Given: A cuboid box with

• Length (L) = 20 cm
• Breadth (B) = 12 cm
• Height (H) = 10 cm

Surface area of vertical faces :

Vertical faces are the 4 side faces except the top and bottom.

Vertical faces are:

• Two faces of size L x H
• Two faces of size B x H

Area of two L x H faces = 2(L × H) = 2(20 × 10) = 2 × 200 = 400 cm²

Area of two B x H faces = 2(B × H) = 2(12 × 10) = 2 × 120 = 240 cm²

Total surface area of all vertical faces = 400 + 240 = 640 cm²

Total Surface Area (TSA) of cuboid :

Formula: TSA = 2(L.B + B.H + L.H)

• B = 20 x 12 = 240
• H = 12 x 10 = 120
• H = 20 x 10 = 200

∴ TSA = 2(240 + 120 + 200) = 2(560)

∴ TSA = 2 x 560 = 1120 cm²

Final Answers:

• Surface area of vertical faces = 640 cm²
• Total surface area of the box = 1120 cm²

Given:

• Total Surface Area (TSA) = 500 sq units
• Breadth b = 6 units
• Height h = 5 units

Formula for TSA of a cuboid:

TSA = 2(lb + bh + hl)

Substitute the values:

500 = 2(l x 6 + 6 x 5 + 5 x l)

500 = 2(6l + 30 + 5l)

500 = 2(11l + 30)

Divide both side by 2:

∴ 250 = 11l + 30

250 - 30 = 11l

11l = 220

∴ l = 220/11 = 20

∴ Length of the box = 20 units

Given:

• Side of the cube a = 4.5 cm

(i) Surface area of all vertical faces

A cube has 4 vertical faces, each a square of side 4.5 cm.

∴ Area of one face = a² = 4.5² = 20.25 cm²

∴ Area of 4 vertical faces = 4 × 20.25 = 81 cm²

Surface area of all vertical faces = 81 cm²

(ii) Total Surface Area (TSA) of cube

Formula: TSA = 6a²

TSA = 6 × 20.25 = 121.5 cm²

Total surface area = 121.5 cm²

Given:

• Total Surface Area (TSA) of cube = 5400 cm²

Formula: TSA = 6a²

∴ 6a² = 5400

∴ a² = 5400/6 = 900

∴ a = 30 cm

Surface area of all vertical faces :

A cube has 4 vertical faces, each of area a².

∴ Surface area of all vertical faces = 4a² = 4 x 900 = 3600 cm²

Surface area of all vertical faces = 3600 cm²

Given:

• Volume V = 34.50 m³
• Breadth b = 1.5 m
• Height h = 1.15 m

Formula for volume of a cuboid: V = l x b x h

Substitute the values:

34.50 = l x 1.5 x 1.15

34.50 = l x 1.725

l = \(\frac{34.50}{1.725}\) = 20 m

Length of the cuboid = 20 metres

The edge of a cube is the side of the cube.

So l = 7.5 cm.

The volume of a cube = l³ = (7.5)³ =7.5 × 7.5 × 7.5 = 421.875

= 421.88 cm³

The volume of the cube is 421.88 cm³.

Given:

• Radius r = 20 cm
• Height h = 13 cm
• π = 3.14
• curved surface area = ?,
• total surface area = ?

The curved surface area of a cylinder = 2πrh = 2 × 3.14 × 20 × 13 = 1632.80 cm²

The total surface area of a cylinder = 2πr(r + h) = 2 × 3.14 × 20(20 + 13)

= 2 × 3.14 × 20 × 33 = 4144.80 cm².

The curved surface area of the cylinder is 1632.80 cm2

Total surface area is 4144.80 cm².

Given:

• Curved Surface Area = 1980 cm²
• Radius r = 15 cm
• height of the cylinder h = ?

Formula : The curved surface area of a cylinder = 2πrh

∴ 1980 = 2 x \(\frac{22}{7}\) x 15 x h

∴ h = \(\frac{1980×7}{2×22×15}\) = 21 cm

The height of the cylinder is 21 cm.

For a right circular cone: l² = h² + r²

Given:

• Perpendicular height h = 12 cm
• Slant height l = 13 cm

Substitute:

13² = 12² + r²

169 = 144 + r²

r² = 169 – 144 = 25

r = 5 cm

Radius of the base of the cone = 5 cm

Let

• Radius r = 28 cm
• Total Surface Area TSA = 7128 cm²
• Use π = \(\frac{22}{7}\)

TSA formula of a cone : TSA = πr(r +l)

Substitute values:

7128 = \(\frac{22}{7}\) x 28(28 + l)

7128 = 88(28 + l)

28 + l = \(\frac{7128}{88}\) = 81

l = 81 – 28 = 53 cm

Find height h using l² = h² + r²

532 = h² + 28²

2809 = h² + 784

h² = 2025

h = 45 cm

Volume of the cone V = \(\frac{1}{3}\)πr²h = \(\frac{1}{3}\) × \(\frac{22}{7}\) x 28² x 45

= \(\frac{1}{3}\) × \(\frac{22}{7}\) x 784 x 45

= \(\frac{110880}{3}\) = 36960

Volume of the cone = 36,960 cm³

Given:

• Curved Surface Area (CSA) = 251.2 cm²
• Radius r = 8 cm
• Use π = 3.14

(i) Find Slant Height l

Formula: CSA = πrl

251.2 = 3.14 × 8 x l

 251.2 = 25.12l

l = \(\frac{251.2}{25.12}\) = 10

Slant height = 10 cm

(ii) Find Perpendicular Height h

Using Pythagoras theorem for a cone:

l² = h² + r²

10² = h² + 8²

100 = h² + 64

h² = 100 – 64 = 36

h² = 36

h = 6

Perpendicular height = 6 cm

Final Answers

• Slant height = 10 cm
• Perpendicular height = 6 cm

Given:

• Radius r = 6 m,
• height l = 8 m,
• the total surface area = ?

The total surface area of a cone = πr(r +l) = \(\frac{22}{7}\) × 6(8 + 6)

= \(\frac{22}{7}\)  × 6 × 14 = 264 m²

The cost of making the closed cone = rate x the total surface area of the cone

= ₹ 10 × 264 = ₹2640

The cost of making the closed cone is ₹ 2640.

Given:

• Volume V = 6280 cm³
• Radius r = 30 cm
• π = 3.14

Formula for volume of a cone: V = \(\frac{1}{3}\)πr²h

Substitute values:

6280 = \(\frac{1}{3}\) x 3.14 x 30² x h

6280 = \(\frac{1}{3}\) x 3.14 x 900 x h

6280 = 3.14 x 300 x h

6280 = 942h

∴  h = \(\frac{6280}{942}\)  = 6.67 cm (approx)

Perpendicular height ≈ 6.67 cm

We assume surface area here means curved surface area (CSA), because only then radius and height can be found.

Given:

• Curved Surface Area (CSA) = 188.4 cm²
• Slant height l = 10 cm
• π = 3.14

Find radius r :

Formula: CSA = πrl

Substitute values:

188.4 = 3.14 x r x 10

188.4 = 31.4r

r = \(\frac{188.4}{31.4}\) = 6

So: Radius = 6 cm

Find perpendicular height h :

Using Pythagoras theorem: l² = h² + r²

10² = h² + 6²

100 = h²+ 36

h²= 100-36 = 64

∴  h = 8

Perpendicular height = 8 cm

To find the surface area of the cone, we first need its radius and slant height.

Use volume to find radius

Given:

• Volume V = 1212 cm³
• Height h = 24 cm

Formula: V = \(\frac{1}{3}\)πr²h

Substitute values:

1212 = \(\frac{1}{3}\) × \(\frac{22}{7}\)  x  r² x 24

1212 = \(\frac{176}{7}\) x r²

1212 x 7 = 176r²

8484 = 176r²

r² = \(\frac{8484}{176}\) = 48.2

r = \(\sqrt{48.2}\) ≈ 6.94 cm

Find slant height l

Using Pythagoras theorem: l² = h² + r²

l = \(\sqrt{h^2+r^2}=\sqrt{(24)^2+(6.94)^2}\) = \(\sqrt{624.2}\) ≈ 25 cm

Find Total Surface Area (TSA)

Formula: TSA = πr (r + l)

TSA = \(\frac{22}{7}\) x 6.94 × (6.94 + 25) = \(\frac{22}{7}\) x 6.94 × 31.94 ≈ 695.8

Surface area of the cone ≈ 696 cm²

Given:

• curved surface area = 2200 cm²,
• l = 50 cm,
• total surface area = ?

Find the radius (r) :

Formula: The curved surface area of a cone = πrl

Substitute values:

∴  2200 = \(\frac{22}{7}\) × r ×50

r = \(\frac{2200×7}{22×50}\) = 14 cm.

Find Total Surface Area (TSA) :

TSA = πr(l + r)

TSA = \(\frac{22}{7}\) × 14 × (50 + 14) = 22 x 2 x 64 = 2816 cm²

Total Surface Area = 2816 sq.cm

Given:

• Number of persons = 25
• Area needed per person = 4 m²
• Therefore, base area of tent = 25 x 4 = 100 m²
• Height (h) of tent = 18 m

Since the tent is conical,

Base Area = πr² = 100

But for volume, we do not need the radius because:

Volume of cone V= \(\frac{1}{3}\) x Base Area x Height

Substitute values:

V = \(\frac{1}{3}\) × 100 × 18 = 1800/3 = 600 m³

Volume of the tent = 600 cubic metres

Given:

• h = 2.1 m,
• d =7.2m,
• r = d =7.2/2 = 3.6m

The volume of conical-shaped fodder = \(\frac{1}{3}\)πr²h

Substitute values:

\(\frac{1}{3}\) × \(\frac{22}{7}\) × 3.6 × 3.6 ×  2.1 = 28.512 m³

l² = h² + r²

= (3.6)² + (2.1)²

= 12.96 + 4.41 = 17.37

∴ l = \(\sqrt{17.37}\) = 4.17 m

The polythene required = the curved surface area of the conical-shaped fodder = πrl

= \(\frac{22}{7}\) x 3.6 x 4.17 ≈ 47.18 m²

The volume of the conical-shaped fodder is 28.512 m³

The polythene required is 47.18 m².

To solve this, we'll use the formulas for a sphere:

Formulas :

Surface Area (S) = 4πr²

Volume (V) = \(\frac{4}{3}\)πr³

Given:

• Radius r = 4 cm

(a) Surface Area

S = 4πr² = 4 x 3.14 x (4²) = 4 x 3.14 x 16 = 200.96 cm²

Surface Area = 200.96 cm²

(b) Volume :

V = \(\frac{4}{3}\)πr³  = \(\frac{4}{3}\) x 3.14 x (4³) = \(\frac{4}{3}\) x 3.14 x 64 = 267.95 cm³

Final Answers

• Surface Area = 200.96 cm²
• Volume = 267.95 cm³

Given:

• Radius r = 9 cm

(a) Surface Area :

S = 4πr² = 4 x 3.14 x (9²) = 4 x 3.14 x 81 = 1017.36 cm²

Surface Area = 1017.36 cm²

(b) Volume :

V = \(\frac{4}{3}\)πr³  = \(\frac{4}{3}\) x 3.14 x (9³) = \(\frac{4}{3}\) x 3.14 x 9 x 9 x 9 = 3052.08 cm³

Final Answers

• Surface Area = 1017.36 cm²
• Volume = 3052.08 cm³

Given:

• Radius r = 9 cm

(a) Surface Area :

S = 4πr² = 4 x 3.14 x (3.5²) = 4 x 3.14 x 3.5 x 3.5 = 153.86 cm²

Surface Area = 1017.36 cm²

(b) Volume :

V = \(\frac{4}{3}\)πr³  = \(\frac{4}{3}\) x 3.14 x (3.5³) = \(\frac{4}{3}\) x 3.14 x 3.5 x 3.5 x 3.5 = 179.50 cm³

Final Answers

• Surface Area = 153.86 cm²
• Volume = 179.50 cm³

Given:

• r = 5 cm, π = 3.14

(i) Curved Surface Area (CSA)

CSA = 2πr²= 2 × 3.14 × 5² = 2 × 3.14 × 25 = 157 cm²

(ii) Total Surface Area (TSA)

TSA = 3πr² = 3 x 3.14 x 25 = 235.5 cm²

Final Answers

• Curved Surface Area = 157 cm²
• Total Surface Area = 235.5 cm²

Given:

• Surface Area of sphere = 2826 cm²

Find the radius :

Surface Area = 4πr²

4πr² = 2826

4× 3.14 x r² =2826

12.56r² = 2826

r² = 2826/12.56 = 225

∴ r = \(\sqrt{225}\) = 15 cm

Find the Volume :

Formula: V =  πr³ 

V = \(\frac{4}{3}\) x 3.14 × 15³ = \(\frac{4}{3}\) × 3.14 x 3375 = 14130 cm³

Final Answer :

• Volume = 14130 cm³

Given:

V = 38808 cm³,

Formula: V = \(\frac{4}{3}\)πr³ 

Find the radius :

38808 = \(\frac{4}{3}\) × \(\frac{22}{7}\) × r³ 

38808 = \(\frac{88}{21}\) x r³ 

r³ = \(\frac{38808×21}{88}\) = 441 × 21 = 9261 = (21)³

∴ r = 21 cm

Surface Area of the Sphere :

Surface Area = 4 πr²

= 4 × \(\frac{22}{7}\) × 21² = 4 × \(\frac{22}{7}\) × 21 x 21 = 4 × 22 × 3 x 21 = 5544

Final Answer :

• Surface Area = 5544 cm²

Given:

• V = 18000π cm³

Volume of a hemisphere: V = \(\frac{2}{3}\)πr³ 

Solve for r³

18000 π = \(\frac{2}{3}\)πr³ 

Cancel π on both sides:

18000 = \(\frac{2}{3}\)r³ 

r³ = \(\frac{18000×3}{2}\) = 27000 = (30)³

∴ r = 30 cm

Diameter = 2r = 2 × 30 =60 cm

Final Answer: Diameter = 60 cm

To find how much area the road roller presses, we use the curved surface area of a cylinder, because the roller is a cylinder.

Given

• Diameter = 0.9 m -> Radius r = 0.45 m
• Length (height of cylinder) h = 1.4 m
• Number of rotations = 500
• Use π = \(\frac{22}{7}\)

Find area pressed in 1 rotation :

Area pressed = Curved Surface Area of roller per rotation

CSA = 2πrh = 2 × \(\frac{22}{7}\) × 0.45 × 1.4 = 3.96 m²

So, area pressed in 1 rotation = 3.96 m²

Area pressed in 500 rotations :

Total area = 500 × 3.96 =1980 m²

Final Answer:

• The roller presses 1980 m² of the field.

The gauge (i.e. thickness) of the glass sheet = 2 mm = 0.2 cm.

To find the volume of water that can be contained in the fish tank, we find the inner dimensions of the fish tank.

The length and breadth will be reduced by 0.2 cm from each side.

∴ l = (60.4 - 0.2 - 0.2) cm = 60 cm,

b = (40.4 - 0.2 - 0.2) cm = 40 cm.

The tank is open.

∴ h = (40.2 - 0.2) cm = 40cm.

The volume of the inner side of the fish tank = l x b x h

= 60 × 40 × 40 = 96000 cm3

Final Answer:

• 96,000 cm³ of water will be contained in the fish tank.

Let the radius = 5k and height = 12k.

Volume of a cone:

V = \(\frac{1}{3}\)πr²h 

314 = \(\frac{1}{3}\) × 3.14 × (5k)² × (12k)

314 = \(\frac{1}{3}\) x 3.14 x 25k² x 12k

314 = \(\frac{1}{3}\) x 3.14× 300k³

314 = 3.14 x 100k³

314 = 314k³

k³ = 1

∴ k = 1

Find radius and height :

r = 5k = 5m

h = 12k = 12 m

Slant height :

l = \(\sqrt{r^2+h^2}\)

l = \(\sqrt{5^2+12^2}\) = \(\sqrt{25+144}\) = \(\sqrt{169}\) = 13 m

Final Answers :

• Perpendicular height = 12 m
• Slant height = 13 m

To find the radius, use the volume of a sphere formula:

V = \(\frac{4}{3}\)πr³ 

Given:

• V = 904.32 cm³, π = 3.14

Substitute the values :

904.32 = \(\frac{4}{3}\) × 3.14 x r³

904.32 = 4.1867r³

r³ = \(\frac{904.32}{4.1867}\) = 216 = (6)³

∴ r = 6 cm

Final Answer: Radius = 6 cm

For a cube:

Total Surface Area (TSA) = 6a²

Given:

• 6a² = 864

Find the side length a

a² = 862/6 = 144

a = \(\sqrt{144}\) = 12 cm

Volume = a³ = 12³ = 1728 cm³

Final Answer: Volume = 1728 cm³

Here, the surface area = 154 cm², V = ?

The surface area of a sphere = 4πr²

154 = 4 x \(\frac{22}{7}\) x r²

22 x 7 = 4 x \(\frac{22}{7}\) x r²

r² = \(\frac{7×7}{4}\)

r = \(\frac{7}{2}\) cm

Volume of a sphere formula:

V = \(\frac{4}{3}\)πr³ = \(\frac{4}{3}\)×\(\frac{22}{7}\)×\((\frac{7}{2})^3\) ≈ 179.67 cm³

Final Answer: Volume ≈ 179.7 cm³

Given:

• Total surface area of cone (TSA) = 616 cm²
• Slant height l = 3r
• Use π = \(\frac{22}{7}\)

Use the TSA formula

TSA = πr(l + r)

Substitute l = 3r:

616 = πr(3r + r)

616 = πr(4r)

616 = 4πr² = 4 x \(\frac{22}{7}\)  x r² = r²

r² = \(\frac{616×7}{88}\) = \(\frac{4312}{88}\) = 49

∴ r = 7 cm

Slant height l = 3r = 3 × 7 = 21 cm

Final Answer: Slant height = 21 cm

Given

• Inner diameter = 4.20 m
• Radius r = 4.20/2 = 2.1 m
• Depth (height) h = 10 m
• Plaster rate = ₹52 per m²

Inner surface area of the well = 2πrh

= 2 × \(\frac{22}{7}\) × 2.1 × 10 = 132 m²

Cost of Plastering :

Cost = 132 × 52 = ₹ 6864

Final Answers :

• Inner surface area of the well = 132 m²
• Cost of plastering = ₹ 6864

Given :

Length of roller (height of cylinder) h = 2.1 m

Diameter = 1.4 m -> Radius r = 1.4/2 = 0.7 m

Number of rotations = 500

Cost of levelling = ₹7 per m²

Use π = \(\frac{22}{7}\)

A road roller levels the ground by its curved surface area (CSA).

Area levelled in 1 rotation

CSA = 2πrh

Substitute values:

= 2 × \(\frac{22}{7}\)  × 0.7 x 2.1 = 9.24 m²

∴ The area levelled in 500 revolutions of the road roller = 500 x 9.24 m² = 4620 m².

The cost of levelling the ground = rate x area of the ground = ₹ 7 x 4620 =₹ 32340

Final Answers :

• Area of the ground levelled is 4620 m²
• The cost of levelling the ground is ₹ 32,340.