Surface Areas And Volumes

Length of each side of cube = 5cm

According to question,

Three cubes are joined end to end to form a cuboid

So, the length of the resulting cuboid = 5 + 5 + 5 = 15cm

But, the breadth and height will remain the same

So, Breadth = 5cm

and height = 5cm

Surface Area of the Cuboid

= 2(Length × Breadth + Breadth × Height + Height × Length)

= 2(lb + bh + hl)

= 2(15×5 + 5×5 + 5×15)

= 2(75 + 25 + 75)

= 2(175)

= 350cm²

Hence, the surface area of cuboid is 350cm².

For bigger box:

Length of the bigger box = 20cm

Breadth of the bigger box = 15cm

Height of the bigger box = 5cm

So, Total Surface Area of bigger box = 2(lb + bh + hl)

= 2(20×15 + 15×5 + 5×20)

= 2(300 + 75 + 100)

= 2(475)

= 950cm²

For Smaller box:

Length of the smaller box = 16cm

Breadth of the smaller box = 12cm

Height of the smaller box = 4cm

So, Total Surface Area of bigger box = 2(lb + bh + hl)

= 2(16×12 + 12×4 + 4×16)

= 2(192 + 48 + 64)

= 2(304)

= 608cm²

Total surface area of 200 boxes of each type = 200(950 +608)

= 200 × 1558

= 311600cm²

Extra Area Required = 5% of (950 + 608)

= 15580cm²

So, Total cardboard Required = 311600 + 15580

= 327180cm²

Cost of Cardboard for 1m² = Rs 20

Cost of Cardboard for 327.18 m² = 20 × 327.18

= Rs 654.36

Hence, the cost of the cardboard for supplying 200 boxes of each kind is Rs 654.36.

Given: Area = 108m², Height = 3m

and Length of a cold storage is double its breadth

So, Let breadth of a cold storage = x

∴ length of the cold storage = 2x

Area of four walls = 108m²

∴ 2(l + b)× h = 108

⇒ 2 (x + 2x)×3 = 108

⇒ 6(3x) = 108

⇒ x = 6

∴ Breadth of a cold storage = 6m

and Length of a cold storage = 2×6 = 12m

Hence, Volume of the cold storage = l × b × h

= 12 × 6 × 3

= 216m³

Given: Radius of base, r = 7cm

Height , h = 13.5cm

(i) Lateral Surface Area of right circular cylinder = 2πrh

= 2×22×13.5

= 594cm²

(ii) Total Surface Area of cylinder = 2πrh + 2πr²

= 2πr(h + r)

= 2×22×20.5

= 902cm²

(iii) Volume of cylinder = πr²h

= 22×7×13.5

= 2079cm³

Given: Volume of a right circular cone = 314cm³

and r : h = 5 : 12

Let r = 5x and h = 12x

The volume of a right circular cone

⇒ x³ = 1

Hence, x = 1

∴ r = 5x = 5×1 = 5cm

and h = 12x = 12×1 = 12cm

Now, we have to find the slant height, l

Slant height, l = √(r² + h²)

= √{(5)² + (12)²}

= √25+144

= √169

= ±13

= 13cm

[taking positive root, because slant height can’t be negative]

∴ Slant Height = 13cm

Let height of a cylinder = h

and diameter = d

According to question,

Given: Radius of Sphere, R = 4cm

Volume of Cylinder = Volume of Sphere (Given)

⇒ r³ = R³

⇒ r³ = 4³

⇒ r = 4cm

Hence, Radius of base of the cylinder is 4cm.

Given: Diameter of a hemisphere = 7cm

So,

Height of the cone, h = 143-.5 – 3.5 = 11cm

The volume of a toy = Volume of cone + Volume of a hemisphere

= 231cm³

Original volume of water in the cylindrical tub

= Volume of Cylinder

= πr²h

= 22×25×1.4

= 770cm³

Given that Radius of hemisphere, R = 2.1cm

and height of cone, h = 4cm

Volume of a solid = Volume of cone + Volume of hemisphere

= 37.884cm³

∴ Volume of water displaced (removed) = 37.884cm³

Hence, the required volume of the water left in the cylindrical tub = 770 – 37.884

= 732.116 cm³

Given:

Diameter of cylinder = 5mm

The radius of cylinder

Height of cylinder = 14 – 5 = 9mm

Here, Diameter of hemisphere =. 5mm

So,

The total area of the capsule = CSA of cylinder + CSA of 2 hemispheres

= 2πrh + 2×2πr²

=220 mm²

Let r be the radius of hemisphere and cylinder

and height of the cylinder = h

Given:

∴ Diameter of the building = 2r

Height of the building (H) = diameter of the building

∴ Height of the cylinder + Radius of hemispherical dome = 2r

⇒ h + r = 2r

⇒ h = 2r – r

⇒ h = r

Volume of air inside the building = Volume of hemispherical portion

+ Volume of cylindrical portion

⇒ r³ = 8

⇒ r = 2

⇒ Height of building = 2r = 2×2 = 4m

Hence, the total height of the building is 4m.

The diameter of a cylindrical portion BCDE of building = 4.3m

∴ The radius of a cylindrical portion

Height = 3.8m

Lateral Surface Area of Cylindrical Portion BCDE = 2πrh

= 51.3543 m²

Let AB be the slant height of the conical portion of the building = l = AB = AC

Now, the Lateral surface of conical portion = πrl

= 20.5417 m²

So,

The total surface area of the building = Surface area of cylindrical portion + Surface area of the conical portion

= 51.3543 + 20.5417

= 71.8960

= 71.90 m² (approx.)

Now, In right ΔBAC,

BC² = AB² + AC²

⇒ BC² = l² + l²

⇒ (4.3)² = 2l²

⇒ 18.49 = 2l²

⇒ l² = 9.245

⇒ l = 3.04 m

Here, r is the radius of the cone and l is the slant height of the cone

⇒ l² = h² + r²

⇒ 9.245 = h² + (2.15)²

⇒ 9.245 = h² + 4.6225

⇒ h² = 9.245 – 4.6225

⇒ h² = 4.6225

⇒ h = 2.15 m

Now, Volume of the conical portion

= 10.4116 m³

Volume of cylindrical portion BCDE = πr²h

= 55.2059 m³

So,

The total volume of the building = Volume of the cylindrical portion

+ Volume of the conical portion

= 55.2059 + 10.4116

= 65.6175 m³

= 65.62 m³ (approx.)

Given: Height of the tent = 77dm

The height of the conical portion = 44dm

∴ The height of the cylindrical portion

= Height of the tent – Height of the conical portion

= 7.7 – 4.4

= 3.3m

Given Diameter of the cylinder, d = 36m

CSA of cylindrical portion = 2πrh

= 497.828

= 497.83 m² (approx.)

Firstly, we have to find the slant height (l) of the conical portion

⇒ l² = h² + r²

⇒ l² = (3.3)² + (18)²

⇒ l² = 10.89 + 324

⇒ l² = 334.89

⇒ l = √334.89

⇒ l = 18.3m

∴ CSA of the conical portion = πrl

= 1035.257

= 1035.26 m² (approx.)

So,

Total Surface Area of the tent = Surface area of conical portion + surface Area of cylindrical portion

= 1035.26 + 497.83

= 1533.09 m²

∴ Canvas required to make the tent = 1533.09 m²

Cost of 1m² canvas = Rs 3.50

Cost of 1533.09m² canvas = Rs 3.50 × 1533.09

= Rs 5365.815

= Rs 5365.82

Given: Height of the tent = 3.3 m

The height of the cylindrical portion = 2.2 m

∴ The height of the conical portion

= Height of the tent – Height of the cylindrical portion

= 3.3 – 2.2

= 1.1m

Given Diameter of the cylinder, d = 12m

CSA of cylindrical portion = 2πrh

= 82.971 m²

Firstly, we have to find the slant height (l) of the conical portion

⇒ l² = h² + r²

⇒ l² = (1.1)² + (6)²

⇒ l² = 1.21 + 36

⇒ l² = 37.21

⇒ l = √37.21

⇒ l = 6.1m

∴ CSA of the conical portion = πrl

= 115.029 m²

So,

Total Surface Area of the tent = Surface area of conical portion + surface Area of cylindrical portion

= 115.029 + 82.971

= 198 m²

∴ Canvas required to make the tent = 198 m²

Cost of 1m² canvas = Rs 500

Cost of 198 m² canvas = Rs 500 × 198

= Rs 99000

Given:

Diameter of cylinder = 5mm

Height of cylinder = 12 – 5 = 7mm

Here, Diameter of hemisphere = 5mm

So,Radius of Hemisphere =

Total area of the capsule = CSA of cylinder + CSA of 2 hemispheres

= 2πrh + 2×2πr²

Given: Diameter of the hemisphere = 14cm

∴ Radius of the hemisphere

Curved Surface Area of Hemisphere = 2πr²

= 308cm²

Now,

Radius of cylinder = radius of hemisphere = 7cm

Height of cylinder = Total height – Radius of hemisphere

= 13 – 7

= 6cm

So, CSA of cylinder = 2πrh

= 264cm²

The inner Surface area of the vessel

= CSA of hemisphere + CSA of cylinder

= 308 + 264

= 572cm²

Hence, Inner surface area of vessel is 572cm²

Given that Diameter of circular cylinder = 10cm

∴ Radius of the cylinder

and Height of the cylinder = 12cm

So,

Volume of the cylinder = πr²h

= 942.857 cm³

= 314.285 cm³

Remaining Volume = Volume of the cylinder – Volume of cone

= 942.857 – 314.285

= 628.572 cm³

Given: Height of the cone = 14cm

and diameter of the circular top = 5cm

Now,

= 91.6666cm³

Now,

= 32.738 cm³

So,

The volume of ice cream in the cone

= Volume of cone + Volume of a hemisphere

= 91.6666 + 32.738

= 124.404cm³

Given that diameter of cylinder = 3cm

So,

Given: Height of cone = 2cm

and

Height of the cylinder + height of cone + height of cone = 10cm

Height of the cylinder + 2 + 2 = 10

Height of the cylinder = 6cm

So,

Volume of the cylinder = πr²h

= 42.428 cm³

= 42.43 cm³ (approx.)

= 4.71 cm³

Volume of 1^st cone = Volume of 2^nd cone = 4.71cm³

So, Total volume of the model

= Volume of cylinder + volume of 1^st cone + volume of 2^nd cone

= 42.43 + 4.71 + 4.71

= 51.85cm³

Given that side of a cube = 10cm

Total Surface Area of cube = 6(side)²

= 6 × 10 × 10

= 600cm²

Now, Radius of hemisphere = 3.5cm

Curved Surface Area of hemisphere = 2πr²

= 77cm²

Base area of hemisphere = Area of circle = πr²

= 38.5cm²

Therefore,

Total Surface Area of block = Total Surface area of a cube + CSA of the hemisphere - Base area of a hemisphere

= 600 + 77 – 38.5

= 638.5cm²

Cost of painting the block of 1cm² = Rs 0.50

Cost of painting the block of 638.5cm² = Rs 0.50 × 638.5

= Rs 319.25

(i) Volume of godown = Volume of cuboid +volume of cylinder

Volume of cuboid = l × b × h = 10 × 7 × 3 = 210 m³

Volume of cylinder = πr²h = = 385 m³

∴ Volume of godown

= 210 + 192.5

= 402.5 cm³

(ii) Internal Surface Area of a godown excluding the floor

the total surface area of the cylinder

= 6 × 17 + 22 × 0.5 × 13.5

= 102 + 148.5

= 250.5 cm²

Hence, internal surface area of godown excluding floor is 250.5cm²

Given: Diameter of cone = 12cm

So, Radius of cone = 6cm

and Height of cone = 9cm

= 339.428 cm³

Radius of the cylinder, R = Radius of cone = 6cm

Height of the cylinder, H = 110cm

Volume of cylinder = πR²H

= 12445.714cm³

Hence, the volume of the pole

= Volume of conical part + Volume of the cylindrical part

= 339.43 + 12445.714

= 12785.142 cm³

Required mass of pole = 8 × 12785.142

= 102281.13 gm

= 102.281kg

The area to be painted orange

= CSA of cone + Base area of Cone – Base area of the cylinder

Curved Surface Area of Cone

Given that diameter of conical portion = 5cm

The radius of conical portion

Height of the conical part = h = 6cm

We need to find the ‘l’ first

We know that

l² = h² + r²

⇒ l² = 6² + (2.5)²

⇒ l² = 36 + (6.25)

⇒ l² = 42.25

⇒ l = √42.25

⇒ l = 6.5cm

So, CSA of conical portion = πrl

=3.14×2.5×6.5

= 51.025 cm²

Base area of the cone = πr²

=3.14×2.5×2.5

=19.625 cm²

Diameter of the cylinder = 3cm

So, Radius of the cylinder = 1.5cm

Base area of the cylinder = π(r’)²

=3.14×1.5×1.5

=7.065 cm²

So, the area to be painted orange = 51.025 + 19.625 – 7.065

= 63.585 cm²

Now, the area to be painted yellow

= CSA of the cylinder + Area of one bottom base of the cylinder

= 2πr’h’ + π(r’)²

= 2 × 3.14 × 1.5 × 20 + 7.065

= 188.4 + 7.065

= 195.465 cm²

Given the inner diameter of the glass = 7cm

So, the radius of the glass =

Height of the glass = 16cm

The volume of the cylindrical glass = πr²h

= 616cm³

Now, radius of the hemisphere = Radius of the cylinder =r = 3.5cm

Volume of the hemisphere

= 89.83 cm³

Now,

Apparent capacity of the glass = Volume of cylinder = 616cm³

The actual capacity of the glass

= Total volume of cylinder – Volume of hemisphere

= 616 – 89.83

= 526.17cm³

Hence,

Apparent Capacity of the glass = 616cm³

and actual capacity of the glass = 526.17cm³

Radius of 1^st cylindrical vessel = 15cm

and height = 25cm

Radius of 2^nd cylindrical vessel = 10cm

and height = 18cm

So, Volume of 1^st cylindrical vessel = πr²h

= π × (15)² × 25

= 5625π cm³

Volume of 2^nd cylindrical vessel = π(r’)²h’

= π × (10)² × 18

= 1800π cm³

Height of the third vessel = 30cm

and let its radius be R

So,

Volume of third cylindrical vessel = πR²H

= πR²×30

= 30πR²

Volume of 1^st cylindrical vessel + Vol. of 2^nd Cylindrical vessel

= Volume of the third cylindrical Vessel

⇒ 5625π + 1800π = 30πR²

⇒ 7425π = 30πR²

⇒ R² = 247.5

⇒ R = 15.73cm

Hence, radius of the required cylinder is 15.73cm

Let the edge of the third smaller cube be x cm.

Three small cubes are formed by melting the cube of edge 12 cm.

Edges of two small cubes are 6 cm and 8 cm.

Now, volume of a cube = (side)³

Volume of the big cube = sum of the volumes of the three small cubes

⇒ (12)³ = (6)³ + (8)³ + (x)

⇒ 1728 = 216 + 512 + x³

⇒ 1728 = 728 + x³

⇒ x³ = 1728 – 728

⇒ x³ = 1000

⇒ x = 10 cm

Therefore, the edge of the third cube is 10 cm

Radius of hemisphere = 8cm

and Radius of right circular cone = 6cm

According to question,

Volume of the hemisphere = Volume of cone

⇒ h = 28.44cm

Given that Radius of bigger sphere (R) = 3cm

and diameter of smaller spherical balls = 0.6cm

So, radius of small spherical balls (r) = 0.3cm

Let the number of small balls = n

According to the question,

n × volume of small balls = Volume of bigger sphere

⇒ nr³ = R³

⇒ n = (10)³

⇒ n = 1000

Hence, the number of small balls = 1000

Given that Radius of cone = 12cm

Height of the cone = 24cm

So,

= 1152π cm³

It is also given that diameter of each spherical balls = 6cm

so, Radius of each ball = 3cm

= 36π cm³

Total number of balls formed by melting the cone

= 32

Hence, 32 balls are formed by melting the cone.

Let the number of cones formed be n

Diameter of a metallic sphere = 21cm

So,

=

= 11 × 21 × 21

= 4851cm³

Now, Diameter of cone = 3.5cm

So,

and height = 3cm

=

cm³

According to question,

n × volume of cone = volume of sphere

⇒ n = 504

Hence, the number of cones formed = 504

Diameter of sphere = 21 cm

So,

=

= 11 × 21 × 21

= 4851cm³

Volume of cube = a³ = 1³

Let number of cubes formed be n, then

Volume of sphere = n × (volume of cube)

⇒ 4851 =n×1

⇒ n = 4851

Hence, number of cubes is 4851.

Given:

Internal Diameter of a hollow hemispherical shell = 6cm

So, internal radius, (r) = 3cm

External Diameter of a hollow hemispherical shell = 10cm

So, external radius (R) = 5cm

Given that diameter of cone =14cm

So, radius = 7cm

According to the question.

Volume of a shell = Volume of a cone

⇒

⇒ h = 4cm

Let inner radius of the cylinder be x cm

Given that radius of sphere = 6cm

=

= 4 × π × 2 × 36 cm³

It is also given that external radius of base of cylinder (R) = 5cm

and height = 32cm

∴ Volume of hollow cylinder = πh[R² – r²]

= π × 32 × [(5)² – x²]

= 32π [25 – x²]

Solid sphere is melted and casted into hollow cylinder

∴ Volume of sphere = Volume of hollow cylinder

⇒ 4 × π × 2 × 36 = 32π [25 – x²]

⇒ 9 – 25 = -x²

⇒ x² = 16

⇒ x = ±4

⇒ x = 4cm

Hence, the inner radius of the cylinder is 4cm

So, the thickness of the cylinder = external radius – inner radius

= 5 – 4

= 1cm

Diameter of a copper sphere = 6 cm

So, Radius of copper sphere = 3cm

=

= 36π cm³

Length of wire = 36cm

Let the radius of wire be R cm

Volume of wire = πR²h

=πR²×36

But the volume of wire = volume of sphere

⇒ 36πR² = 36π

⇒ R² = 1

⇒ R = 1cm [taking positive root, because radius can’t be negative]

Hence, the radius of wire is 1cm

Let the radius of the base of conical ice cream = x cm

Then, height of the conical ice cream = 2 × diameter

= 2 × (2x)

= 4x cm

Volume of ice- cream cone

= Volume of conical portion + Volume of hemispherical portion

= 2πx³ cm³

Now,

Diameter of cylindrical container = 12cm

So, radius = 6cm

and height = 15cm

∴ Volume of cylindrical container = πr²h

= π × (6)² × (15)

= 540π cm³

⇒ x³ = 27

⇒ x = 3

So, the radius of the base of conical ice cream is 3cm

Hence, the diameter of the base of conical ice-cream = 2×3 = 6cm

Frustum = difference of two right circular cones OAB and OCD

Let the height of the cone OAB be h₁ and its slant height l₁

i.e. OA = OB = l₁ and OP = h₁.

Let h₂ be the height of cone OCD and l₂ its slant height

i.e. OC = OD = l₂ and OQ = h₂

We have , r₁ = 28cm and r₂ = 7cm

and height of frustum (h) = 45cm

Also,

h₁ = 45 + h₂ …(i)

Now, we first need to determine the h₁ and h₂

∵ ΔOPB and OQD are similar, we have

⇒ h₁ = 4h₂ …(ii)

From (i) and (ii), we get

4h₂ = 45 + h₂

⇒ 3h₂ = 45

⇒ h₂ = 15cm

Putting the value of h₂ in eq. (ii), we get

h₁ = 4×15 = 60cm

So, h₁ = 60cm and h₂ = 15cm

Now, the volume of frustum = Vol. of cone OAB – Vol. of cone OCD

= 5 × 22 × 441

= 48510 cm³

Now, we first have to find the slant height l₁ and l₂

l₁ = √{(28)²+(60)²}

⇒ l₁ = √{(4×7)² + (4×15)²

⇒ l₁ = 4√{(7)² + (15)²}

⇒ l₁ = 4√49 + 225

⇒ l₁ = 4√(274)

⇒ l₁ = 4 × 16.55

⇒ l₁ = 66.20cm

and l₂ = √{(7)² + (15)²}

⇒ l₂ = √49 + 225

⇒ l₂ = √(274)

⇒ l₂ = 16.55 cm

So, CSA of frustum = CSA of cone OAB – CSA of cone OCD

= πr₁l₁ – πr₂l₂

= 22 [4×66.20 – 16.55]

= 22 × 248.25

= 5461.5cm²

Given: Height of frustum = 14cm

Diameter of 1^st circular end = 4cm

So, radius (R) = 2cm

Diameter of 2^nd circular end = 2cm

So, radius (r) = 1cm

Capacity of the glass = Volume of frustum

= 102.66 cm³

Hence, the volume of glass = 102.66cm³

Greater radius = R = 33 cm

Smaller radius = r = 27 cm

Slant height = l = 10 cm

Total Surface area of the frustum = πR² + πr² + πl(R+r)

= π[(33)² + (27)²+ 10×(33+27)]

= π[1089 + 729 + 600]

= 7599.428 cm²

= 7599.43 cm²

We first need to find the height, h

l² = (R – r)² + h²

⇒ (10)² = (33 – 27)² + h²

⇒ 100 = (6)² + h²

⇒ 100 = 36 + h²

⇒ h² = 100 – 36

⇒ h² = 64

⇒ h = ±8

⇒ h = 8cm

Now, Capacity of a solid frustum of a cone = Volume of frustum

= 22704 cm³

Let radii of the circular ends of the frustum are R and r respectively.

Given that Perimeter of one end = 96cm

⇒ 2πR = 96

⇒ R = 15.27 cm

Perimeter of other end = 68cm

⇒ 2πr = 68

⇒ r = 10.82 cm

It is given that height of the frustum = 20cm

So, Slant height, l = √{h² + (R – r)²}

= √{(20)² +(15.27 – 10.82)²

= √400 + (4.45)²

= √400 + 19.80

= √419.80

= 20.49 cm

Now,

= 10800.25 cm³

Total Surface Area of the frustum = πR² + πr² + πl(R+r)

= π[(15.27)² + (10.82)²+ 20.49×(15.27+10.82)]

= π[233.1729 + 117.0724 + 534.5841]

= 2780.89 cm²

We have,

Diameter of one end = 10cm

So, radius (R) = 5cm

Diameter of other end = 8cm

So, radius (r) = 4cm

and slant height, l = 8cm

So, height, h = √{l² – (R – r)²}

= √{(8)² – (5 – 4)²

= √64 – (1)²

= √64 – 1

= √63

= 7.937 cm

Bearing surface of the clutch = CSA of the frustum

= πl (R + r)

= 3.14 × 8 (5 + 4)

= 25.12 (9)

= 226.08 cm²

= 506.75 cm³

Hence, Volume of frustum = 506.75 cm³

Greater diameter of the frustum = 56 cm

So, radius of the frustum = R = 28 cm

Smaller diameter of the frustum = 42 cm

So, Radius of the smaller end of the frustum = r = 21 cm

and Height of the frustum = h = 15 cm

Capacity of the frustum

= 28490 cm³

= 28.49 litres

Diameter of bigger circular end = 45cm

So,

Diameter of smaller circular end = 25cm

So,

Now,

Height of frustum = Total height of bucket – Height of cylinder

= 40 – 6

= 34cm

Also,

Slant height of the frustum, l = √{h² + (R – r)²}

= √1156 + 100

= √ 1256

= 35.44cm

Curved Surface Area of frustum = πl(R + r)

= 22 × 35.44 × 5

= 3898.4 cm²

Curved Surface area of cylinder = 2πrh

= 471.428 cm²

Area of circular base = πr²

= 491.07 cm²

Now, Area of metallic sheet used

= CSA of frustum + Area of circular base + CSA of cylinder

= 3898.4 + 471.428 + 491.07

= 4860.89

= 4860.9 cm²

= 33615.47 cm³

We know that 1 cm³ = 0.001 litre

∴ Volume of water that bucket can hold = 33.62 litres (approx.)

Greater diameter of the bucket = 30 cm

Radius of the bigger end of the bucket = R = 15 cm

Diameter of the smaller end of the bucket = 10 cm

Radius of the smaller end of the bucket = r = 5 cm

Height of the bucket = 24 cm

Slant height, l = √{h² + (R – r)²}

= √{(24)² +(15 – 5)²

= √576 + (10)²

= √576 + 100

= √676

= 26 cm

Now,

= 3.14 × 8 [225 + 25 + 75]

= 3.14 × 8 ×325

= 8164 cm³

= 8.164 Litres

A litre of milk cost Rs 20

So, total cost of filling the bucket with milk = Rs 20 × 8.164

= Rs 163.28

Surface Area of the bucket

= CSA of the frustum + Area of circular base

= πl(R+r) + πr²

= π [{26×(15+5)} + (5)²]

= π[520 + 25 ]

= 3.14 × 545

= 1711.3 cm²

Cost of 100cm² of metal sheet = Rs 10

= Rs 171.13

For the lower portion of the tent:

Diameter of the base = 14 m

Radius, R, of the base = 7 m

Diameter of the top end of the frustum = 7 m

Radius of the top end of the frustum =

Height of the frustum = h = 8 m

Slant height = l = √{h² + (R – r)²}

= √{(8)² +(7 – 3.5)²

= √64 + (3.5)²

= √64 + 12.25

= √76.25

= 8.73 m

For the conical part

Radius of the cone base = r = 3.5 m

Height of the cone = Height of the tent – height of frustum

= 12 – 8

= 4 m

Slant height of cone = L = √(4)²+(3.5)²

= √16 + 12.25

=√28.25

= 5.3 m

Total quantity of canvas = CSA of frustum + CSA of conical top

= πl(R + r) + πrL

= π [8.73(7 + 3.5) + 3.5 × 5.3]

= 346.5 m²

Diameter of top of funnel = 10cm

So, Radius (R) = 5cm

Diameter of cylindrical portion = 1cm

So,

Height of frustum = Total height – Height of cylindrical part

= 15 – 8

= 7cm

and Slant Height, l = √{h² + (R – r)²}

= √{(7)² +(5 – 0.5)²

= √49 + (4.5)²

= √49 + 20.25

= √69.25

= 8.32 cm

Now, CSA of frustum = πl (R + r)

= 143.81 cm²

Height of cylinder, H = 8cm

Now, CSA of cylinder = 2πrH

= 25.14 cm²

Area of tin required = CSA of frustum + CSA of cylinder

= 143.81 + 25.14

= 168.95 cm²