NCERT Solutions For Class 10 Maths Chapter 12 Surface Areas and Volumes, Download PDF

NCERT Solutions Class 10 Chapter 12 Maths: Surface Areas and Volumes is the second chapter of Unit-6 Mensuration of the NCERT Class 10 Maths textbook. This unit’s weightage is 10. Thus, the surface area and volume chapter will be around 5 marks. The CBSE experts design questions as per NCERT language and its exercise questions, and thus solving them is a must. Get here NCERT Solutions for Chapter 12 Surface Areas and Volumes for Exercises 12.1 and 12.2. Download PDFs for a better understanding.

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• CBSE Class 10 Maths CBQs Chapter-Wise PDFs
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NCERT Solutions 10th Maths Chapter 12 Surface Areas And Volumes

Exercise 12.1

Q1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid. 

Answer: 

• Volume of each cube = 64 cm³
• Volume of a cube = side³, so side = ∛64 = 4 cm
• When the two cubes are joined end to end, the dimensions of the resulting cuboid will be:
  • Length = 4 cm + 4 cm = 8 cm
  • Width = 4 cm
  • Height = 4 cm

The surface area of a cuboid is given by:
Surface area = 2(lb + bh + hl)

Substitute the values:
Surface area = 2(8 × 4 + 4 × 4 + 4 × 8)
= 2(32 + 16 + 32)
= 2 × 80
= 160 cm²

Q2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

Answer: 

• Radius of the hemisphere = 14 cm ÷ 2 = 7 cm
• Height of the cylinder = 13 cm - 7 cm = 6 cm

The surface area of the hemisphere (inner) = 2πr²
The surface area of the cylinder (inner) = 2πrh

Total inner surface area of the vessel = surface area of hemisphere + surface area of cylinder
= 2πr² + 2πrh
= 2π × 7² + 2π × 7 × 6
= 2π × 49 + 2π × 42
= 2π(49 + 42)
= 2π × 91
= 182π

Using π = 22/7:
Total inner surface area = 182 × 22 ÷ 7
= 572 cm²

Q3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy. 

Answer: 

• Radius of the hemisphere = 3.5 cm
• Total height of the toy = 15.5 cm
• Height of the cone = 15.5 cm - 3.5 cm = 12 cm

The total surface area of the toy = curved surface area of the cone + curved surface area of the hemisphere

Curved surface area of cone = πrl
where l is the slant height = √(r² + h²)
l = √(3.5² + 12²) = √(12.25 + 144) = √156.25 = 12.5 cm

Curved surface area of cone = π × 3.5 × 12.5
= 22/7 × 3.5 × 12.5
= 137.5 cm²

Curved surface area of the hemisphere = 2πr²
= 2π × 3.5²
= 2π × 12.25
= 2 × 22/7 × 12.25
= 77 cm²

Total surface area = 137.5 + 77 = 214.5 cm²

Check: NCERT Class 10 Exemplers

Q4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid. 

Answer: 

• Side of the cubical block = 7 cm
• The greatest diameter the hemisphere can have is equal to the side of the cube = 7 cm, so the radius of the hemisphere = 7 cm ÷ 2 = 3.5 cm

The total surface area of the solid = surface area of the cube + curved surface area of the hemisphere - area of the base of the hemisphere

Surface area of the cube = 6a²
= 6 × 7²
= 6 × 49
= 294 cm²

Curved surface area of the hemisphere = 2πr²
= 2 × 22/7 × 3.5²
= 77 cm²

Area of the base of the hemisphere = πr²
= 22/7 × 3.5²
= 38.5 cm²

Total surface area = 294 + 77 - 38.5
= 332.5 cm²

Q5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Answer: 

• Let the edge of the cube = l
• The diameter of the hemisphere = l, so the radius of the hemisphere = l ÷ 2

The surface area of the remaining solid = surface area of the cube + curved surface area of the hemisphere - area of the base of the hemisphere

Surface area of the cube = 6l²
Curved surface area of the hemisphere = 2πr² = 2π(l ÷ 2)² = 2πl² ÷ 4 = πl² ÷ 2
Area of the base of the hemisphere = πr² = π(l ÷ 2)² = πl² ÷ 4

Total surface area = 6l² + (πl² ÷ 2) - (πl² ÷ 4)
= 6l² + πl² ÷ 4
= l²(6 + π ÷ 4)

Substituting π = 22/7:
Total surface area = l²(6 + 22 ÷ 28)
= l²(6 + 11 ÷ 14)
= l²(95 ÷ 14)

For l = 7 cm:
Total surface area = 7² × 95 ÷ 14
= 49 × 95 ÷ 14
= 332.5 cm²

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Check: NCERT Class 10 Maths Lab Manual PDF

Exercise 12.2

Q1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π. 

Answer: 

• Radius of the hemisphere = Radius of the cone = 1 cm
• Height of the cone = 1 cm (since height = radius)

Volume of the solid = Volume of the hemisphere + Volume of the cone

1. Volume of the hemisphere
   The volume of a hemisphere is given by:

   Volume of hemisphere=2/3πr³

   Substituting r=1

   Volume of hemisphere=2/3π(1)³=2/3π

2. Volume of the cone
   The volume of a cone is given by:

   Volume of cone=1/3πr²h

   Since r=1 and h=1:

   Volume of cone=1/3π(1)²(1)=1/3π

Total volume of the solid

Total volume=2/3π+1/3π=π

Answer: The volume of the solid is π cm³

Q2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

Answer: 

• Diameter of the model = 3 cm, so the radius = 3 cm ÷ 2 = 1.5 cm
• Total length of the model = 12 cm
• Height of each cone = 2 cm
• Height of the cylinder = Total length - Height of both cones = 12 cm - 2 × 2 cm = 8 cm

Volume of the model = Volume of the cylinder + 2 × Volume of a cone

1. Volume of the cylinder
   The volume of a cylinder is given by:

   Volume of cylinder=πr²h

   Substituting r=1.5 cm and h=8 cm

   Volume of cylinder=π(1.5)²(8)=π(2.25)(8)=18π cm³

2. Volume of one cone
   The volume of a cone is given by:

   Volume of cone=1/3πr²h

   Substituting r=1.5 cm and h=2 cm

   Volume of cone=1/3π(1.5)²(2)=1/3π(2.25)(2)=4.5π/3=1.5π cm³

Total volume of the model

Total volume=18π+2×1.5π=18π+3π=21π cm³

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Also Check:

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• NCERT Class 10 Maths Solutions PDF
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• CBSE Class 10 Sample Paper 2024-25