Surface Areas and Volumes (NCERT Ch. 12)

• Find the surface area and volume of combinations of solids (e.g., a cone on top of a cylinder, a hemisphere on a cube).
• Convert one solid shape into another of equal volume (e.g., melting a sphere to form a cylinder).
• Find the volume of a frustum of a cone.

1. When solids are combined, add/subtract the relevant curved/flat surface areas carefully — don't double count hidden joining faces.
2. Conservation of volume: when a solid is melted and recast into another shape, the volume stays the same (used to find new dimensions).
3. Frustum: the part of a cone left after cutting off the top with a plane parallel to the base; volume = (πh/3)(r₁² + r₂² + r₁r₂).
4. Standard formulas: Sphere volume = (4/3)πr³; Cone volume = (1/3)πr²h; Cylinder volume = πr²h.

A solid is a cylinder of radius 3 cm and height 8 cm with a cone of the same radius and height 4 cm on top. Find the total volume.

Cylinder volume = πr²h = π(9)(8) = 72π
Cone volume = (1/3)πr²h = (1/3)π(9)(4) = 12π
Total volume = 72π + 12π = 84π cubic cm

• Adding the base area of a solid that sits on top of another (the joining face should not be counted as exposed surface area).
• Forgetting to use the SAME radius when combining a cone and cylinder or hemisphere and cylinder.
• Confusing surface area conservation with volume conservation when melting/recasting (only volume is conserved, not surface area).

• A cylinder and a cone share the same base radius and are joined at their circular faces. Which face's area should NOT be included in the total exposed surface area?
• If a sphere is melted and recast into a cylinder, which quantity remains the same?

Open the Practice tab for graded questions on Surface Areas and Volumes (NCERT Ch. 12).