm²

Cost of painting 1000 cm² is Rs 2

Thus, cost of painting 55400 cm² =

= 55.4 × 2

= 110.8 Rs

Therefore, cost of painting the inner surface = 110.8 Rs

Let the length, breadth and height be l, b and h respectively

Given:

l + b + h = 19 cm

length of diagonal = 11 cm

length of diagonal of a cuboid =

⇒ = 11 cm

Squaring both sides we get

⇒ l² + b² + h² = 121

Using identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac

⇒ (l + b + h)² = l² + b² + h² + 2lb + 2bh + 2hl

Substituting given values

⇒ 19² = 121 + 2 × (lb + bh + hl)

⇒ 361 = 121 + 2 × (lb + bh + hl)

⇒ 2 × (lb + bh + hl) = 361 – 121

⇒ 2 × (lb + bh + hl) = 240 cm²

Therefore, total surface area of cuboid is 240 cm²

Length of room= l = 6 m

Breadth of room = b = 6 m

Let height of room be h

Room contains 180 cubic metre of air which means volume of room is 180 m³

Room is in the shape of a cuboid

Volume of cuboid = lbh

⇒ 180 = 6 × 6 × h

⇒ 180 = 36h

⇒ h =

⇒ h = 5 m

Height of room is 5 m

Length of brick = 22 cm

Breadth of brick = 10 cm

Height of brick = 7 cm

Brick is in the shape of a cuboid

Volume of cuboid = length × breadth × height

⇒ volume of 1 brick = 22 × 10 × 7

= 1540 cm³

Let n bricks be required to make the wall

Length of wall = 44 m = 4400 cm

Breadth of wall = 85 cm

Height of wall = 1.5m = 150 cm

Volume of wall = 4400 × 85 × 150

= 56100000 cm³

The wall is completely made from bricks therefore

Volume of n bricks = volume of wall

⇒ n × volume of one brick = 56100000

⇒ n × 1540 = 56100000

⇒ n =

⇒ n =

Dividing numerator and denominator by 22, we get

⇒ n =

⇒ n = 36428.57 ~ 36429

Number of brick required = 36429

The longest edge of a cuboid is the body diagonal

So the length of longest rod will be same as the length of body diagonal

Length of room = l = 10 m

Breadth of room = b = 8 m

Height of room = h = 6 m

Length of body diagonal =

⇒ length of rod =

=

=

Therefore, length of longest rod = 10√2 m

Volume = 512 m³

Let a be the side of cube

Volume of cube = a³

⇒ a³ = 512

⇒ a =

⇒ a = 8 m

Length of brick = 20 cm

Breadth of brick = 10 cm

Height of brick = 7.5 cm

Brick is in the shape of a cuboid

Volume of cuboid = length × breadth × height

⇒ volume of 1 brick = 20 × 10 × 7.5

= 1500 cm³

Let n bricks be required to make the wall

Length of wall = 5 m = 500 cm

Breadth of wall = 30 cm

Height of wall = 3 m = 300 cm

Volume of wall = 500 × 30 × 300

= 4500000 cm³

The wall is completely made from bricks therefore

Volume of n bricks = volume of wall

⇒ n × volume of one brick = 56100000

⇒ n × 1500 = 4500000

⇒ n =

⇒ n =

⇒ n = 3000

Number of brick required = 3000

Let

Length of cuboid = l = 5x

Breadth of cuboid = b = 3x

Height of cuboid = h = 2x

Such that the ratio is 5:3:2 as mentioned in question where x is any positive number

Total surface area of cuboid = 558 cm²

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

⇒ 558 = 2 × [(5x)(3x) + (3x)(2x) + (2x)(5x)]

⇒ = 15x² + 6x² + 10x²

⇒ 279 = 31x²

⇒ x² =

⇒ x² = 9

⇒ x = ± 3

x is length therefore x cannot be negative therefore x = 3

therefore,

length of cuboid = 5x = 5 × 3 = 15 cm

breadth of cuboid = 3x = 3 × 3 = 9 cm

height of cuboid = 2x = 2 × 3 = 6 cm

length of edges of cuboid are 15 cm, 9 cm and 6 cm

Diameter of cylinder = d = 14 cm

⇒ radius of cylinder = r = = = 7 cm

Height of cylinder = h = 15 cm

Total surface area will be the sum of curved surface area and the area of two circles flat surfaces

Total surface area of cylinder = 2πrh + 2πr²

= 2 × × 7 × 15 + 2 × × 7²

= 30 × 22 + 2 × 22 × 7

= 660 + 308

= 968 cm²

Volume of cylinder = πr²h

= × 7² × 15

= 22 × 7 × 15

= 22 × 105

= 2310 cm³

Therefore, total surface area of cylinder is 968 cm² and volume is 2310 cm³

Height of cylinder = h = 7 cm

Radius of base of cylinder = r = 3 cm

Curved surface area of cylinder = 2πrh

= 2 × × 3 × 7

= 6 × 22

= 132 cm²

Total surface area will be the sum of the curved surface area and the area of two circles flat surfaces

Total surface area of cylinder = 132 + 2πr²

= 132 + 2 × 3.14 × 3²

= 132 + 6.28 × 9

= 132 + 56.52

= 188.52 cm²

Volume of cylinder = πr²h

= × 3² × 7

= 9 × 22

= 198 cm³

Therefore, curved surface area of cylinder is 132 cm², total surface area is 188.52 cm² and volume is 198 cm³

Area of circular base = 154 cm²

Height = h = 21 cm

Let r be radius of circular base

Area = πr²

⇒

⇒ = r²

⇒ r² = 7 × 7 = 49

⇒ r = ± 7

Radius cannot be negative hence r = 7

Volume of cylinder = πr²h

= × 7² × 21

= 7 × 21 × 22

= 3234 cm³

Curved surface area of cylinder = 2πrh

= 2 × × 7 × 21

= 14 × 22 × 3

= 924 cm²

Therefore, volume of cylinder is 3234 cm³ and curved surface area is 924 cm²

Let the radii of two circles r₁ and r₂

⇒ =

Let the heights of two cylinders h₁ and h₂

⇒ =

Curved surface area of cylinder = 2πrh

Ratio of their surface areas =

= ×

= ×

=

Volume of cylinder = πr²h

Ratio of their volumes =

= ×

= ×

= ×

=

Therefore, ratio of their curved surface areas is 5:6 and the ratio of their volumes is 5:9

Total surface area of cylinder = 462 m²

Curved surface area of cylinder = × Total surface area of cylinder

= × 462

= 154 m²

Curved surface area of cylinder = 154 m²

We know

Total surface area of cylinder = curved surface area + area of circles at top and bottom of cylinder

⇒ 462 = 154 + area of circles at top and bottom of cylinder

⇒ area of circles at top and bottom of cylinder = 462 – 154 = 308 m²

Let r be the radius of to and bottom circle of cylinder

area of circles at top and bottom of cylinder = 2πr²

⇒ 2πr² = 308

⇒

⇒ r² =

⇒ r² = 7 × 7

⇒ r = ± 7

r is radius and radius cannot be negative hence r = 7 m

now to find volume we need to find one more parameter about the cylinder which is the height

let us assume the height to be h

curved surface area of cylinder = 2πrh

⇒ 2πrh = 154

⇒ 2 × × 7 × h = 154

⇒ h =

⇒ h = m

volume of cylinder = πr²h

= × 7² ×

= 11 × 7 × 7

= 539 m³

Therefore, volume of cylinder is 539 m³

Height = 15 cm

lateral curved surface area of a cylinder = 660 sq. cm

let r be the radius

lateral curved surface area of a cylinder = 2πrh

⇒ 2πrh = 660

⇒ 2 × × r × 15 = 660

⇒ 30 × r =

⇒ 30 × r = 30 × 7

⇒ r = 7 cm

Volume of cylinder = πr²h

= × 7² × 15

= 22 × 7 × 15

= 2310 cm³

Therefore, volume is 2310 cm³

Let the radius of base of cylinder be r

Area of base = 6π cm²

Area of base = πr²

⇒ πr² = 6π

⇒ r² = 6

Volume of cylinder = 30π cm³

Let h be the height of cylinder

Volume of cylinder = πr²h

⇒ πr²h = 30π

⇒ 6h = 30

⇒ h = 5 cm

Height of cylinder is 5 cm

Let r be the radius of base and h be the height of cylinder

Curved surface area of cylinder = 660 cm²

Curved surface area = 2πrh

⇒ 2πrh = 660

⇒ 2 × × r × h = 660

⇒ r × h =

⇒ r × h =

⇒ r × h = 15 × 7

⇒ r × h = 105

⇒ rh = 105

Volume of cylinder = 1650 cm³

Volume of cylinder = πr²h

⇒ πr²h = 1650

⇒ × r × r × h = 1650

Using r × h = 105 we get

⇒ × r × 105 = 1650

⇒ 22 × 15 × r = 1650

⇒ r =

⇒ r =

⇒ r = 5 cm

Substitute r = 5 in r × h = 105 to get h

⇒ 5 × h = 105

⇒ h =

⇒ h = 21 cm

Therefore, radius is 5 cm and height is 21 cm

Let r be the radius and h be the height of cylinder

h = 7.5 cm = cm

r = 3.5 cm = cm

total surface area = 2πrh + 2πr²

curved surface area = 2πrh

ratio of total surface area to curved surface area =

⇒ ratio =

⇒ ratio =

⇒ ratio = 1 +

⇒ ratio = 1 +

⇒ ratio = 1 +

⇒ ratio =

⇒ ratio =

Therefore, ratio between the total surface area and curved surface area of the cylinder is 22:15

The volume of earth removed by digging for cylindrical well is the same volume used to make platform which is cuboidal in shape

Therefore, volume of well dug = volume of platform formed

Parameters of well

Diameter = d = 7 m

⇒ radius = r = = m

The well is 20 m deep so we can say that its height is 20 m

Height = h = 20 m

Volume of cylinder = πr²h

⇒ volume of well dug = × × 20

= × × 20

= 22 × 7 × 5

= 770 m³

Parameters for platform

Length = l = 22 m

Breadth = b = 14 m

Let height be h

Volume of cuboid = l × b × h

⇒ Volume of platform = 22 × 14 × h

= 308h m³

As volume of well dug = volume of platform formed

⇒ 770 = 308h

⇒ h =

Divide numerator and denominator by 7 we get

⇒ h =

Divide numerator and denominator by 22 we get

⇒ h = = 2.5

Therefore, height of platform is 2.5 m

Volume of water filled in cylinder = volume of cylindrical container

Radius = r = 14 cm

Let Height = h

Volume of cylindrical container = 30800

Volume of cylinder = πr²h

⇒ πr²h = 30800

⇒ × 14² × h = 30800

⇒ × 14× 14 × h = 30800

⇒ 28 × 22 × h = 30800

⇒ h =

⇒ h =

Divide numerator and denominator by 14

⇒ h = = 50 cm

Curved surface area of cylinder = 2πrh

⇒ inner curved surface area = 2 × × 14 × 50

= 100 × 22 × 2

= 4400 cm²

Therefore, inner curved surface area of the container is 4400 cm²

Cross section of cylinder i.e. top/bottom of cylinder will look like as shown

Inner diameter of cylinder = 14 cm

⇒ Inner radius = r_i = = 7 cm

As the thickness is 2 cm it can be seen from figure that external radius = inner radius + 2

⇒ external radius = r_e = 7 + 2 = 9 cm

Height of cylinder = h = 26 cm

Curved surface area of cylinder = 2πrh

⇒ external curved surface area of cylinder = 2πr_eh

⇒ inner curved surface area of cylinder = 2πr_ih

As seen the figure the area between the external circle and inner circle will also be counted in total surface area

That area is given by subtracting inner circle area from external circle area

Let us call that area as ring area

⇒ ring area = πr_e² – πr_i²

= π × (r_e² – r_i²)

Using identity a² – b² = (a + b)(a – b)

⇒ ring area = π × (r_e+ r_i) × (r_e– r_i)

We have two such ring areas at the top and bottom of cylinder

So total ring area = 2 × π × (r_e+ r_i) × (r_e– r_i)

Now,

Total surface area of hollow cylinder = external curved surface area of cylinder + inner curved surface area of cylinder + total ring area

⇒ Total surface area of hollow cylinder = 2πr_eh + 2πr_ih + 2 × π × (r_e+ r_i) × (r_e– r_i)

⇒ Total surface area of hollow cylinder = 2πh(r_e + r_i) + 2π(r_e+ r_i)(r_e– r_i)

⇒ Total surface area of hollow cylinder = 2π(r_e + r_i)[h + r_e– r_i]

Substituting values

⇒ Total surface area of hollow cylinder = 2 × × (9 + 7)[26 + 9– 7]

= 2 × × 16 × 28

= 32 × 22 × 4

= 2816

Therefore, total surface of the hollow cylinder is 2816 cm²

Height of cylinder = h = 20 cm

Inner diameter = 26 cm

⇒ Inner radius = r_i = = 13 cm

Inner diameter = 30 cm

⇒ Inner radius = r_o = = 15 cm

Volume of hollow cylinder = π(r_i² – r_o²)h

⇒ volume of hollow cylinder = × (15² – 13²) × 20

= × (225 – 169) × 20

= × 56 × 20

= 22 × 8 × 20

= 3520 cm³

Therefore, volume of cylinder is 3520 cm³

Height = h = 28 cm

Radius of base = r = 21 cm

Curved surface area of cone = πrl

Where l =

⇒ l =

⇒ l =

⇒ l =

⇒ l = 35 cm

Therefore,

Curved surface area = × 21 × 35

= 22 × 21 × 5

= 2310 cm²

Total surface area of cone = curved surface area + area of base

= 2310 + πr²

= 2310 + × 21 × 21

= 2310 + 22 × 3 × 21

= 2310 + 1386

= 3696 cm²

Volume of cone = πr²h

= × × 21 × 21 × 28

= 22 × 21 × 28

= 12936 cm³

Therefore, curved surface area is 2310 cm², total surface area is 3696 cm² and volume is 12936 cm³

Height of cone = h = 24 cm

Volume of cone = 1232 cm³

Let r be the radius of base of cone

Volume of cone = πr²h

⇒ × × r² × 24 = 1232

⇒ × r² × 8 = 1232

⇒ × r² =

⇒ × r² = 154

⇒ r² =

⇒ r² = 7 × 7

⇒ r = 7

Slant height l =

⇒ l =

⇒ l =

⇒ l =

⇒ l = 25 cm

Therefore, slant height of cone = 25 cm

Diameter of base = 14 m

⇒ radius of base = r = = 7 m

Let h be the height of cone

Slant height = l = 25 m

Total surface area of cone = curved surface area of cone + base area of cone

⇒ total surface area = πrl + πr²

= πr(l + r)

= × 7 × (25 + 7)

= 22 × 32

= 704 m²

Therefore, total surface area of the cone is 704 m²

Radius of base of cone = r = 14 cm

Slant height = l = 50 cm

Curved surface area of cone = πrl

= × 14 × 50

= 22 × 2 × 50

= 2200 cm²

Total surface area of cone = curved surface area of cone + base area of cone

⇒ total surface area = 4400 + πr²

= 4400 + × 14 × 14

= 2200 + 22 × 2 × 14

= 2816 cm²

Therefore, curved surface area is 2200 cm² and total surface area is 2816 cm²

Height of cone = 8 cm

Radius of base = r = 6 cm

Volume of cone = πr²h

= × × 6 × 6 × 8

= × 2 × 48

=

=

= 301.714 cm³

Therefore, volume is 301.714 cm³

Curved surface area of cone = 1884.4 m²

Slant height = l = 12 m

Let r be the radius of base of cone

Curved surface area of cone = πrl

⇒ 1884.4 = × r × 12

⇒ 1884.4 = × r × 12

⇒ r =

⇒ r =

⇒ r = 49.96 ~ 50 m

Therefore, radius of base of cone is 50 m

Area of base of cone = 154 cm²

Let r be the radius of cone

Area of base of cone = πr²

⇒ πr² = 154

⇒ × r² = 154

⇒ r² =

⇒ r² = 7 × 7

⇒ r = 7 cm

Slant height = 25 cm

Let h be the height of cone

Slant height =

⇒ 25 =

Squaring both sides we get

⇒ 625 = 7² + h²

⇒ 625 = 49 + h²

⇒ h² = 625 – 49

⇒ h² = 576

⇒ h = ± 24

Height cannot be negative therefore h = 24 cm

Height of cone is 24 cm

As the diameters of the cones are equal, their radius is also equal

Let the radius of both cones be r

Let the slant height of bigger cone be l₁ and that of smaller cone be l₂

l₁:l₂ = 5:4

⇒ =

Curved surface area of smaller cone = 400 cm²

Curved surface area of cone = πrl

Curved surface area of bigger cone = πrl₁

Curved surface area of smaller cone = πrl₂

Consider ratio of curved surface area of bigger cone to smaller cone

⇒ =

⇒ =

⇒ =

⇒ Curved surface area of bigger cone =

⇒ Curved surface area of bigger cone = 5 × 100

⇒ Curved surface area of bigger cone = 500 cm²

Therefore, curved surface area of bigger cone is 500 cm²

Let the slant height of cone be l and radius of base of cone be r

Ratio of slant height to radius = 7:4

⇒ =

⇒ l = × r

Curved surface area of cone = 792 cm²

⇒ πrl = 792

put l = × r

⇒ × r × × r = 792

⇒ × r² = 792

⇒ r² =

⇒ r² = 36 × 4

⇒ r² = 144

⇒ r² = √144

⇒ r = ± 12

r is radius and radius cannot be negative thus r = 12 cm

Height of cone = h = 9 m

Circumference of base of cone = 44 m

Let r be the radius of base of cone

Circumference of base of cone = 2πr

⇒ 2πr = 44

⇒ πr = 22

⇒ × r = 22

⇒ r =

⇒ r = 7 m

Volume of air present inside the cone = volume of cone

Volume of cone = πr²h

= × × 7² × 9

= 22 × 7 × 3

= 462 m³

Therefore, volume of air inside cone is 462 m³

Radius of base of conical vessel = r = 10 cm

Height of conical vessel = h = 18 cm

Volume of water in conical vessel = volume of cone

Volume of cone = πr²h

⇒ Volume of water in conical vessel = × π × 10² × 18

= π × 100 × 6

= 600π

Now this volume of water is transferred in cylindrical vessel whose radius is 5 cm

Let h₁ be the height up to which the water rises in the cylindrical vessel

Volume of water in cylindrical vessel = π(radius)²h

= π × 5² × h

= 25hπ

As water is transferred from conical vessel to cylindrical vessel volume of water is same

⇒ Volume of water in conical vessel = Volume of water in cylindrical vessel

⇒ 600π = 25hπ

⇒ 25h = 600

⇒ h =

⇒ h = 24 cm

Therefore, the height to which the water rises is 24 cm

Volume of cone = πr²h

For volume to be largest the radius of base of cone ‘r’ and height of cone ‘h’ should be maximum

In a cube whose edge is 14 cm

As it can be seen from the figure that the maximum height of cone can be 14 cm and also the maximum diameter can be 14 cm

Radius = r = = 7 cm

Height = h = 14 cm

⇒ Volume of cone = × × 7² × 14

= × 22 × 49 × 2

=

=

= 718.67 cm²

Therefore, volume of the largest right circular cone that can be cut out of the cube is 718.67 cm²

Base radius of cone = r = 7 cm

Height of cone = h = 24 cm

Slant height = l =

⇒ l =

⇒ l =

⇒ l =

⇒ l = 25 cm

Curved surface area = πrl

= × 7 × 25

= 22 × 25

= 550 cm²

Total surface area of cone = curved surface area + area of base

= πrl + πr²

= πr(l + r)

= × 7 × (25 + 7)

= 22 × 32

= 704 cm²

Volume of cone = πr²h

= × × 7² × 24

= 22 × 7 × 8

= 1232 cm³

Therefore, slant height of cone is 25 cm, curved surface area is 550 cm², total surface area is 704 cm² and volume is 1232 cm³

The sector is as shown in the figure with TU = 12 as radius and ∠VTU = 120°

Now edges VT and UT are joined to form the cone as shown and hence points U and V become the same

For the cone formed TU or TV is the slant height of cone

TU = l = 12 cm

To find volume of cone we need to find base radius of cone and height of cone

Radius of base of cone as seen in figure is OV or OU let us assume that to be r_c

To find base radius of cone we can use the fact that the cone is made by folding the sector which means the curved surface area of cone will be same as area of sector thus we will find those areas and equate them

Area of sector =

r is the radius of sector which is TU = 12 cm and θ is the angle which is ∠VTU = 120°

⇒ area of sector =

=

=

= 48π

Curved surface area of cone = πr_cl

= π × r_c × 12

= 12πr_c

Therefore,

⇒ 48π = 12πr_c

⇒ r_c = 4 cm

Now we have found the base radius of cone we need to find one more parameter which is the height to find volume

Let the height of the cone be h which is TO in the diagram

using the formula

l =

⇒ 12 =

Squaring both the sides

⇒ 144 = 16 + h²

⇒ h² = 144 – 16

⇒ h² = 128

⇒ h² = 64 × 2

⇒ h = 8√2 cm

Volume of cone = πr_c²h

= × × 4² × 8√2

= × × 16 × 8√2

=

=

= 134.095 × √2

= 134.095 × 1.414

= 189.61 cm³

Therefore, volume of cone is 189.61 cm³

Radius of sphere = r = 1.4 cm

Surface area of sphere = 4πr²

= 4 × × 1.4 × 1.4

= 5.6 × 22 × 0.2

= 24.64 cm²

Volume of sphere = πr³

= × × 1.4 × 1.4 × 1.4

= × 22 × 0.2 × 1.4 × 1.4

= × 8.624

= 11.49 ~ 11.5 cm³

Therefore, surface area of sphere is 24.64 cm² and volume of sphere is 11.5 cm³

Surface area of sphere = 616 cm²

Let the radius of sphere be r

Surface area of sphere = 4πr²

⇒ 4πr² = 616

⇒ πr² =

⇒ πr² = 154

⇒ × r² = 154

⇒ r² =

⇒ r² = 7 × 7

⇒ r = 7 cm

Volume of sphere = πr³

= × × 7³

= × 22 × 7²

=

=

= 1437.33 cm³

Therefore, volume of sphere is 1437.33 cm³

Hemisphere is half of a sphere

So the volume and area will be halved

But for total surface area there would be one more area which is the flat circle of a hemisphere

Radius of hemisphere = r = 4.5 cm =

Surface area of hemisphere = 3πr²

= 3 × ×

= 3 × ×

=

=

= 190.92 cm²

Volume of hemisphere = πr³

= × ×

=

=

= 190.92 cm³

Therefore, surface area is 190.92 cm² and volume is 190.92 cm³

Volume of sphere = 38808 cm³

Let r be the radius of sphere

Volume of sphere = πr³

⇒ πr³ = 38808

⇒ × × r³ = 38808

⇒ r³ =

⇒ r³ =

⇒ r³ = 441 × 21

⇒ r³ = 9261

⇒ r = 21 cm

Surface area of sphere = 4πr²

= 4 × × 21 × 21

= 4 × 22 × 3 × 21

= 5544 cm²

Therefore, surface area is 5544 cm²

Radius of each lead shot = r_s = 2 cm

Radius of cylinder = r_c = 4 cm

Height of cylinder = h = 10 cm

volume of sphere = πr_s³

Volume of cylinder= πr_c²h

The lead shots area made by melting the cylinder which means same melted volume is used to make the lead shots thus the volume remains the same

Let ‘n’ be the number of lead shots formed

i.e. volume of n lead shots = volume of cylinder

⇒ n × πr_s³ = πr_c²h

⇒ 4 × r_s³ × n = 3 × r_c² × h

Substituting values

⇒ 4 × 2³ × n = 3 × 4² × 10

⇒ 4 × 8 × n = 3 × 16 × 10

⇒ 4 × n = 3 × 2 × 10

⇒ 4 × n = 3 × 20

⇒ n = 3 × 5

⇒ n = 15

Number of lead shots made = 15

Outer radius = r₁ = 8 cm

The shell is 2 cm thick therefore we must subtract 2 cm from outer radius to get the inner radius

Inner radius = r₂ = 8 – 2 = 6 cm

Volume of metal required to make shell = volume of shell

Volume of shell we can get by subtracting the volume of hollow part of sphere from the complete volume of sphere with 8 cm radius

⇒ Volume of shell = πr₁³ – πr₂³

= π × (r₁³ – r₂³)

Substituting values

⇒ Volume of shell = × × (8³ – 6³)

= × × (512 – 216)

= × × 296

=

= 1240.38 cm³

Therefore, volume of the metal to make the shell is 1240.38 cm³

base radius of cone= r_c = 3 cm

Height of cone = h = 6 cm

Radius of sphere = r_s = 9 cm

volume of cone = πr_c²h

Volume of sphere = πr_s³

Let n be the number of cones made

As the sphere is melted and then the cones are made from the same amount of melted sphere therefore the volume remains same

i.e. volume of n cones made = volume of sphere

⇒ n × πr_c²h = πr_s³

⇒ nr_c²h = 4r_s³

Substituting values

⇒ n × 3² × 6 = 4 × 9³

⇒ n × 6 = 4 × 9 × 9

⇒ n = 6 × 9

⇒ n = 54

Number of cones formed = 54

Let the radius of small spheres formed be r₂

Radius of sphere which is melted = r₁ = 10 cm

As the sphere is melted and then the small spheres are made from the same amount of melted sphere, therefore, the volume remains same

8 small spheres are made

Volume of sphere = πr³

⇒ 8 × πr₂³ = πr₁³

⇒ 8r₂³ = r₁³

Substituting values

⇒ 8 × r₂³ = 10³

⇒ 8 × r₂³ = 1000

⇒ r₂³ =

⇒ r₂ = = 5 cm

Surface area of sphere = 4πr²

⇒ surface area of small spheres made = 4πr₂²

= 4 × π × 5²

= 4 × π × 25

= 100π cm²

Therefore, surface area of each sphere obtained is 100π cm²

Surface area of sphere = 5544 cm²

Let r be the radius of sphere

⇒ 4πr² = 5544

⇒ πr² = 1386

⇒ × r² = 1386

⇒ r² = 63 × 7

⇒ r² = 441

⇒ r = ± 21

r is radius and radius cannot be negative thus r = 21

volume of sphere = πr³

= × × 21 × 21 × 21

= 4 × 22 × 3 × 7 × 21

= 38808 cm³

Therefore, volume of the sphere is 38808 cm³

Diameter of spherical lead shot = 4.2 cm

⇒ radius of spherical lead shots = r = = 2.1 cm

Length of solid lead = l = 66 cm

Breadth = b = 42 cm

Height = h = 21 cm

Volume of sphere = πr³

Volume of rectangular lead = lbh

As the spherical lead shots are obtained from rectangular lead volume remains same

Let n spherical lead shots are made

⇒ n × πr³ = lbh

Substituting values

⇒ n × × × 2.1³ = 66 × 42 × 21

⇒ n × × × 2.1² = 3 × 42 × 10

⇒ n × × × 2.1 = 3 × 20 × 10

⇒ n × × 0.3 = 600

⇒ n × 4 × 0.1 = 600

⇒ n × 4 × = 600

⇒ n × 4 = 6000

⇒ n = 1500

Therefore, 1500 spherical lead shots are formed

Diameter of sphere = 6 cm

Radius of sphere = r_s = = 3 cm

The diameter of cylindrical vessel = 12 cm

Radius of cylindrical vessel = r_c = = 6 cm

The volume of sphere submerged in water is equal to the volume of water rise

Let h be the rise of water level

it is given that the sphere is submerged completely

thus, volume of sphere = volume of water rise

volume of sphere = πr_s³

volume of cylinder = πr_c²h

⇒ πr_s³ = πr_c²h

⇒ 4r_s³ = 3r_c²h

⇒ 4 × 3³ = 3 × 6² × h

⇒ 4 × 9 × 3 = 3 × 36 × h

⇒ 4 × 3 = 3 × 4 × h

⇒ 3 = 3 × h

⇒ h = 1 cm

Therefore, rise in water level in the cylindrical vessel is 1 cm

Radius of hemispherical bowl = r_h = 9 cm

Diameter of cylindrical bottle = 3 cm

Radius of cylindrical bottle = r_c = cm

Height of cylindrical bottle = h = 4 cm

The volume of water is transferred from hemispherical bowl to cylindrical bottles

Suppose there are n bottles then volume of water filled in n bottles will be same as volume of water in hemispherical bowl

Volume of hemisphere = πr_h³

Volume of cylinder = πr_c²h

⇒ πr_h³ = n × πr_c²h

⇒ 2r_h³ = 3nr_c²h

Substitute values

⇒ 2 × 9³ = 3n × × 4

⇒ 2 × 9³ = 3n × 9

⇒ 2 × 9² = 3n

⇒ 162 = 3n

⇒ n = 54

Therefore, 54 bottles are required to empty the bowl

Total volume of sphere = 3000 × volume of one sphere

diameter of sphere = 0.7 cm

radius of sphere = r = = = cm

Volume of sphere = πr³

⇒ volume of one sphere = × × × ×

= × cm³

= × cm³

= cm³

3000 such spheres are submerged

Total volume of sphere = 3000 × volume of one sphere

= 3000 ×

= 539 cm²

The total volume of spheres submerged in tank is the total amount of water that overflows

Therefore, 539 cm² volume of water overflows

The hemispherical bowl will look like as shown

To find the cost required to paint we first need to calculate its total surface area

Total surface area will include the area of outter curved surface the inner curved surface and the flat surface as shown

Diameter of outer surface = 43 cm

Radius of outer surface = r_o = cm = 21.5 cm

Diameter of inner surface = 42 cm

Radius of inner surface = r_i = cm = 21 cm

Curved surface area of hemisphere = 2πr²

⇒ outter curved surface area = 2πr_o²

= 2 × π ×

= π cm²

= 924.5π cm²

⇒ inner curved surface area = 2πr_i²

= 2 × π × 21²

= 882π cm²

The flat area can be found by subtracting the inner circle area from outter circle area

⇒ flat area = πr_o² – πr_i²

= π(21.5)² – π(21)²

= 462.25π – 441π

= 21.25π cm²

Total surface area of hemispherical vessel = outter curved surface area + inner curved surface area + flat area

⇒ Total surface area of hemispherical vessel = 924.5π + 882π + 21.25π = 1827.75π cm²

Now cost of colouring 1 cm² is 7 paise

So, cost of colouring 1827.75π cm² = 7 × 1827.75 × π

= 7 × 1827.75 ×

= 1827.75 × 22

= 40210.5 paise

100 paise is 1 Rs

⇒ 40210.5 paise = = 402.105 Rs

Therefore, cost of colouring the vessel is 402.105 Rs

Let the side of cube be a

Total surface area of cube = 6a²

⇒ 6a² = 486

⇒ a² = 81

⇒ a = 9 cm

Length of cuboid = l = 9 m

Breadth of cuboid = b = 2 m

Height of cuboid = h = 1 m

Lateral surface area of cuboid is surface area without the top and bottom surfaces

So here we won’t consider the area l × b

Lateral surface area = 2lh + 2bh

= 2h(l + b)

= 2 × 1 × (9 + 2)

= 22 m²

Diameter of sphere = 6 cm

Radius of sphere = r = = 3 cm

Volume of sphere = πr³

= × π × 3³

= 4 × 3² × π

= 36π cm³

Radius of cylinder = r = 14 cm

Height of cylinder = h = 10 cm

Curved surface area of cylinder = 2πrh

= 2 × × 14 × 10

= 2 × 22 × 2 × 10

= 880 cm²

Let r be the base radius of cone

Volume of cone = 308 cm³

Height = h = 6 cm

Volume of cone = πr²h

⇒ 308 = × × r² × 6

⇒ = r² × 2

⇒ 14 × 7 = r² × 2

⇒ r² = 7 × 7

⇒ r = 7 cm

Diameter of hemisphere = 42 cm

Radius of hemisphere = r = = 21 cm

Total surface area of hemisphere = 3πr²

= 3 × × 21²

= 3 × 22 × 3 × 21

= 4158 cm²

Cost of polishing 1 sq. cm is 20 paise

⇒ cost of polishing 4158 cm² = 4158 × 20

= 83160 paise

100 paise is 1 Rs

⇒ 83160 paise = = 831.6 Rs

Therefore, cost of polishing its total surface is 831.6 Rs

The cone, hemisphere and cylinder stand on equal bases which means their base radius is same

Let that base radius be r

It is given that height = r

Volume of cone = πr²h = πr³

Volume of hemisphere = πr³

Volume of cylinder = πr²h =πr³

Ratio of cone to hemisphere = =

Ratio of hemisphere to cylinder = =

⇒ ratio of volumes of all three solid = 1:2:3

The solid will look like as shown

Diameter of cylinder = 14 cm

Radius of cylinder = r = = 7 cm

Height of cylinder = h = 40 cm

Volume of cylinder = πr²h

= × 7² × 40

= 22 × 7 × 40

= 6160 cm³

Diameter of base of cone = 14 cm

Radius of base of cone = r₁ = = 7 cm

Height of cone = 12 cm

Volume of cone = πr²h

= × × 7² × 12

= 22 × 7 × 4

= 616 cm³

Volume of solid = Volume of cylinder + Volume of cone

= 6160 + 616

= 6776 cm³

Therefore, volume of solid is 6776 cm³

Radius of metal sphere = r_s = 9 cm

Radius of cone formed = r_c = 3 cm

Height of cone = h = 6 cm

The sphere is melted and the volume of sphere is used to make cones

Volume of sphere = πr_s³

Volume of cone = πr_c²h

Let n cones are made then the volume of n cones is same as the volume of sphere

⇒ πr_s³ = n × πr_c²h

⇒ 4r_s³ = nr_c²h

Substitute values

⇒ 4 × 9³ = n × 3² × 6

⇒ 4 × 9² = n × 6

⇒ 4 × 9² = 6n

⇒ 324 = 6n

⇒ n = 54

Therefore, 54 cones are made

Length of tank = l = 20 m

Breadth of tank = b = 15 m

Height of tank = h = 6 m

Volume of water in tank = volume of cuboidal tank

Volume of cuboid = lbh

⇒ volume of cuboidal tank = 20 × 15 × 6

= 1800 m³

1 m³ is 1000 litres

⇒ 1800 m³ = 1800 × 1000 = 1800000 litres

150 litres of water is required per head per day and the population of village is 4000

⇒ volume of water consumed by 4000 people per day = 4000 × 150 = 600000 litres

In one day village consumes 600000 litres of water

⇒ days require to consume 1800000 litres of water = = 3

Thus water of tank will last 3 days

Three spheres are melted and one big sphere is made out of the melted spheres

Thus volume of sphere formed will be same as volume of 3 melted spheres

Let the radius of spheres be r₁ = 6 cm, r₂ = 8 cm and r₃ = 10 cm

⇒ Volume of spheres melted = πr₁³ + πr₂³ + πr₃³

= π(r₁³ + r₂³ + r₃³)

Let r be the radius of sphere formed

⇒ Volume of sphere formed = πr³

As volumes are same

⇒ Volume of spheres melted = Volume of sphere formed

⇒ π(r₁³ + r₂³ + r₃³) = πr³

⇒ r₁³ + r₂³ + r₃³ = r³

Substituting values

⇒ 6³ + 8³ + 10³ = r³

⇒ 216 + 512 + 1000 = r³

⇒ r³ = 1728

⇒ r = 12 cm

Therefore, radius of sphere made is 12 cm

Base radius of conical vessel = r₁ = 10 cm

Height of conical vessel = h₁ =18 cm

Base radius of cylindrical vessel = r₂ = 5 cm

Let h be the height of water in the cylindrical vessel

(note that we does not need to know the height of cylindrical vessel just assume that the height will be till the water level)

As the water is transferred from conical vessel to cylindrical vessel volume is unchanged

Volume of conical vessel = πr₁²h₁

Volume of cylindrical vessel = πr₂²h

⇒ πr₁²h₁ = πr₂²h

⇒ r₁²h₁ = 3r₂²h

Substituting values

⇒ 10² × 18 = 3 × 5² × h

⇒ 100 × 6 = 25× h

⇒ h = 4 × 6

⇒ h = 24 cm

Therefore, height of water in the cylindrical vessel is 24 cm

Length of wax cuboid = l = 11 cm

Breadth of wax cuboid = b = 3.5 cm = cm

Height of wax cuboid = h = 2.5 cm = cm

Diameter of candle = 2.8 cm

Radius of candle = r = = 1.4 cm

Let the length of candle be l₁

The candle is cylindrical so the length of candle will be its height

Candle is made from wax cuboid hence volume of cuboid wax is equal to volume of cylindrical candle candle

Volume of wax cuboid = lbh

Volume of cylindrical candle = πr²l₁

⇒ lbh = πr²l₁

⇒ 11 × × = × 1.4² × l₁

⇒ × = 2 × 0.2 × 1.4 × l₁

⇒ × = 0.56 × l₁

⇒ × = 0.08 × l₁

⇒ × = × l₁

⇒ l₁ =

⇒ l₁ =

⇒ l₁ =

⇒ l₁ = 15.625 cm

Therefore, candle is 15.625 cm long

Diameter of sphere = 6 cm

Radius of sphere = r_s = = 3 cm

Length of wire = h = 36 m = 3600 cm

Let the radius of wire be r_w

Wire is cylindrical

Wire is formed by melting the sphere thus volume of wire is same as volume of sphere

Volume of sphere = πr_s³

Volume of cylindrical wire = πr_w²h

⇒ πr_s³ = πr_w²h

⇒ × r_s³ = r_w² × h

Substitute values

⇒ × 3³ = r_w² × 3600

⇒ 4 × 3² = r_w² × 3600

⇒ 36 = r_w² × 3600

⇒ r_w² =

⇒ r_w = 0.1 cm

Therefore, radius of wire is 0.1 cm