Volume And Surface Area

(i) length = 12 cm, breadth = 8 cm and height = 4.5 cm

Volume of cuboid = (length × breadth × height) = (12 × 8 × 4.5) = 432 cm³

Lateral surface area of cuboid = 2(length + breadth) × height = 2(12 + 8) × 4.5 = 180 cm²

Total surface area of cuboid = 2(length × breadth + breadth × height + height × length)

= 2(12 × 8 + 8 × 4.5 + 4.5 × 12) = 2(96 + 36 + 54) = 2 × 186 = 372 cm²

(ii) length = 26 m, breadth = 14 m and height = 6.5 m

Volume of cuboid = (length × breadth × height) = (26 × 14 × 6.5) = 2366 m³

Lateral surface area of cuboid = 2(length + breadth) × height = 2(26 + 14) × 6.5 = 520 m²

Total surface area of cuboid = 2(length × breadth + breadth × height + height × length)

= 2(26 × 14 + 14 × 6.5 + 6.5 × 26) = 2 × 624 = 1248 m²

(iii) length = 15 m, breadth = 6 m and height = 5 dm = (0.5m)

Volume of cuboid = (length × breadth × height) = (15 × 6 × 0.5) = 45 m³

Lateral surface area of cuboid = 2(length + breadth) × height = 2(15 + 6) × 0.5 = 21 m²

Total surface area of cuboid = 2(length × breadth + breadth × height + height × length)

= 2(15 × 6 + 6 × 0.5 + 0.5 × 15) = 2(90 + 3.0 + 7.5) = 2 × 100.5 = 201 m²

(iv) length = 24 m, breadth = 25 cm and height = 6 m

Volume of cuboid = (length × breadth × height) = (24 × 0.25 × 6) = 36 m³

Lateral surface area of cuboid = 2(length + breadth) × height = 2(24 + 0.25) × 6 = 291 m²

Total surface area of cuboid = 2(length × breadth + breadth × height + height × length)

= 2(24 × 0.25 + 0.25 × 6 + 6 × 24) = 303 m²

_Given,

_{Dimensions of closed rectangular cistern = 8m × 6m × 2.5 m}

_{∴Capacity of tank = volume of tank = (l × b × h) = 8 × 6 × 2.5 = 120 m}³

_{Area of iron sheet required to make the tank =} 2(lb + bh + hl) = 2(8 × 6 + 6 × 2.5 + 2.5 × 8)=2(48 + 15 + 20)= 2 × 83 = 166m²

Given,

Dimensions of room = 9m × 8m × 6.5m

Area of 4 walls = 2 (length + breadth) × height = 2 (9 + 8) × 6.5 = 13 × 17 = 221 m²

Dimensions of one door = 2m × 1.5m

Area of door = length × breadth = 2 × 1.5 = 3.0 m²

Dimensions of windows = 1.5m × 1m

Area of 2 windows = 2 (l × b) = 2 (1.5 × 1) = 3.0 m²

Hence,

Area required for white-washing = Area of 4 walls – (area of door + area of 2 windows)

= 221 – (3 + 3) = 221 – 6 = 215 m²

∵ cost of white-washing 1 m² area = Rs. 6.40

∴ cost of white-washing 215 m² = 6.40 × 215 = Rs. 1376.

Given,

Dimensions of plank;

l = 5m

b = 25cm = 0.25 m

h = 10cm = 0.10 m

Dimensions of pit;

l = 20m

b = 6m

h = 80m

Given,

Dimensions of wall = 8m × 6m × 22.5 cm = 800 cm × 600 cm × 22.5 cm

Dimensions of each brick = 25 cm × 11.25cm × 6 cm

Hence,

Number of bricks required =

Given,

Dimensions of wall = 15m × 30cm × 4m = 1500 cm × 30 cm × 400 cm

Dimensions of each brick = 22 cm × 12.5 cm × 7.5 cm

Volume of wall = l × b × h = 1500 × 30 × 400 = 180000000 cm³

Area of mortar = 1/12 × volume of wall =

Hence,

Area occupied by bricks only = 180000000 – 15000000 = 165000000 cm³

Number of bricks required =

Given,

External Dimensions of cistern = 1.35m × 1.08m × 90cm = 135cm × 108cm × 90cm

External volume of cistern = l × b × h = 135 × 108 × 90 = 1312200 cm³

Internal dimensions of cistern = length = 135 – (2.5 × 2) = 130 cm

Breadth = 108 – (2.5 × 2) = 103 cm

Height = 90 – 2.5 = 87.5 cm

∴ internal volume of cistern = 130 × 103 × 87.5 = 1171625 cm³

Volumeof iron used = (External volume – Internal volume)

= 1312200 – 1171625 = 140575 cm³

Given,

Depth of river (h) = 2 m

Breadth of river (b) = 45 m

Rate of flowing = 3 km/h

∴ Length = meter/min.

Volume of water = l × b × h = × 2 × 45 = 90 × 50 = 4500 m³

Given,

Total cost of box made of sheet metal = Rs. 1620

Cost of per square meter metal = Rs. 30

∴ Area of box = = 54 m²

Dimensions of box = 5m × 3m × height

Let height of box = h meter

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

= 54 = 2 (5 × 3 + 3h + 5h)

= = 15 + 8h

= 8h = 27 – 15 = 12

= h =

Height of box = 1.5 meter.

Given,

Dimensions of room = 10m × 10m × 5m

∴ length of longest pole can be put in room = diagonal of room

=

Given,

Dimensions of dining hall = 20m × 16m × 4.5m

Volume of hall = 20 × 16 × 4.5 = 1440 m³

Volume of air required by one person = 5 m³

∴ Number of persons in hall =

Given,

Dimensions of classroom = 10m × 6.4m × 5m

Area of room = length × breadth = 10 × 6.4 = 64 m²

Area of floor required by one student = 1.6 m²

∴ Number of students can sit in classroom =

Volume of classroom = 10 × 6.4 × 5 m³

Air required by each student =

Given,

Volume of cuboid = 1536 m³

Length of cuboid = 16 m

Ratio of breadth and height = 3 : 2

Let breadth = 3x

Let breadth = 2x

∴ Volume of cuboid = l × b × h

= 1536 = 16 × 3x × 2x

= 6x² =

= x² =

= x = = 4

Hence,

Breadth of cuboid = 3x = 3 × 4 = 12m

Height of cuboid = 2x = 2 × 4 = 8m

Given,

Surface area of cuboid = 758 cm²

Length of cuboid = 14 cm

Breadth of cuboid = 11 cm

Let height of cuboid = h cm

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

= 758 = 2 (14 × 11 + 11h + 14h)

= 154 + 25h =

= 25h = 379 – 154 = 225

= h =

Height of cuboid = 9 meter.

Given,

a) Edge of cube (a) = 9m

Volume of cube = a³ = 9³ = 729 m³

Lateral surface area of cube = 4a² = 4 × 9² = 4 × 81 = 324 m²

Total surface area of cube = 6a² = 6 × 9² = 6 × 81 = 486 m²

Diagonal of cube = a = × 9 = 1.73 × 9 = 15.57 m

b) Edge of cube (a) = 6.5 cm

Volume of cube = a³ = 6.5³ = 274.625 cm³

Lateral surface area of cube = 4a² = 4 × 6.5² = 4 × 42.25 = 169 cm²

Total surface area of cube = 6a² = 6 × 6.5² = 6 × 42.25 = 253.5 cm²

Diagonal of cube = a = × 6.5 = 1.73 × 6.5 = 11.245 cm

Given,

Total surface area of cube = 1176 cm²

Let edge of cube = a cm

= 6 a² = 1176

= a² =

= a = = 14 cm

∴ Volume of cube = a³ = 14³ = 2744 cm³

Given,

Lateral surface area of cube = 900 cm²

Let edge of cube = a cm

4a² = 900

= a² =

= a =

Volume of cube = a³ = 15³ = 3375 cm³

Given

Volume of cube = 512 cm³

Let edge of cube = a cm

So,

= a³ = 512

= a = = 8 cm

Total surface area of cube = 6 a² = 6 × 8 × 8 = 384 cm²

Given,

Edge of three cubes a₁ = 3 cm , a₂ = 4 cm , a₃ = 5 cm

Let edge of single cube formed = A cm

Sum of volume of three cubes = volume of single cube formed

= a₁³ + a₂³ + a₃³ = A³

= 3³ + 4³ + 5³ = A³

A³ = 27 + 64 + 125 = 216

A =

Lateral surface area of new cube = 4a² = 4 × 6 × 6 = 144 cm²

Given,

Area of field = 2 hectare = 20000 m²

Height of rainfall = 5 cm = 0.05 m

Volume of water that falls = Area × height

= 20000 × 0.05 = 1000 m³

Given,

Height of cylinder = 21 cm

Radius of base = 5 cm

∴ volume of right circular cylinder = πr²h = × 5 × 5 × 21 = 1650 cm³

Curved surface area = 2πrh = 2 × × 5 × 21 = 660 cm²

Given,

Diameter of cylinder = 28 cm

Height of cylinder = 40 cm

Radius of cylinder =

∴ Curved surface area of cylinder = 2πrh = 2 × × 14 × 40 = 44 × 40 × 2 = 3520 cm²

∴ total surface area of cylinder = 2πrh + 2πr² = 2πr(h + r) = 2 × × 14 × 54 = 88 × 54 = 4752 cm²

∴ Volume of cylinder = πr²h = × 14 × 14 × 40 = 24640 cm³

Given,

Radius of cylinder = 10.5 cm

Height of cylinder = 60 cm

∴ Volume of cylinder = πr²h = × 10.5 × 10.5 × 60 = 20790 cm³

∴ Weight of cylinder = volume of cylinder × wt. of cylinder per gram

= 20790 × 5 g = 103950 g = 103.95 kg

Given,

Curved surface area of cylinder = 1210 cm²

Diameter of cylinder = 20 cm

Radius of cylinder = = 10 cm

Let height of cylinder = h cm

Curved surface area = 2πrh

= 2πrh = 1210

= 2 × × 10 × h = 1210

= h = 19.25 cm

∴ Volume of cylinder = πr²h = × 10 × 10 × 19.25 = 6050 cm³

Given,

Curved surface area of cylinder = 4400 cm²

Circumference of its base = 110 cm

2πr = 110

= r =

Let height of cylinder = h cm

C.S.A = 4400

2πrh = 4400

= 2 × × × h = 4400

= h =

∴ Volume of cylinder = πr²h =

Given,

Volume of cylinder = 1617 cm³

Ratio of radius of base and height = 2 : 3

Let base radius = 2x cm

Let height = 3x cm

Volume = πr²h

=

= x³ =

= x³ = 42.875

= x =

Hence,

Radius of cylinder = 2 × 3.5 = 7 cm

Height of cylinder = 3 × 3.5 = 10.5 cm

Total surface area of cylinder = 2πrh + 2πr² = 2πr (h + r) = 2 × × 7 × 17.5 = 770 cm²

Given,

Total surface area of cylinder = 462 cm²

2πr (h + r) = 462

⇒ r (h + r) =

⇒ r² + rh =

CSA =

2πrh = × 462 = 154

⇒ rh =

Putting value of rh in equation (i)

⇒ r² +

⇒ r² =

⇒ r =

From (ii)

⇒ rh =

⇒ h =

∴ Volume of cylinder = πr²h = × 7 × 7 × = 532 cm³

Given,

Total surface area of solid = 231 cm²

2πr(h + r) = 231

⇒ r (r + h) =

⇒ r² + rh =

CSA =

2πrh =

⇒ rh =

Putting value of rh in (i) we get,

⇒ r² +

⇒ r² =

⇒ r =

From equation (ii)

⇒ rh =

⇒ h =

∴ Volume of cylinder = πr²h = × 3.5 × 3.5 × 7 = 269.5 cm³

Given,

Total surface area of cylinder = 1628 m²

Sum of height and radius = (h + r) = 37 m

2πr (h + r) = 1628

2πr × 37 = 1628

⇒ r =

∵ r + h = 37

⇒ h = 37 – 7 = 30 m

∴ Volume of cylinder = πr²h = × 7 × 7 × 30 = 4620 m³

Given,

Total surface area of cylinder = 616 cm²

⇒ 2h = r + h

⇒ h = r……………………….(i)

2πr (h + r) = 616

⇒ r (r + r) =

⇒ r² =

⇒ r =

H = 7 cm

∴ Volume of cylinder = πr²h = × 7 × 7 × 7 = 22 × 49 = 1078 cm³

Given,

Diameter of wire = 0.1 mm = 0.01 cm

Radius of wire =

Volume of gold = 1 cm³

⇒ πr²h = 1

⇒ h = = 12727.27 cm or 127.27 m

Length of wire = 127.27 meter.

Given,

Ratio of radii of two cylinders = R₁ : R₂ = 2 : 3

Ratio of their heights = H₁ : H₂ = 5 : 3

∴ Ratio of volumes of cylinders =

∴ Ratio of their curved surface area =

Given,

Side of square base = 12 cm

Height = 17.5 cm

Volume of tin = lbh = 12 × 12 × 17.5 = 2520 cm³

Diameter of cylindrical base = 12 cm

Radius =

Height of cylinder = 17.5 cm

Volume of tin in cylinder = πr²h =

Hence,

Capacity of square tin is more by = 2520 – 1980 = 540 cm³

Given,

Diameter of cylindrical bucket = 28 cm

Radius of bucket =

Height of bucket = 72 cm

Volume of water in bucket = πr²h =

Length of rectangular tank = 66 cm

Width of tank = 28 cm

Let rise in water level in rectangular tank = h cm

∵ Volume of cylinder = Volume of rectangular tank

= 66 × 28 × h

⇒ h =

Given,

Weight of 1 cm³ cast iron = 21 g

Length of wire = h = 1 m = 100 cm

Internal radius (r₁) =

Thickness of metal = 1 cm

So, External radius (r₂) = 1.5 + 1 = 2.5 cm

Volume of metal = (External volume – internal volume)

= πr₂²h – πr₁²h = πh (r₂² – r₁²) = × 100 (2.5 + 1.5) (2.5 – 1.5)

= × 100 × 4 × 1 cm³

Weight of metal =

Given,

Internal diameter of tube = 10.4 cm

Internal radius of tube =

Thickness of metal = 8 mm = 0.8 cm

External radius of tube = 5.2 + 0.8 = 6 cm

Length of tube = 25 cm

∴ Volume of metal = (external volume – internal volume)

= πh (6² – 5.2²) =

Given,

Length of cylindrical barrel (h) = 7 cm

Diameter = 5 mm

Radius =

Volume of cylindrical barrel = πr²h =

∵ cm³ volume of barrel is used for writing = 330 words

∴ will be used for writing =

Given,

Diameter of pencil = 7 mm

Radius of pencil =

Diameter of graphite = 1 mm

Radius of graphite =

Volume of graphite = πr²h =

Weight of graphite = volume × specific gravity of graphite

=

Volume of wood = volume of pencil – volume of graphite

=

=

∴ Total weight of the pencil = weight of wood + weight of graphite

= 0.165 + 2.64 = 2.805 g.

Given,

Radius of the cone = 35cm

Height of the cone = 84cm

Curved surface area = πrl

So, we need to find out the l;

l = slant height

l = 91cm

Curved surface area =

= 110 × 91 = 10010cm²

Volume of the cone =

=

= 88 × 1225

= 107800cm²

Total surface area = πrl + πr²

= 10010 + 3850 = 13860

Total surface area = 13860cm²

Given,

Height (h) = 6cm

Slant height (l) = 10cm

r =

r

r = 8cm

Volume of the cone =

= 401.92cm²

Curved surface area = πrl

Curved surface area =

= 251.2cm²

Total surface area = πr(r + l)

= 452.16 cm²

Given,

h = 12 cm

Volume of the cone = 100π cm³

r²h = 100 × 3

r² × 12 = 100 × 3

r² = = 25

r = 5cm

Curved surface area = πrl

= π × 5 × 13

= (65π) cm²

Given,

Circumference of the base of the cone = 44cm

2πr = 44cm

r = 7cm

= 22 × 56 = 1232

Volume = 1232 cm³

Curved surface area = πrl

= 550 cm²

Given,

Curved surface area = 550cm²

πrl = 550

r = 7cm

h = 24cm

Volume =

= 24 × 56

Volume = 1232cm³

Given,

r = 35cm

l = 37cm

Volume of the cone =

Volume = 1540 cm³

Given,

Curved surface area = 4070

πrl = 4070

= 37cm

Given,

Radius = 7cm

h = 24cm

Curved surface area of the conical tent = πrl

= 25m

Curved surface area of the tent = πrl

= 550 m²

Length of cloth =

When we melt any shape, and recast into another shape than volume of both shapes remain same.

Radius of the circular cone (r₁) = 1.6cm

Height of the circular cone (h₁) = 3.6cm

Radius of the new circular cone (r₂) = 1.2 cm

Let height of the new circular cone be h₂

Volume of the circular cone = volume of the new circular cone

(1.6)² × (3.6) = (1.2)² × h₂

h₂ = 64 cm

So, the height of the new circular cone will be 64cm

Given,

Ratio of the heights = h₁ : h₂ = 1:3

Let the heights of the cones be x and 3x,

Ratio of radius of base of the two cones = r₁:r₂ = 3:1

So,

Let the radius be 3x and x for the cones and volume will be v₁ and v₂

So, ratio of the volume of the two cones will be 3:1

Given,

Cylindrical height of the tent = 3m

Diameter of the base of the cone = 105

Radius =

Height of the conical portion = 53m

Area of canvas = curved surface area of conical part + curved surface area of cylindrical part

= πrl + 2πrh

= 8745 + 990

= 9735m²

Length of canvas = area/width

Hence the length of the canvas will be 1947m

Given,

Number of person in the room = 11

Each person covers area = 4m²

Total area covered by all = 44m²

πr² = 44

Volume of the cone = 220m³

We know that,

Volume of the cone =

Hence, the height of the cone will be 15m

Let the radius of the heap be r and the slant height h,

So, we have

Height of the cylindrical bucket = 32cm

Radius of the base of cylindrical bucket = 18cm

Height of the conical heap = 24cm

Volume of cylinder = volume of cone

πr²h =

r² = 18 × 8 × 4

r = 18 × 2

r = 36cm

slant height l =

l = 43.27cm

Given,

Curved surface area of cylinder = curved surface area of cone = 8:5

Let C.S.A of cylinder = 8x

C.S.A of cone = 5x

As mention above cone and cylinder have equal radius and equal height

100h² = 64h² + 64r²

100h² – 64h² = 64r²

36h² = 64r²

Given,

Height of cylinder R = 2.8m = 280cm

Radius =

Height of cone = 42cm

Volume of pillar = volume of cone + volume of cylinder

=

Given that,

Weight of 1 cm³ iron = 7.5gm

Weight of the pillar = 92400 × 75

Weight of the pillar = 693000g

= 693kg

Let’s suppose the smaller cone have the radius r and height h cm

And radius of the given cone be R cm

Height of the original cone = 30cm

In triangle ∆OAB and ∆OCD

∠COD = ∠AOB (common angle)

∠OCD = ∠OAB (90°)

؞ ∆ OAB ∼ ∆OCD [ by A-A criteria]

Then,

………..(i)

Volume of small cone = volume of original cone

From equation (i)

h³ = 1000

h = 10cm

Height of the small cone = 10cm

AC = OA-OC

AC = 30-10 = 20

Hence selection has been made at height of 20cm above the base.

Height of the cylinder = 10cm

Radius of the cylinder = 6cm

The height and base of the cone is equals to the height and base of the cylinder.

Volume of the remaining solid = volume of cylinder – volume of cone

Volume of remaining solid = 753.6cm³

Given,

Radius of the cylindrical pipe = = 2.5mm

= 0.25cm [as we know 10mm = 1cm]

Water flowing per minute through cylindrical pipe = π(0.25)² × 1000

Radius of the conical vessel =

Depth of the vessel = 24cm

Volume of the vessel =

Let the time to fill the conical vessel be x minute,

Water flowing per minute through cylindrical pipe x x = volume of conical vessel

x =

x =

x = 51min 12 sec.

Hence the required time to fill a conical vessel is 51min 12 sec

(i) Radius of sphere = 3.5cm

Volume =

= 179.67cm³

Surface area = 4πr²

=

= 2 × 22 × 3.5 = 154 cm²

(ii) R = 4.2cm

Volume =

=

= 310.464cm²

Surface area = 4πr²

=

= 4 × 22 × .6 × 4.2

= 221.76cm²

(iii) R = 5cm

Volume =

= 523.80cm³

Surface area = 4πr²

=

= 314.28cm² +

Volume of sphere = 38808cm³

r³ = 441 × 21

r³ = 21 × 21 × 21

r = 21cm

surface area = 4πr²

=

= 5544cm³

Given,

Volume = 606.375cm³

r³ = 144.703

r = 5.25m

Surface area = 4πr²

=

= 346.5m²

Given,

Surface area = 394.24m²

4πr² = 394.24

r² = 31.36

r = 5.67cm

Volume =

= 735.91cm³

Given,

Surface area = 576π

4πr² = 576π

r = 12cm

Volume =

= 2304cm³

Given,

Outer Diameter of spherical shell = 12cm

Radius of the outer sphere r₁ = 6cm

Inner diameter of spherical shell = 8cm

Radius of the inner sphere r₂ = 4cm

Volume of metal = outer volume – inner volume

=

= 636.95cm³

Surface area of outer surface = 4πr²

=

= 452.571cm²

Given,

Dimensions of cuboid l = 12cm

b = 11cm

h = 9cm

Diameter of sphere (d) = 3mm

r = = 1.5mm

r = 0.15 cm

When we melt any object, and convert it into another then the volume of both the object will be same.

So,

Volume of cuboid = n × volume of sphere

n = no. of sphere

l × b × h =

12 × 11 × 9 =

n =

n = 84000

Given,

Radius of big sphere (R) = 8cm

Radius of small sphere (r) = 1cm

Volume of big sphere = 2 × volume of small sphere

n = no. of sphere

512 = n

n = 512 ball

Given,

Radius of big ball = 3cm

Diameter of small ball = 0.16cm

Volume of big ball = n × volume of small ball

n = 1000

Given,

Sphere radius = 10.5cm

Cone radius = 3.5cm

h = 3cm

When any object is melt and recast into another so the volume of both the object will be same

Volume of sphere = n × volume of cone

n = 126

Given,

Diameter of cylinder = 8cm

Radius = 4cm

Height = 90cm

Diameter of sphere = 12cm

Radius = 6cm

When we convert any object into another shape the volume will remain same.

Volume of cylinder = n × volume of sphere

n = 5

Given,

Diameter sphere = 6cm

r = 3cm

radius of wire =

r = 0.1cm

let us consider length of wire = h cm

When we convert any object into another shape the volume will remain same.

Volume of sphere = volume of cylinder

h = 36 × 100 =3600

h = 36m

Given,

Radius of sphere = 9cm

Let us consider diameter at cylinder = d cm

Radius = r cm

Height = 108 m = 10800 cm

When we convert any object into another shape the volume will remain same.

Volume of sphere = volume of cylinder

r² = 0.09

r = 0.03 cm

Diameter = 2 × 0.03 = 0.06 cm

Given,

When we convert any object into another shape the volume will remain same.

Radius of sphere =

Radius of cone = r cm

Volume of sphere = volume of cone

r² = 60.84

r = 7.8cm

d = 2 × r = 2 × 7.8 =15.6 cm

Given,

Radius of sphere (r₃) = 14cm

Diameter of cone = 35cm

When we convert any object into another shape the volume will remain same.

Volume of sphere = volume of cone

Given,

Radius of big ball (R) = 3cm

Radius of smaller ball (r₁) = 1.5 cm

Radius of second smaller ball (r₂) = 2 cm

Let r₃ be the radius of 3^rd smaller ball

V = v₁ + v₂ + v₃

(3)³ = (1.5)³ + (2)³ + (r₃)³

27 = 2.817 + 8 + (r₃)³

(r₃)³ = 27- (2.817 + 8) = 16.875

r₃ = 2.5 cm

Given,

Ratio of radii of spheres = R₁ : R₂ = 1 : 2

Ratio of their surface areas = = =

Given,

Ratio of Surface area of two spheres = A₁: A₂ = 1 : 4

Let radius of these sphere are resp. = R₁ and R₂

= =

=

=

=

Ratio of their volumes =

Given,

Radius of cylinder = 12 cm

Height = 20 cm

Before drop a ball volume of water = v₁ = πr²h = πr² × 20 cm³

After droping rise in water level = 6.75 cm

New height = 20 + 6.75 = 26.75 cm

New volume = πr² × 26.75 cm³

Volume of spherical ball = πr² (26.75 – 20)

= πr² × 6.75 = cm³

= πR³ = 3054.85

= R³ =

= R =

Given,

Radius of spherical ball = 9 cm

Volume of spherical ball = πr³ =

Radius of cylinder = 15 cm

Let the increase in level = h cm

= π × 729 = π × 15 × 15 × h

= h =

Given,

Radius of hemisphere = (R) = 9 cm

Height of cone = 72 cm

Let radius of cone = r cm

We know that,

Volume of hemisphere = volume of cone

= πR³ = πr²h

= = r² × 72

= r² =

= r =

Radius of base of cone = 4.5 cm.

Given,

Radius of hemisphere (R) = 9 cm

Radius of cylinder (r) =

Height of cylinder = 4 cm

Volume of hemisphere = n × volume of cylinder

= πR³ = n × πr²h

= = n × 1.5² × 4

= n =

Given,

Internal radius of sphere (r_i) = 8 cm

External radius of sphere (r_e) = 9 cm

Volume of shell = (external volume – internal volume)

=

Weight of sphere = 909.33 × 4.5 = 4092 gm = 4.092 kg

Given,

In-radius of bowl (r_i) = 4 cm

Thickness of steel = 0.5 cm

External radius of bowl (r_e)= 4 + 0.5 = 4.5 cm

Volume of metal = πr_e³ – r_i³

= (r_e³ – r_i³) = ³ – 4³) =

= = 56.83 cm³

Given: Length = 15 cm

Breadth = 12 cm

Height = 4.5 cm

Volume of a cuboid = Length × Breadth × Height

Volume = 15 cm ×12 cm ×4.5 cm = 810 cm³

Given: Length = 12 cm

Breadth = 9 cm

Height = 8 cm

Total surface area of a cuboid = 2[(Length ×Breadth) + (Breadth ×Height) + (Height ×Length)]

Total surface area = 2[(12×9) + (9×8) + (8×12)] cm² = 2(108+72+96) cm²

= 2(276) cm² = 552 cm²

Given: Length = 15 m

Breadth = 6 m

Height = 5 m

Lateral surface area of a cuboid = 2(Length +Breadth) × Height

1 m = 10dm

⇒ 5dm =0.5m

Lateral surface area = 2(15+6) × 0.5 m² = 1× 21 m²

= 21 m²

Given: Length = 9 cm

Breadth = 40 cm

Height = 20 cm

Volume of a cuboid = Length × Breadth × Height

1 m =100 cm

⇒ 40 cm= 0.4 m and 20 cm =0.2 m

Volume = 9 m × 0.4m × 0.2m = 0.72 m³

Given that 1 cm³ weighs 50 kg

⇒ 0.72 m³ weighs 50 × 0.72 kg = 36 kg

Longest rod = diagonal of the cuboid =

Length of longest rod = =

= = 15m

Maximum length of a pencil = diagonal of the cuboid

Now, the diagonal of cuboid is =

Thus,

Length of longest rod

=

=

=

= 5√5 cm

= 5(2.24) cm

= 11.2 cm

Volume of a cuboid = Length × Breadth × Height

Volume of pit = 40 m × 12 m × 16 m = 7680 m³

Volume of plank = 4 m × 5 m × 2 m = 40 m³

No. of planks =

= 192

Volume of a cuboid = Length × Breadth × Height

1 m =100 cm

Volume of pit = 20 m × 6 m × 0.5 m = 60 m³

Volume of plank = 5 m × 0.25 m × 0.1 m = 0.125 m³

No. of planks =

= 480

Volume of a cuboid = Length × Breadth × Height

1 m =100 cm

Volume of wall = 8 m × 6 m × 0.225 m = 10.8 m³

Volume of a brick = 0.25 m × 0.1125 m × 0.06 m = 0.0016875 m³

No. of bricks =

= 6400

Volume of a cuboid = Length × Breadth × Height

Volume of hall = 20 m × 15 m × 4.5 m = 1350 m³

Volume of air required by 1 person = 5 m³

No. of persons =

=270

Volume of a cuboid = Length × Breadth × Height

Length of the river = Speed of river = 3km (in an hr)

1km =1000 m and 1 hour =60 min

Speed in m per minute =

Volume of water that runs in a minute = 1.5 m × 30 m × 50 m = 2250 m³

Lateral surface area of a cube = 4(side)²

Given Lateral surface area = 256 m²

⇒ 4(side)² = 256m²

⇒ (side)² = m²

⇒ (side) = √64m= 8m

Volume of a cube = (side)³

⇒ Volume = (8)³ m³ = 512 m³

Total surface area of a cube = 6(side)²

Given Total surface area = 96 cm²

⇒ 6(side)² = 96m²

⇒ (side)² = cm²

⇒ (side) = √16cm= 4cm

Volume of a cube = (side)³

⇒ Volume = (4)³ cm³ = 64 cm³

Volume of a cube = (side)³

Given volume = 512 cm³

⇒ (side)³ = 512 cm³

⇒ side = = 8 cm

Total surface area of a cube = 6(side)²

⇒ Total surface area = 6(8)²cm² = 384 cm²

Length of the longest rod = diagonal of the cube = side

Length of longest rod =10 cm

Diagonal of the cube = side

Given diagonal =8 cm = side

⇒ side = 8 cm

Total surface area of a cube = 6(side)²

⇒ Surface area = 6(8)² = 6×64 = 384 cm²

Let original side be x, on increasing it by 50% i.e.

New side will be x + x = x

Total surface area of a cube = 6(side)²

Original surface area = 6 (x)²

New surface area = 6 (x)² = 6× x² = x²

Change in surface area = x² − 6 (x)²

Taking LCM of 2 and 1 = 2

⇒

The percentage increase in its surface area is

Here, the volume of three cubes = volume of the new cube

Volume of a cube = (side)³

Volume of three cubes = (3)³ + (4)³ + (5)³= (27 +64+ 125) cm³ =216 cm³

⇒ Volume of new cube = 216 cm³= (side)³

⇒ (side)³ = (6 cm)³

⇒ side = 6cm

Lateral surface area = 4(side)² = 4(6)² = 144cm²

1 hectare = 10000 m²

2 hectares = 20000 m²

1 cm = 0.01 m ⇒ 5cm= 0.05 m

Volume of water that falls on 2 hectares of ground = 20000× 0.05 m³ = 1000 m³

Volume of a cube = (side)³

Let the sides be x and y

Ratio of volumes =

⇒

Surface area of a cube = 6(side)²

Ratio of surface areas = = 1: 9

Let original side be x, New side will be 2x

Volume of a cube = (side)³

Original volume = (x)³

New volume = (2x)³ = 8x³

So, the volume is 8 times of the original volume

Volume of a cylinder =

Diameter = 6cm ⇒ radius = 3cm

⇒ Volume = × 3² × 14

= 22 × 9 × 2 = 396cm³

Curved surface area of a cylinder =

Diameter = 28 cm ⇒ radius = 14 cm

⇒ Curved surface area = 2× × 14 × 20

= 44 × 40 =1760 cm²

Curved surface area of a cylinder =

⇒ Curved surface area =

⇒

Curved surface area of a cylinder =

⇒ Curved surface area = 1760 cm²

× r × 14 = 1760 cm²

⇒

Volume of a cylinder =

Volume =

= 17,600 cm³

Curved surface area of a cylinder =

⇒ Curved surface area =

Volume of a cylinder =

Volume =

Let the radii be 2x and 3x respectively and heights be 5y and 3y respectively.

Curved surface area of a cylinder =

⇒ Ratio of their Curved surface area =

Let the radii be 2x and 3x respectively and heights be 5y and 3y respectively.

Volume of a cylinder =

⇒ Ratio of their Volumes =

Let the radius be 2x and height be 3x respectively.

Volume of a cylinder =

⇒ Volume =

So, radius = 2× 3.5 =7 cm and height = 3× 3.5 = 10.5 cm

Total surface area of a cylinder =

⇒ T.S.A.

Let the heights be h = x and H=2x respectively of the two cylinders.

Volume of a cylinder =

Given that

Total surface area of a cylinder =

Curved surface area of a cylinder =

⇒ 2h=r + h ⇒ h = r

Given that total surface area = 616 cm ²

⇒

⇒ r² = 7× 7

So, r = h = 7 cm

Volume of a cylinder =

Let the radius be r and height be h

Volume of a cylinder =

When radius = and height = 2h

Volume =

The volume will be halved.

Volume of a cylinder =

Volume of the coin =

Volume of the cylinder =

Number of coins

Let radius and length of a wire be r and h respectively

Volume of a wire =

If radius = and new length = H

Volume of the wire =

⇒ H = 9h i.e. 9 times

Curved surface area of a cylinder =

1m= 100cm, radius = 42 cm = 0.42m

Curved surface area

Area of the playground = 500 × 2.64 m² = 1320 m²

Given volume of the cylindrical wire is 2.2dm³

Volume of a wire =

1 dm = 10 cm ⇒ 0.50 cm = 0.05 dm

Volume of the wire =

1 m= 10 dm

⇒ 11.2 dm = 112 m

The curved surface area of a cylinder is only the lateral surface area

And, we know that the curved surface area = 2πrh

Curved surface area of a cone =

where

Here, r=7cm and h=24cm

Volume of a cone

Curved surface area of a cone =

where

Here, r=7m and h=24m

The cloth required

= 220 m

Volume of a cone

Given volume = 1570 cm³

Volume of a cone

where

Here, l=28 cm and h=21 cm

Volume

7546 cm³

Given:

Volume of cone = 1232 cm³

As we know, Volume of a cone

⇒ r² = 49

⇒ r =7 cm

Here, r =7 cm and h = 24 cm

Curved surface area of a cone =

Volume of a cone

and

Volume of a cone

If height is doubled,

volume

Thus, there will be 100% increase in the volume.

Curved surface area of a cone =

Given that curved surface area of 1^st = 2× curved surface area of 2^nd

And slant height of 2^nd = 2× slant height of 1^st

⇒ L= 2l

⇒ r : R = 4 : 1

Given that heights and radii of cone and cylinder are equal

Volume of a cone

Volume of a cylinder =

Ratio of their volumes

{because h=H and r=R}

Ans – 3:1

Let height of cylinder and cone be H and h respectively

Given that radii of cone and cylinder are equal

Volume of a cone

Volume of a cylinder =

Given

Ans: 1 : 3

Given that radii of cone and cylinder are 4x and 3x respectively and

height of cylinder and cone are 2y and 3y respectively

Volume of a cone

Volume of a cylinder

Volume of a cone

If height and radius are doubled, volume

The volume of the cone becomes 8 times.

Volume of a cone

Volume of a cylinder

Volume of solid metallic cylinder

1cm=10mm

Volume of solid coin

No. of coins

As each person needs 4 m ² spaces on ground, so 11 persons will need 44 m ² space on the ground.
Therefore, Area of ground = 44 m ²

⇒ r² = 14
Each person needs of air
Therefore volume of tent = 220 m³

Volume of a cone

⇒ h = 15cm

Volume of a sphere

Volume

Volume of a sphere

Volume

Surface area of a sphere

Surface area

Surface area of a sphere

Given Surface area = 1386 cm²

Volume of a sphere

Volume

Surface area of a sphere

Given Surface area =

⇒ r= 6m

Volume of a sphere

Volume

Volume of a sphere

Given Volume=

⇒ r=21cm

Surface area of a sphere

Surface area

Volume of a sphere

Given that

Surface area of a sphere

Volume of a sphere

Volume of the solid metal ball

Volume of smaller ball

No. of balls

Volume of a cone

On recasting a cone into sphere, the volume will remain same

Volume of a sphere

Volume of sphere =12.348π cm³

⇒ r = 2.1cm

Volume of a sphere

On recasting a sphere into cylinder, the volume will remain same

Volume of a cylinder

Radius = 0.1 cm

(∵ 1m =100 cm)

⇒ h = 288 m

Volume of a sphere

Volume of a cone

No. of cones

Volume of a cuboid = l× b× h = 9× 11× 12 cm³

Radius of a lead shot = 0.15 cm

Volume of a lead shot

No. of lead shot

Radius of the sphere = 3 cm

Volume of a sphere

On recasting a sphere into cylinder wire, the volume will remain same

Volume of a cylinder

1cm=10mm

⇒ 2mm =0.2cm

Radius = 0.1 cm

(∵ 1m =100 cm)

Radius of the sphere = 6.3 cm

Volume of a sphere

Volume of a cone

On recasting a sphere into a cone, volume will remain same

⇒ r = 6.3 cm

Volume of a sphere

Volume of spherical ball

Volume of three balls

On recasting this sphere into three spherical balls, volume will remain same

⇒ r = 2.5 cm

Surface area of a hemisphere = 2πr²

Radii are 6cm and 12 cm respectively

Ratio of surface areas

Ans 1:4

Volume of a sphere

Given Ratio of volumes of two spheres

So, r = 4x and R = 3x

Also given that the sum of radii = 7

⇒ r +R = 4x +3x =7x =7

⇒ x =1

So r = 4cm and R = 3cm

Surface area of a sphere

Difference in total surface area =

Volume of a hemisphere

Volume of a cylinder

Volume of a cylindrical bottle

No. of bottles required

Thus, total 54 bottles are required.

Given that Radius of the hemisphere = Radius of cone

And Volume of hemisphere = Volume of cone

Volume of a hemisphere

Volume of a cone

Given that Radius of the hemisphere = Radius of cone = Radius of cylinder

And Height of the hemisphere = Height of cone = Height of cylinder

Volume of a hemisphere

Volume of a cone

Volume of a cylinder

=

Volume of a sphere

Surface area of a sphere

Given that volume = surface area

Inner curved surface area of a hollow cylinder =

Curved surface area of a hemisphere = 2πr²

A) Curved surface area of a cylinder =

B) Total surface area of a cylinder =

C) Volume of a cylinder =

D) Total area of the end faces = 2× πr² {Because there are two circular faces}

A) Inner curved surface area =

B) Outer curved surface area =

C) Surface area of the end face = π(R ² – r²){Because there are two circular faces}

D) R = 2.2 cm, r = 2 cm and h = 63 cm

Total surface area of a hollow cylinder

Slant height

Here, r=7cm and l=25cm

Volume of a cone

Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

Surface area of a sphere

Given Surface area = 2464 cm²

Volume of a sphere

Volume

Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

The volume of a hollow cylinder with external and internal radii R and r respectively and height h

Thus, the volume is

Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

Volume of a sphere

Volume

Ratio = 1:8

Reason is wrong. Assertion (A) is true and Reason (R) is false.

Curved surface area of a cone =

⇒ l=25 cm

Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

True

Curved surface area of a sphere

Radius of cylinder = r + r = 2r

Curved surface area of a cylinder = =

True

The dimensions of the cone are
diameter = r ; radius = r/2 height = r
Volume of a cone

True

Let the radius of sphere be r so the edge of cube = 2r

Volume of a sphere

Volume of a cube

False

Diagonal of the cube =

Length of longest rod = cm

Side = 6 cm

Let the radii be 2x and 3x respectively and heights be 5y and 3y respectively.

Volume of a cylinder =

⇒ Ratio of their Volumes =

Thus, the ratio of two cylinders

Total surface area of a cone =

=

Volume of a cone

On recasting a cone into sphere, the volume will remain same

Volume of a sphere

Volume of sphere =12.348π cm³

⇒ r = 2.1cm

Surface area of a hemisphere = 2πr²

Radii are 6cm and 12 cm respectively

Ratio of surface areas

Radius of the sphere = 3 cm

Volume of a sphere

On recasting a sphere into cylinder wire, the volume will remain same

Volume of a cylinder

Total surface area of a cube = 6(side)²

⇒ Total surface area = 6(8)²cm² = 384 cm²

Lateral surface area of a cube = 4(side)²

⇒ Total surface area = 4(8)²cm² = 256 cm²

Total surface area of a cuboid = 2[(Length ×Breadth) + (Breadth ×Height) + (Height ×Length)]

Total surface area = 2[(40×30) + (30×20) + (20×40)] cm² = 2(1200+600+800) cm²

= 2(2600) cm² =5200 cm²

Lateral surface area of a cuboid = 2(Length +Breadth) ×Height

Lateral surface area = 2(40+30) × 20 cm² = 140× 20 cm²

= 2800 cm²

Total surface area of a cylinder =

Curved surface area of a cylinder =

⇒ 3h=r+h ⇒ 2h=r

Given that total surface area = 462 cm ²

⇒

⇒ r² = 7× 7

So, r = 7 cm , h =3.5cm

Volume of a cylinder =

Area of the floor

Given that length and breadth are in ratio 3:2, so l = 3x and b = 2x

⇒ l= 9 m and b =6 m

Lateral surface area of a cuboid = 2(Length +Breadth) ×Height

Lateral surface area

Adding door and window, Lateral surface area = 180 m²

⇒ h = 6 m

Volume of a sphere

Let the radius be 'r'
Increased Radius = 1.1r

Volume

Change in volume

Curved surface area of a sphere

Curved Surface area of a cone

Given that

l= 5 cm and r= 4 cm

Volume

Volume = l× b × h

Volume = 5× 4.5× × 2.1=47.25 m³

Area over which it is spread = 13.5 x 25 - 5 x 4.5
= 33.75 - 220 = 11.75 m
Rise in level = 4.2 m

Curved surface area of a cone =

where

Here, r=7cm and h=24cm

Area of 10 such caps = 5500 cm²

Volume of a cone

Given volume = 9856 cm³

Radius of cone = 14 cm

Volume of a cylinder =

Volume of the cylinder =

Number of bags

Volume of a cylinder =

Radius of well = 5m , Height of well = 14m

Volume of the well =

For embankment, radius = 5+5 =10 m and let height be h m

Volume of well = Volume of embankment

⇒ h = 4.67 m

Curved surface area of a cone =

where

Here, r=7m and h=24m

The cloth required

Volume of a cylinder , Given volume =

⇒ r² = 36 ⇒ r =6 cm

Total surface area of a cylinder =

Volume of a sphere

Given Ratio of volumes of two spheres

So, r = 4x and R = 3x

Also given that the sum of radii = 7

⇒ r +R = 4x +3x =7x =7

⇒ x =1

So, r = 4cm and R = 3cm

Surface area of a sphere

Difference in total surface area =

Since the radius and height of a cone is 4:3 so let radius = 4x and height = 3x

Volume of a cone , Given volume =

⇒ ⇒ x

So , r =14 cm and height = 10.5 cm

Here, r=14 cm and h=10.5 cm

Curved surface area of a cone =

Curved surface area of a cone =

where

Here, r=14cm and h=24cm

Total surface area of a cone =

Volume of a cone

Volume of a cylinder

Volume of first vessel =

Volume of second vessel =

The volume of the third vessel = volume of first vessel + volume of second vessel

Total surface area of a cylinder =

Curved surface area of a cylinder =

⇒ 2h=r + h

⇒ h=r

Given that total surface area = 616 cm ²

⇒

⇒ r² = 7× 7

So, r = h = 7 cm

Volume of a cylinder =