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Ch 12Surface Areas and Volumes

Mensuration — 10 marks

Chapter 12: Surface Areas and Volumes

This chapter deals with finding the surface areas and volumes of solids that are combinations of two basic solids — for example, a toy shaped like a cone mounted on a hemisphere, or a medicine capsule shaped like a cylinder with two hemispherical ends. In the rationalised CBSE 2026-27 syllabus, all problems are restricted to combinations of at most two different solids.

1. Basic Solids — Complete Formula Table

Solid Curved / Lateral Surface Area Total Surface Area Volume
Cube (edge a) 4a2 6a2 a3
Cuboid (l × b × h) 2h(l + b) 2(lb + bh + hl) l × b × h
Cylinder (radius r, height h) 2πrh 2πr(r + h) πr2h
Cone (radius r, height h, slant height l) πrl πr(l + r) (1/3)πr2h
Sphere (radius r) 4πr2 4πr2 (4/3)πr3
Hemisphere (radius r) 2πr2 3πr2 (2/3)πr3
Slant height of a cone: l2 = r2 + h2, so l = √(r2 + h2)
Diagonal of a cube: a√3    Diagonal of a cuboid: √(l2 + b2 + h2)
Exam Tip: Unless the question says otherwise, take π = 22/7. Questions are almost always designed so that the radius is a multiple of 7 (or 3.5, 1.4, 2.1 etc.) to make 22/7 cancel neatly. Always convert the diameter given in the question to the radius before substituting — this is the most common careless mistake.

2. Surface Area of a Combination of Solids

When two solids are joined, part of each solid gets hidden at the joint. Therefore:

TSA of combined solid = sum of the VISIBLE (exposed) surface areas of the parts
It is NOT equal to the sum of the total surface areas of the individual solids.

Method (step-by-step)

  1. Draw a rough figure and identify the two basic solids in the combination.
  2. Mark which surfaces are exposed and which are hidden at the joint (usually the flat circular faces get hidden).
  3. Write the surface area of the combination as the sum of the curved surface areas of the parts, adding any flat face that still remains exposed (e.g. the base of a tent's cylinder is on the ground, so it is not counted as canvas).
  4. Substitute values, keeping the same units throughout, and simplify with π = 22/7.

3. Volume of a Combination of Solids

Volume of combined solid = Volume of part 1 + Volume of part 2

Unlike surface area, volumes simply add up, because joining solids does not destroy any space. If a solid is scooped out or removed (e.g. a hemisphere hollowed out of a cylinder, or a cone removed from a cube), then the volume of the removed part is subtracted.

4. Standard Board-Exam Combinations

  • Toy = cone on a hemisphere (same radius r; cone height h, slant height l):
    TSA = CSA of cone + CSA of hemisphere = πrl + 2πr2
    Volume = (1/3)πr2h + (2/3)πr3
  • Capsule = cylinder with a hemisphere on each end (radius r, cylinder length h):
    Surface area = 2πrh + 2(2πr2) = 2πrh + 4πr2
    Volume = πr2h + (4/3)πr3. Note: total length of capsule = h + 2r.
  • Tent = cylinder surmounted by a cone (common radius r):
    Canvas required = CSA of cylinder + CSA of cone = 2πrH + πrl (no base, no top circle).
    Cost of canvas = area × rate per m2.
  • Ice-cream cone = cone with hemispherical top: Volume = (1/3)πr2h + (2/3)πr3.
  • Article with hemispheres scooped out (cylinder of height h with a hemisphere carved out of each flat end):
    TSA = CSA of cylinder + 2 × CSA of hemisphere = 2πrh + 4πr2
    Volume = πr2h − 2 × (2/3)πr3
  • Cubical block surmounted by a hemisphere (cube edge a, hemisphere of greatest possible diameter, so r = a/2):
    TSA = 6a2 − πr2 + 2πr2 = 6a2 + πr2
  • Two cubes joined end to end (each of edge a): the result is a cuboid 2a × a × a with surface area 2(2a·a + a·a + a·2a) = 10a2, not 12a2.
  • Glass with raised hemispherical bottom: apparent capacity = πr2h; actual capacity = πr2h − (2/3)πr3.
Exam Tip: In a "cone on hemisphere" toy, if the total height of the toy is given, first find the height of the cone: height of cone = total height − radius of hemisphere. Similarly, in a capsule, length of the cylindrical part = total length − 2r. Writing this one line earns a method mark even if the arithmetic slips.

5. Units — Quick Reference

  • Surface area is in square units (cm2, m2); volume is in cubic units (cm3, m3).
  • 1 m = 100 cm, so 1 m2 = 10000 cm2 and 1 m3 = 1000000 cm3.
  • 1 litre = 1000 cm3; 1 m3 = 1000 litres (kilolitre).
Exam Tip: Board papers usually place one 2-mark direct-formula question, one 3-mark combination question and one 4–5-mark case study (tent / toy / capsule) from this chapter. Always end word problems with a statement sentence including the correct unit — "Hence, the canvas required is 44 m2." Missing units costs marks.

6. Solved Pattern (Model)

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

Solution. r = 3.5 cm; height of cone h = 15.5 − 3.5 = 12 cm.
Slant height l = √(r2 + h2) = √(12.25 + 144) = √156.25 = 12.5 cm.
TSA of toy = CSA of cone + CSA of hemisphere = πrl + 2πr2
= (22/7)(3.5)(12.5) + 2(22/7)(3.5)2 = 137.5 + 77 = 214.5 cm2.

Deleted from the rationalised syllabus (2026-27) — do NOT study for boards:
  • Frustum of a cone — all formulas and problems on the frustum (bucket-shaped solids) are removed.
  • Conversion of one solid into another — problems on melting/recasting (e.g. a metallic sphere melted into wires or smaller spheres) are removed.
Problems in the exam are restricted to combinations of not more than two different solids.
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