Chapter 12 · CBSE Class 10 Maths (NCERT)

Surface Areas and Volumes

Calculate surface area and volume of combined solids, and solve problems involving conversion of one solid shape into another.

📦 Chapter 12

12.1

Surface Area of Combined Solids

When solids are joined (like a cone on a cylinder), the total surface area excludes the internal shared face, using only exposed surfaces.

1

A cylinder (radius 3.5cm, height 10cm) has a hemisphere (same radius) on top. Find the total surface area (π=22/7).

Cylinder CSA=2×22/7×3.5×10=220; base=22/7×3.5²=38.5; hemisphere CSA=2×22/7×3.5²=77; total=220+38.5+77=335.5 cm²

2

Why is the base of the hemisphere not included in the total surface area calculation above?

Because it's internal — joined directly to the cylinder's top, so it's not an exposed surface

12.2

Converting One Solid into Another

When a solid is melted and recast into another shape, the volume remains constant even though the surface area changes.

1

A metallic sphere of radius 6cm is melted and recast into a cylinder of radius 4cm. Find the height.

Sphere volume=4/3π×216=288π; cylinder volume=π×16×h; 16h=288; h=18cm

2

Why does volume (not surface area) stay constant when melting and recasting a solid?

Because the total amount of material is conserved, and volume measures the material's quantity, while surface area depends on the new shape's geometry

Practice

Exercises

1. A cylinder (radius 7cm, height 10cm) has a cone (same radius, height 5cm) on top. Find the total volume (π=22/7).

2. A cube of side 6cm is melted and recast into a cuboid with base 4cm×3cm. Find the height.

3. A metallic sphere of radius 9cm is melted into a cylinder of radius 6cm. Find the height.

4. A solid consists of a cone (radius 3cm, height 4cm) on a hemisphere (radius 3cm). Find the total height of the solid.

5. Find the volume of a hemisphere with radius 6cm (π=3.14).

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