Three guided lessons with explanations, visual steps, practice and a final check.
Original learning companion, not full textbook exercise solutions. Covers selected combinations of solids and exposed surfaces. The final remoulding lesson is labelled an extension; it is not a claim about the current prescribed exercise list. Hollow-object and every textbook configuration coverage remain pending.
Bottom area = πr² = 9π. Total exposed area = 30π + 18π + 9π = 57π cm².
Worked example: For a cone attached to a hemisphere with shared radius 4 cm and cone slant height 5 cm, the join is hidden. Exposed area = πrl + 2πr² = 20π + 32π = 52π cm².
Watch for this: Adding the total surface areas of the separate pieces counts the joined circles twice. Decide what is exposed first.
Discuss: Sketch a cylinder with both ends capped by hemispheres. List the surfaces that remain exposed and derive an area expression.
LESSON 2
Add volumes of solid parts
Use each part’s formula and one consistent unit.
Volume measures occupied three-dimensional space and uses cubic units.
Partition the object
A solid is a cylinder of radius 3 cm and height 6 cm joined to a hemisphere of radius 3 cm.
Compute the cylinder volume
V = πr²h = π × 9 × 6 = 54π cm³.
Compute the hemisphere volume
V = (2/3)πr³ = (2/3)π × 27 = 18π cm³.
Add and label cubic units
The interiors do not overlap, so total volume = 54π + 18π = 72π cm³. No face-area subtraction is needed for this zero-thickness join.
Worked example: A cylinder of radius 2 cm and height 9 cm has a cone of the same radius and height 3 cm attached. Total volume = 36π + (1/3)π×4×3 = 40π cm³.
Watch for this: Surface-area rules and volume rules differ. A joined circular face has area but no thickness, so it does not remove volume.
Discuss: Draw a composite object, label radius versus diameter and convert every dimension to centimetres before calculating volume.
LESSON 3
Use volume conservation in an extension
Distinguish conserved volume from surface area during remoulding.
Changing a shape without losing material preserves its volume, not generally its surface area.
State the modelling assumption
A sphere of radius 6 cm is remoulded into spheres of radius 2 cm without material loss. This is an extension application.
Write both volumes
Large sphere volume = (4/3)π×6³ = 288π. One small sphere volume = (4/3)π×2³ = 32π/3.
Take the volume ratio
Number = 288π ÷ (32π/3) = 27, equivalently (6/2)³.
Avoid an area assumption
The small spheres have more total exposed surface. Volume conservation does not mean equal surface areas. Real manufacturing losses would change the model.
Worked example: A cylinder of radius 3 cm and height 8 cm contains volume 72π cm³. If remoulded without loss into cones each with radius 3 cm and height 4 cm, each cone has 12π cm³, so 6 cones are made.
Watch for this: Dividing radii gives a length ratio, not a number of equal-volume pieces. Cube the radius ratio for similar spheres.
Discuss: Compare total area before and after splitting one radius-4 sphere into radius-2 spheres. Explain why the area grows although volume stays equal.
Original learning companion, not full textbook exercise solutions. Covers selected combinations of solids and exposed surfaces. The final remoulding lesson is labelled an extension; it is not a claim about the current prescribed exercise list. Hollow-object and every textbook configuration coverage remain pending.
Original HelioLearn teaching and questions. Source-aligned companion; not an official NCERT/CBSE publication or full textbook solution bank. Independent teacher review and narrated video are pending.
