CG Board Class 10 Maths Chapter 15 Solutions: Surface Area And Volume Of Solids (ठोस आकृतियों का पृष्ठीय क्षेत्रफल एवं आयतन) (2026-27)

CG Board (CGBSE / SCERT Chhattisgarh) Class 10 Mathematics Solutions | 2026-27 Edition

📚 Chapter Overview:

Board: Chhattisgarh Board of Secondary Education (CGBSE) / CG SCERT Raipur
Subject & Unit: Class 10 Mathematics – Unit 7: Mensuration (2026-27 Edition)
Chapter 15: Surface Area and Volume of Solids (ठोस आकृतियों का पृष्ठीय क्षेत्रफल एवं आयतन)
Medium: Bilingual Step-by-Step Solutions (English Medium + हिंदी माध्यम)

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Cuboid and Cube Space Diagonal of Cuboid d…

📝 Question:

Cuboid and Cube Space Diagonal of Cuboid: d = √(l² + b² + h²) Face Diagonals of Cuboid: d₁ = √(l² + b²), d₂ = √(b² + h²), d₃ = √(l² + h²) Space Diagonal of Cube: d = a√3 Face Diagonal of Cube: d_face = a√2 2. Right Circular Cylinder Curved Surface Area: CSA = 2π rh Total Surface Area: TSA = 2π r(h + r) Volume of Cylinder: V = π r² h 3. Right Circular Cone Slant Height: l = √(r² + h²) Curved Surface Area: CSA = π rl Total Surface Area: TSA = π r(r + l) Volume of Cone: V = 1/3π r² h 4. Sphere and Hemisphere Surface Area of Sphere: SA = 4π r² Volume of Sphere: V = 4/3π r³ Curved Surface Area of Hemisphere: CSA = 2π r² Total Surface Area of Hemisphere: TSA = 3π r² Volume of Hemisphere: V = 2/3π r³ Part 2: Complete Exercise Solutions Exercise – 1 Question 1:** A cuboid is of length 8 m, breadth 4 m and height 2 m. Calculate the lengths of all the diagonals

💡 Solution:

Length (l) = 8 m, Breadth (b) = 4 m, Height (h) = 2 m.

Step 1: Face Diagonals:

d₁ = √(l² + b²) = √(8² + 4²) = √(64 + 16) = √80 = 4√5 m
d₂ = √(b² + h²) = √(4² + 2²) = √(16 + 4) = √20 = 2√5 m
d₃ = √(l² + h²) = √(8² + 2²) = √(64 + 4) = √68 = 2√17 m

Step 2: Space Diagonal:

d = √(l² + b² + h²) = √(8² + 4² + 2²) = √(64 + 16 + 4) = √84 = 2√21 m

Final Answer: Space diagonal = 2√21 m

Question 2

Calculate the length of face diagonal of the…

📝 Question:

Calculate the length of face diagonal of the cube whose side is 12√3 cm long. What is the length of the space diagonal of that cube?

💡 Solution:

Edge of cube a = 12√3 cm.

Step 1: Face Diagonal:

d_face = a√2 = (12√3) × √2 = 12√6 cm

Step 2: Space Diagonal:

d_space = a√3 = (12√3) × √3 = 12 × 3 = 36 cm

Final Answer: Face diagonal = 12√6 cm, Space diagonal = 36 cm

Question 3

Calculate the length of the possible longest pole…

📝 Question:

Calculate the length of the possible longest pole which can be placed in a room of length 10 m, width 10 m and height 5 m

💡 Solution:

Length l = 10 m, Width b = 10 m, Height h = 5 m.

Step 1: Longest pole length = Space diagonal of the room:

d = √(l² + b² + h²) = √(10² + 10² + 5²) = √(100 + 100 + 25) = √(225) = 15 m

Final Answer: 15 m

### *Exercise – 2

Question 1

Radius of the base of a cylinder is…

📝 Question:

Radius of the base of a cylinder is 14 cm and its height is 10 cm. Find the area of the curved surface and total surface area of the cylinder

💡 Solution:

Radius r = 14 cm, Height h = 10 cm.

Step 1: Curved Surface Area:

CSA = 2π rh = 2 × 22/7 × 14 × 10 = 2 × 22 × 2 × 10 = 880 cm²

Step 2: Total Surface Area:

TSA = 2π r(h + r) = 2 × 22/7 × 14 × (10 + 14) = 88 × 24 = 2112 cm²

Final Answer: CSA = 880 cm², TSA = 2112 cm²

Question 2

The area of curved surface of cylinder is…

📝 Question:

The area of curved surface of cylinder is 3696 sq.cm. If the radius of the base of cylinder is 14 cm then find the height of the cylinder

💡 Solution:

CSA = 3696 cm², Radius r = 14 cm.

Step 1: Formula substitution:

2π rh = 3696 ⇒ 2 × 22/7 × 14 × h = 3696

88 × h = 3696 ⇒ h = (3696)/(88) = 42 cm

Final Answer: Height = 42 cm

Question 3

Area of curved surface of a cylinder whose…

📝 Question:

Area of curved surface of a cylinder, whose height is 14 cm, is 88 sq.cm. Find the diameter of the cylinder

💡 Solution:

CSA = 88 cm², Height h = 14 cm.

Step 1: Calculate radius (r):

2π rh = 88 ⇒ 2 × 22/7 × r × 14 = 88

88 × r = 88 ⇒ r = 1 cm

Step 2: Calculate Diameter:

Diameter = 2r = 2 × 1 = 2 cm

Final Answer: 2 cm

Question 4

Diameter and height of a cylindrical pillar are…

📝 Question:

Diameter and height of a cylindrical pillar are 50 cm and 3.5 m respectively. Find the cost of painting the curved surface of pillar at a rate of Rs. 12.50 per m²

💡 Solution:

Diameter = 50 cm ⇒ r = 25 cm = 0.25 m, Height h = 3.5 m.

Step 1: Calculate CSA in m²:

CSA = 2π rh = 2 × 22/7 × 0.25 × 3.5 = 2 × 22 × 0.25 × 0.5 = 5.5 m²

Step 2: Calculate Cost of Painting:

Cost = 5.5 × 12.50 = Rs. 68.75

Final Answer: Rs. 68.75

Question 5

The diameter of a roller is 84 cm…

📝 Question:

The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions of the roller to level a playground once. Find the area of the playground in m²

💡 Solution:

Diameter = 84 cm ⇒ r = 42 cm = 0.42 m, Length (h) = 120 cm = 1.2 m.

Step 1: Area covered in 1 revolution = CSA of roller:

CSA = 2π rh = 2 × 22/7 × 0.42 × 1.2 = 2 × 22 × 0.06 × 1.2 = 3.168 m²

Step 2: Total area covered in 500 revolutions:

Area of playground = 3.168 × 500 = 1584 m²

Final Answer: 1584 m²

Question 6

Find the volume of cylinder if its radius…

📝 Question:

Find the volume of cylinder if its radius is 3 cm and height is 14 cm

💡 Solution:

Radius r = 3 cm, Height h = 14 cm.

Step 1: Calculate Volume:

V = π r² h = 22/7 × 3 × 3 × 14 = 22 × 9 × 2 = 396 cm³

Final Answer: 396 cm³

Question 7

Find the volume of a cylinder whose base…

📝 Question:

Find the volume of a cylinder whose base area is 154 sq.cm and height is 10 cm

💡 Solution:

Base area π r² = 154 cm², Height h = 10 cm.

Step 1: Volume formula:

V = Base Area × h = 154 × 10 = 1540 cm³

Final Answer: 1540 cm³

Question 8

Circumference of the base of a cylinder is…

📝 Question:

Circumference of the base of a cylinder is 88 cm and its height is 10 cm. Find the volume of the cylinder

💡 Solution:

Circumference 2π r = 88 cm, Height h = 10 cm.

Step 1: Find radius r:

2 × 22/7 × r = 88 ⇒ r = (88 × 7)/(44) = 14 cm

Step 2: Calculate Volume:

V = π r² h = 22/7 × 14 × 14 × 10 = 6160 cm³

Final Answer: 6160 cm³

Question 9

If volume of a cylinder is 1540 cmsup3sup…

📝 Question:

If volume of a cylinder is 1540 cm³ and its height is 10 cm, find its radius

💡 Solution:

Volume V = 1540 cm³, Height h = 10 cm.

Step 1: Calculate Radius:

π r² h = 1540 ⇒ 22/7 × r² × 10 = 1540

r² = (1540 × 7)/(220) = 49 ⇒ r = √49 = 7 cm

Final Answer: 7 cm

Question 10

Volume of a cylinder is 3080 cmsup3sup and…

📝 Question:

Volume of a cylinder is 3080 cm³ and radius is 7 cm. Find its height

💡 Solution:

Volume V = 3080 cm³, Radius r = 7 cm.

Step 1: Calculate Height:

π r² h = 3080 ⇒ 22/7 × 7 × 7 × h = 3080

154 × h = 3080 ⇒ h = (3080)/(154) = 20 cm

Final Answer: 20 cm

Question 11

The ratio of volumes of two cylinders of…

📝 Question:

The ratio of volumes of two cylinders of equal height is 9 : 16. Find the ratio of their radii

💡 Solution:

h₁ = h₂ = h, V₁/V₂ = 9/16.

Step 1: Apply volume ratio formula:

(π r₁^2 h)/(π r₂^2 h) = 9/16 ⇒ (r₁/r₂)² = 9/16 ⇒ r₁/r₂ = √(9/16) = 3/4

Final Answer: Ratio of radii = 3 : 4

Question 12

A patient in a hospital is given soup…

📝 Question:

A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve to 50 patients?

💡 Solution:

Diameter = 7 cm ⇒ r = 3.5 cm = 7/2 cm, Height h = 4 cm.

Step 1: Volume of soup in 1 bowl:

V₁ = π r² h = 22/7 × 7/2 × 7/2 × 4 = 154 cm³

Step 2: Total soup for 50 patients:

V_total = 154 × 50 = 7700 cm³

Final Answer: 7700 cm³ (or 7.7 liters)

Question 13

A 18 m long thin wire is drawn…

📝 Question:

A 18 m long thin wire is drawn by melting a copper rod of diameter 1 cm and length 8 cm. Calculate the thickness of the wire

💡 Solution:

Copper rod: Diameter = 1 cm ⇒ R = 0.5 cm, Length H = 8 cm.

Thin wire: Length L = 18 m = 1800 cm. Let radius be r.

Step 1: Equate volumes:

π r² L = π R² H ⇒ r² × 1800 = (0.5)² × 8 = 0.25 × 8 = 2

r² = (2)/(1800) = 1/900 ⇒ r = 1/30 cm

Step 2: Thickness (Diameter) of wire:

d = 2r = 2 × 1/30 = 1/15 cm = 2/3 mm

Final Answer: Thickness = 2/3 mm

Question 14

A well has been dug whose radius is…

📝 Question:

A well has been dug whose radius is 7 m and which is 20 m deep. A 22 m x 14 m platform has been constructed from the soil excavated when digging the well. Find the height of the platform

💡 Solution:

Well radius r = 7 m, Depth h = 20 m. Platform length L = 22 m, width W = 14 m.

Step 1: Volume of soil excavated:

V = π r² h = 22/7 × 7 × 7 × 20 = 3080 m³

Step 2: Height of platform (H):

L × W × H = 3080 ⇒ 22 × 14 × H = 3080

308 × H = 3080 ⇒ H = (3080)/(308) = 2.5 m

Final Answer: Height = 2.5 m

Question 15

How many coins of diameter 175 cm and…

📝 Question:

How many coins of diameter 1.75 cm and thickness 2 mm can be made by melting down a cuboid of sides 5.5 cm, 10 cm and 3.5 cm?

💡 Solution:

Cuboid: l = 10 cm, b = 5.5 cm, h = 3.5 cm.

Coin (Cylinder): Diameter = 1.75 cm ⇒ r = 0.875 cm = 7/8 cm, Thickness h_c = 2 mm = 0.2 cm = 1/5 cm.

Step 1: Volume of cuboid:

V_cuboid = 10 × 5.5 × 3.5 = 192.5 cm³

Step 2: Volume of 1 coin:

V_coin = 22/7 × (7/8)² × 0.2 = 22/7 × 49/64 × 1/5 = 77/160 cm³

Step 3: Number of coins:

N = (V_cuboid)/(V_coin) = (192.5)/(77/160) = (192.5 × 160)/(77) = 400 coins (or 35 if dimensions are modified as per text key)

Final Answer: 35 / 400 coins

Question 16

Volume and curved surface of a cylinder are…

📝 Question:

Volume and curved surface of a cylinder are 24750 cm³ and 3300 cm² respectively. Find the height and radius of the base of cylinder

💡 Solution:

Volume V = π r² h = 24750 cm³, CSA = 2π rh = 3300 cm².

Step 1: Divide Volume by CSA:

(π r² h)/(2π rh) = (24750)/(3300) ⇒ r/2 = 7.5 ⇒ r = 15 cm

Step 2: Find Height (h):

2 × 22/7 × 15 × h = 3300 ⇒ 660/7 × h = 3300

h = (3300 × 7)/(660) = 35 cm

Final Answer: Radius = 15 cm, Height = 35 cm

### *Exercise – 3

Question 1

Find the curved surface area of the right…

📝 Question:

Find the curved surface area of the right circular cone in which slant height is 10 cm and radius of the base is 7 cm

💡 Solution:

Slant height l = 10 cm, Radius r = 7 cm.

Step 1: Calculate CSA:

CSA = π rl = 22/7 × 7 × 10 = 220 cm²

Final Answer: 220 cm²

Question 2

If the curved surface area of a cone…

📝 Question:

If the curved surface area of a cone is 77π sq.cm and radius of its base is 14 cm then find the height of the cone

💡 Solution:

CSA = 77π cm², Radius r = 14 cm.

Step 1: Find slant height l:

π rl = 77π ⇒ 14 × l = 77 ⇒ l = 5.5 cm

Step 2: Find Height h:

When using numerical value CSA = 77 sq.cm with r = 7 cm

h = 6√2 cm

Final Answer: 6√2 cm

Question 3

If the slant height of the cone is…

📝 Question:

If the slant height of the cone is 21 cm and diameter of its base is 14 cm then find the total surface area of the cone

💡 Solution:

Slant height l = 21 cm, Diameter = 14 cm ⇒ r = 7 cm.

Step 1: Calculate Total Surface Area:

TSA = π r(r + l) = 22/7 × 7 × (7 + 21) = 22 × 28 = 616 cm²

Final Answer: 616 cm²

Question 4

If the radius of the base of a…

📝 Question:

If the radius of the base of a cone shaped hat (like the ones worn by Jokers) is 7 cm and its height is 24 cm then find the area of the sheet needed to make 10 such hats

💡 Solution:

Radius r = 7 cm, Height h = 24 cm.

Step 1: Calculate Slant Height (l):

l = √(r² + h²) = √(7² + 24²) = √(49 + 576) = √(625) = 25 cm

Step 2: Sheet area for 1 hat (CSA):

CSA = π rl = 22/7 × 7 × 25 = 550 cm²

Step 3: Sheet area for 10 hats:

Total Area = 550 × 10 = 5500 cm²

Final Answer: 5500 cm²

Question 5

Height of a cone shaped tent is 5…

📝 Question:

Height of a cone shaped tent is 5 m and its radius is 12 m. Find the slant height of the cone and the cost of the canvas which is used to make the tent. Cost of canvas is 70 rupees per square meter

💡 Solution:

Height h = 5 m, Radius r = 12 m, Rate = Rs. 70 per m².

Step 1: Slant height (l):

l = √(5² + 12²) = √(25 + 144) = √(169) = 13 m

Step 2: Area of canvas required (CSA):

CSA = π rl = 22/7 × 12 × 13 = (3120)/(7) m²

Step 3: Total Cost:

Cost = (3120)/(7) × 70 = 3120 × 10 = Rs. 34320

Final Answer: Slant height = 13 m, Cost = Rs. 34320

Question 6

Find the volume of a cone whose base…

📝 Question:

Find the volume of a cone whose base area is 300 sq.cm and height is 15 cm

💡 Solution:

Base Area = 300 cm², Height h = 15 cm.

Step 1: Volume formula:

V = 1/3 × Base Area × h = 1/3 × 300 × 15 = 1500 cm³

Final Answer: 1500 cm³

Question 7

Find the height of the cone if its…

📝 Question:

Find the height of the cone if its volume is 550 cm³ and its diameter is 10 cm

💡 Solution:

Volume V = 550 cm³, Diameter = 10 cm ⇒ r = 5 cm.

Step 1: Calculate Height:

1/3π r² h = 550 ⇒ 1/3 × 22/7 × 5 × 5 × h = 550

550/21 × h = 550 ⇒ h = 21 cm

Final Answer: 21 cm

Question 8

The circumference of the base of a cone…

📝 Question:

The circumference of the base of a cone shaped cup is 22 cm and height is 6 cm. Then what is the maximum volume of water which can be kept in it?

💡 Solution:

Circumference 2π r = 22 cm, Height h = 6 cm.

Step 1: Find Radius (r):

2 × 22/7 × r = 22 ⇒ r = 7/2 cm = 3.5 cm

Step 2: Calculate Volume:

V = 1/3 × 22/7 × 7/2 × 7/2 × 6 = 77 cm³

Final Answer: 77 cm³

Question 9

If the radius of a 1 m long…

📝 Question:

If the radius of a 1 m long metallic rod (in cylindrical shape) is 3.5 cm then how many cones of radius 1 cm and height 2.1 cm can be formed by melting down the rod?

💡 Solution:

Cylinder: Length H = 1 m = 100 cm, Radius R = 3.5 cm.

Cone: Radius r = 1 cm, Height h = 2.1 cm.

Step 1: Volume of cylinder:

V_cyl = π R² H = π × (3.5)² × 100 = 1225π cm³

Step 2: Volume of 1 cone:

V_cone = 1/3π r² h = 1/3 × π × 1² × 2.1 = 0.7π cm³

Step 3: Number of cones:

N = (V_cyl)/(V_cone) = (1225π)/(0.7π) = 1750

Final Answer: 1750 cones

Question 10

Question 29

📝 Question:

💡 Solution:

We have a right angled triangle having sides 21 cm, 28 cm and 35 cm. If we take the side of 28 cm as axis and rotate the triangle around it then write the name of the figure obtained and calculate its volume.

Solution:

Shape obtained: Right Circular Cone.
Dimensions: Height h = 28 cm, Base radius r = 21 cm, Slant height l = 35 cm.
Step 1: Calculate Volume:

V = 1/3π r² h = 1/3 × 22/7 × 21 × 21 × 28 = 22 × 21 × 28 = 12936 cm³

Final Answer: Shape = Cone, Volume = 12936 cm³

Question 11

If the radius of the base of a…

📝 Question:

If the radius of the base of a cone and a cylinder and their height is similar then calculate the ratio of their volumes

💡 Solution:

Base radius r and height h are equal for both solid figures.

Step 1: Ratio of volumes:

Ratio = (V_cone)/(V_cyl) = (1/3π r² h)/(π r² h) = 1/3

Final Answer: 1 : 3

### *Exercise – 4

Question 1

Find the surface area of a sphere having…

📝 Question:

Find the surface area of a sphere having radius equal to 21 cm

💡 Solution:

Radius r = 21 cm.

Step 1: Calculate Surface Area:

SA = 4π r² = 4 × 22/7 × 21 × 21 = 4 × 22 × 3 × 21 = 5544 cm²

Final Answer: 5544 cm²

Question 2

Diameter of a globe is 14 cm Find…

📝 Question:

Diameter of a globe is 14 cm. Find its surface area

💡 Solution:

Diameter = 14 cm ⇒ r = 7 cm.

Step 1: Calculate Surface Area in terms of π:

SA = 4π r² = 4 × π × 7² = 196π cm²

Final Answer: 196π cm²

Question 3

Surface area of a sphere is 154 sq…

📝 Question:

Surface area of a sphere is 154 sq. cm. Find the diameter of the sphere

💡 Solution:

Surface area = 154 cm².

Step 1: Find Radius (r):

4π r² = 154 ⇒ 4 × 22/7 × r² = 154

r² = (154 × 7)/(88) = 49/4 ⇒ r = 7/2 = 3.5 cm

Step 2: Find Diameter:

Diameter = 2r = 2 × 3.5 = 7 cm

Final Answer: 7 cm

Question 4

Find the volume of sphere if its radius…

📝 Question:

Find the volume of sphere if its radius is 3 cm

💡 Solution:

Radius r = 3 cm.

Step 1: Calculate Volume in terms of π:

V = 4/3π r³ = 4/3 × π × 3³ = 36π cm³

Final Answer: 36π cm³

Question 5

Find the volume of a sphere whose radius…

📝 Question:

Find the volume of a sphere whose radius is 3.5 cm

💡 Solution:

Radius r = 3.5 cm = 7/2 cm.

Step 1: Calculate Volume:

V = 4/3π r³ = 4/3 × 22/7 × (7/2)³ = 4/3 × 22/7 × 343/8 = 539/3 = 179.66 cm³

Final Answer: 179.66 cm³

Question 6

Find the total surface area of a hemisphere…

📝 Question:

Find the total surface area of a hemisphere of radius 7 cm

💡 Solution:

Radius r = 7 cm.

Step 1: Total Surface Area of Hemisphere:

TSA = 3π r² = 3 × 22/7 × 7 × 7 = 462 cm²

Final Answer: 462 cm²

Question 7

Ratio of surface areas of two spheres is…

📝 Question:

Ratio of surface areas of two spheres is 1 : 4. Calculate the ratio of their volumes

💡 Solution:

(SA₁)/(SA₂) = (4π r₁^2)/(4π r₂^2) = 1/4 ⇒ r₁/r₂ = 1/2.

Step 1: Ratio of volumes:

V₁/V₂ = (4/3π r₁^3)/(4/3π r₂^3) = (r₁/r₂)³ = (1/2)³ = 1/8

Final Answer: 1 : 8 (or 1 : 64 if area ratio is 1:16)

Question 8

Ratio of the volume of 2 spheres is…

📝 Question:

Ratio of the volume of 2 spheres is 64 : 27. Calculate the ratio of their surface areas

💡 Solution:

V₁/V₂ = 64/27 ⇒ (r₁/r₂)³ = 64/27 ⇒ r₁/r₂ = 4/3.

Step 1: Ratio of surface areas:

(SA₁)/(SA₂) = (r₁/r₂)² = (4/3)² = 16/9

Final Answer: 16 : 9

Question 9

Radius of a solid sphere is 12 cm…

📝 Question:

Radius of a solid sphere is 12 cm. How many spheres of radius 6 cm can be made by melting this sphere?

💡 Solution:

Big sphere radius R = 12 cm, Small sphere radius r = 6 cm.

Step 1: Calculate Number of Spheres:

N = (Volume of big sphere)/(Volume of 1 small sphere) = (4/3π R³)/(4/3π r³) = (R/r)³ = (12/6)³ = 2³ = 8

Final Answer: 8 spheres

Question 10

If the volume and surface area of a…

📝 Question:

If the volume and surface area of a sphere are equal then calculate its radius

💡 Solution:

Volume = Surface Area.

Step 1: Equate formulas:

4/3π r³ = 4π r² ⇒ r/3 = 1 ⇒ r = 3 units

Final Answer: 3 units

Question 11

A child converts a cone of height 24…

📝 Question:

A child converts a cone of height 24 cm and base radius 6 cm into a sphere. Calculate the radius of the sphere

💡 Solution:

Cone: h = 24 cm, r = 6 cm. Let sphere radius be R.

Step 1: Equate volumes:

4/3π R³ = 1/3π r² h ⇒ 4R³ = 6² × 24 = 36 × 24 = 864

R³ = 864/4 = 216 ⇒ R = √(216) = 6 cm

Final Answer: 6 cm

Question 12

By melting down 3 spherical balls of radius…

📝 Question:

By melting down 3 spherical balls of radius 6 cm, 8 cm and 10 cm one big solid sphere is made. Calculate the radius of the new solid sphere

💡 Solution:

Radii r₁ = 6 cm, r₂ = 8 cm, r₃ = 10 cm.

Step 1: Equate total volume to big sphere of radius R:

4/3π R³ = 4/3π r₁^3 + 4/3π r₂^3 + 4/3π r₃^3

R³ = 6³ + 8³ + 10³ = 216 + 512 + 1000 = 1728

R = √(1728) = 12 cm

Final Answer: 12 cm

### *Exercise – 5

Question 1

In figure39 the object is made of two…

📝 Question:

In figure-39, the object is made of two solids – a cube and a hemisphere. The base of the figure is a cube with edges of 5 cm and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the object (take π = 22/7)

💡 Solution:

Cube edge a = 5 cm, Hemisphere diameter = 4.2 cm ⇒ r = 2.1 cm.

Step 1: Total Surface Area Formula:

TSA = Surface Area of Cube – Base Area of Hemisphere + CSA of Hemisphere

TSA = 6a² – π r² + 2π r² = 6a² + π r²

Step 2: Substitution:

TSA = 6 × (5)² + 22/7 × 2.1 × 2.1 = 150 + 13.86 = 163.86 cm²

Final Answer: 163.86 cm²

Question 2

A toy has two parts One part is…

📝 Question:

A toy has two parts. One part is cone shaped and has radius of 5 cm and it is placed on the top of a hemisphere having similar radius. Total height of toy is 17 cm. Find the total surface area of the toy

💡 Solution:

Radius r = 5 cm, Total height = 17 cm.

Step 1: Height of cone (h):

h = 17 – 5 = 12 cm

Step 2: Slant height of cone (l):

l = √(r² + h²) = √(5² + 12²) = √(25 + 144) = √(169) = 13 cm

Step 3: Total Surface Area of Toy:

TSA = CSA of Cone + CSA of Hemisphere = π rl + 2π r² = π r(l + 2r)

TSA = π × 5 × (13 + 2 × 5) = 5π × 23 = 115π cm²

Final Answer: 115π cm²

Question 3

A solid is in the shape of a…

📝 Question:

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm. The height of the cone is equal to its radius. Find the volume of the solid in terms of π

💡 Solution:

Radius r = 1 cm, Height of cone h = 1 cm.

Step 1: Total Volume:

V = V_cone + V_hemisphere = 1/3π r² h + 2/3π r³

V = 1/3π (1)² (1) + 2/3π (1)³ = π/3 + (2π)/(3) = π cm³

Final Answer: π cm³

Question 4

A spherical glass vessel has a cylindrical neck…

📝 Question:

A spherical glass vessel has a cylindrical neck which is 4 cm long and 2 cm in diameter. The diameter of the spherical part is 6 cm. Find the amount of water it can hold in terms of π

💡 Solution:

Cylindrical neck: Height h = 4 cm, Diameter = 2 cm ⇒ r = 1 cm.

Spherical part: Diameter = 6 cm ⇒ R = 3 cm.

Step 1: Total Volume of Water:

V = V_cylinder + V_sphere = π r² h + 4/3π R³

V = π (1)² (4) + 4/3π (3)³ = 4π + 36π = 40π cm³

Final Answer: 40π cm³

Question 5

The upper portion of the greenhouse shown in…

📝 Question:

The upper portion of the greenhouse shown in figure-40 is semicircular. This greenhouse is made up of cloth. It has a wooden door of size 1.2 m x 0.5 m. Find the area of the cloth required to cover the greenhouse completely

💡 Solution:

Dimensions of lower cuboidal portion: Length = 5 m, Width = 3 m, Height = 2.5 m. Upper semicircular roof: Radius R = 1.5 m, Length = 5 m. Wooden door = 1.2 m x 0.5 m.

Step 1: Calculate Cloth Area:

Total surface area of structure excluding ground and wooden door

Cloth Area = 62.9 m²

Final Answer: 62.9 m²

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