Surface Areas and Volumes l Exercise 12.1 l Class 10 l NCERT Solutions

NCERT Class 10 Maths | Exercise 12.1 | Surface Areas and Volumes

Class 10 Mathematics

Chapter 12: Surface Areas and Volumes

Exercise 12.1 – Stepwise Solutions

Instruction: Unless stated otherwise, take π = 22/7.


Q1) Two cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Step 1: Volume of cube = a³

a³ = 64

a = 4 cm

Step 2: When two cubes are joined end-to-end:

Length = 4 + 4 = 8 cm

Breadth = 4 cm

Height = 4 cm

Step 3: Surface area of cuboid = 2(lb + bh + lh)

= 2(8×4 + 4×4 + 8×4)

= 2(32 + 16 + 32)

= 2 × 80

Surface area = 160 cm²


Q2) A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder.

Diameter of hemisphere = 14 cm

Radius r = 7 cm

Total height of vessel = 13 cm

Step 1: Height of cylinder

Height = 13 − 7 = 6 cm

Step 2: Inner surface area = CSA of cylinder + CSA of hemisphere

CSA cylinder = 2πrh

= 2 × 22/7 × 7 × 6

= 264 cm²

CSA hemisphere = 2πr²

= 2 × 22/7 × 7 × 7

= 308 cm²

Total inner surface area

= 264 + 308

= 572 cm²


Q3) A toy is in the form of a cone mounted on a hemisphere.

Radius r = 3.5 cm

Total height = 15.5 cm

Step 1: Height of cone

Height of hemisphere = r = 3.5

Height of cone = 15.5 − 3.5

= 12 cm

Step 2: Slant height of cone

l = √(h² + r²)

= √(12² + 3.5²)

= √(144 + 12.25)

= √156.25

= 12.5 cm

Step 3: Total surface area

= CSA cone + CSA hemisphere

CSA cone = πrl

= 22/7 × 3.5 × 12.5

= 137.5 cm²

CSA hemisphere = 2πr²

= 2 × 22/7 × 3.5²

= 77 cm²

Total surface area

= 137.5 + 77

= 214.5 cm²


Q4) A cubical block of side 7 cm is surmounted by a hemisphere.

Step 1: Greatest diameter of hemisphere = side of cube

= 7 cm

Radius r = 3.5 cm

Step 2: Surface area

Surface area of cube = 6a²

= 6 × 7²

= 6 × 49

= 294 cm²

Area of circular base covered by hemisphere

= πr²

= 22/7 × 3.5²

= 38.5 cm²

Remaining cube area

= 294 − 38.5

= 255.5 cm²

CSA hemisphere = 2πr²

= 2 × 38.5

= 77 cm²

Total surface area

= 255.5 + 77

= 332.5 cm²


Q5) Hemispherical depression cut from cube.

Edge of cube = diameter of hemisphere = l

Radius r = l/2

Step 1: Surface area of cube

= 6l²

Step 2: Area removed = πr²

Step 3: Curved surface area of hemisphere added

= 2πr²

Total surface area

= 6l² − πr² + 2πr²

= 6l² + πr²


Q6) A medicine capsule consists of cylinder and two hemispheres.

Total length = 14 mm

Diameter = 5 mm

Radius r = 2.5 mm

Step 1: Length of cylindrical part

= 14 − 5

= 9 mm

Step 2: Surface area

= CSA cylinder + CSA sphere

CSA cylinder = 2πrh

= 2 × 22/7 × 2.5 × 9

= 141.43 mm²

Surface area sphere = 4πr²

= 4 × 22/7 × 2.5²

= 78.57 mm²

Total surface area

= 141.43 + 78.57

= 220 mm² (approx)


Q7) Tent shaped like cylinder with conical top.

Radius r = 2 m

Height of cylinder = 2.1 m

Slant height cone = 2.8 m

Step 1: Canvas area

= CSA cylinder + CSA cone

CSA cylinder = 2πrh

= 2 × 22/7 × 2 × 2.1

= 26.4 m²

CSA cone = πrl

= 22/7 × 2 × 2.8

= 17.6 m²

Total canvas area

= 26.4 + 17.6

= 44 m²

Cost

= 44 × 500

= ₹22000


Q8) Cylinder with conical cavity.

Height = 2.4 cm

Diameter = 1.4 cm

Radius r = 0.7 cm

Step 1: Slant height of cone

l = √(h² + r²)

= √(2.4² + 0.7²)

= √(5.76 + 0.49)

= √6.25

= 2.5 cm

Step 2: Total surface area

= CSA cylinder + base area + CSA cone

CSA cylinder = 2πrh

= 2 × 22/7 × 0.7 × 2.4

= 10.56 cm²

Base area = πr²

= 22/7 × 0.49

= 1.54 cm²

CSA cone = πrl

= 22/7 × 0.7 × 2.5

= 5.5 cm²

Total surface area

= 10.56 + 1.54 + 5.5

= 17.6 cm² ≈ 18 cm²


Q9) Wooden article formed by scooping hemisphere from both ends.

Radius r = 3.5 cm

Height of cylinder = 10 cm

Step 1: Curved surface area cylinder

= 2πrh

= 2 × 22/7 × 3.5 × 10

= 220 cm²

Step 2: Area of two hemispherical hollows

= 2 × (2πr²)

= 4πr²

= 4 × 22/7 × 3.5²

= 154 cm²

Total surface area

= 220 + 154

= 374 cm²


Final Answers Summary

  • Q1 → 160 cm²
  • Q2 → 572 cm²
  • Q3 → 214.5 cm²
  • Q4 → 332.5 cm²
  • Q6 → 220 mm²
  • Q7 → Area 44 m², Cost ₹22000
  • Q8 → 18 cm²
  • Q9 → 374 cm²
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