1. Chapter at a Glance
Real objects are rarely a single simple solid. A capsule is a cylinder with two hemispheres; an ice-cream cone is a cone topped with a hemisphere; a tent may be a cylinder with a conical roof. In this chapter we find the surface area and volume of such combinations of two solids — cubes, cuboids, cylinders, cones, spheres and hemispheres. The key is to see which surfaces are visible (for area) and that volumes simply add (or subtract when a part is scooped out).
- Recap of surface areas and volumes of basic solids.
- Surface area of combined solids: count only the outer, exposed surfaces.
- Volume of combined solids: add volumes; subtract hollows.
- Problems on toys, capsules, tents, bird-baths, pen stands and more.
2. Recap of Basic Solids
| Solid | CSA | TSA | Volume |
|---|---|---|---|
| Cuboid l × b × h | 2h(l + b) | 2(lb + bh + hl) | lbh |
| Cube a | 4a² | 6a² | a³ |
| Cylinder r, h | 2πrh | 2πr(r + h) | πr²h |
| Cone r, h, l | πrl | πr(l + r) | ⅓πr²h |
| Sphere r | 4πr² | 4πr² | (4/3)πr³ |
| Hemisphere r | 2πr² | 3πr² | (2/3)πr³ |
Slant height l = √(r² + h²).
3. Surface Area of Combinations
Total surface area of a combined solid = sum of the curved/visible surfaces of the parts. Faces that are joined (hidden) are not counted.
A toy is a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height is 15.5 cm. Find its total surface area (π = 22/7).
Solution: Cone height = 15.5 − 3.5 = 12; l = √(12.25 + 144) = 12.5. TSA = πrl + 2πr² = (22/7)(3.5)(12.5 + 7) = 11 × 19.5 = 214.5 cm².
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 (π = 22/7).
Solution: Greatest diameter = 7 cm (r = 3.5). SA = 6a² − πr² + 2πr² = 294 + πr² = 294 + 38.5 = 332.5 cm².
Notice: placing a hemisphere on a face removes a circle πr² from the cube but adds 2πr² of curved surface — a net gain of πr².
4. Volume of Combinations
Volume of combined solid = sum of the volumes of the parts. If a part is removed (hollowed out), subtract its volume.
A capsule is a cylinder with two hemispherical ends. Its total length is 14 mm and diameter 5 mm. Find its volume and surface area (π = 22/7).
Solution: r = 2.5, cylinder height = 14 − 5 = 9. Volume = πr²h + (4/3)πr³ = (22/7)(6.25)(9) + (4/3)(22/7)(15.625) ≈ 176.79 + 65.48 = ≈ 242.26 mm³. SA = 2πrh + 4πr² = (22/7)(45 + 25) = 220 mm².
A wooden pen stand is a cuboid 15 × 10 × 3.5 cm with four conical depressions, each of radius 0.5 cm and depth 1.4 cm. Find the volume of wood (π = 22/7).
Solution: Cuboid = 525. Each cone = ⅓ × (22/7) × 0.25 × 1.4 = 0.367. Four = 1.47. Wood = 525 − 1.47 = 523.53 cm³.
5. More Combinations
A tent is a cylinder of diameter 4.2 m and height 4 m topped by a cone of the same base and height 2.8 m. Find the canvas needed (π = 22/7).
Solution: r = 2.1, l = √(4.41 + 7.84) = 3.5. Canvas = 2πrh + πrl = (22/7)(2.1)(8 + 3.5) = 6.6 × 11.5 = 75.9 m².
Each gulab jamun is a cylinder with two hemispherical ends, length 5 cm and diameter 2.8 cm. About 30% of its volume is syrup. Find the syrup in 45 gulab jamuns (π = 22/7).
Solution: r = 1.4, cylinder h = 2.2. Volume = π(1.96)(2.2) + (4/3)π(2.744) = 13.55 + 11.50 = 25.05 cm³. 45 pieces × 30% ≈ 338 cm³.
6. Solved Problems (Board Pattern)
A bird-bath is a cylinder of height 1.45 m and radius 30 cm with a hemispherical depression at the top. Find its outer surface area (π = 22/7).
Solution: SA = 2πrh + 2πr² = 2π(30)(145 + 30) = 2 × (22/7) × 30 × 175 = 33000 cm² = 3.3 m².
From a solid cylinder of height 2.4 cm and diameter 1.4 cm, a cone of the same height and diameter is hollowed out. Find the total surface area of the remaining solid (π = 22/7).
Solution: r = 0.7, l = √(0.49 + 5.76) = 2.5. TSA = 2πrh + πr² + πrl = π(0.7)(4.8 + 0.7 + 2.5) = (22/7)(0.7)(8) = 17.6 cm².
A cone of radius 3 cm and height 12 cm is filled with ice cream with a hemispherical top. Find the volume of ice cream (π = 3.14).
Solution: ⅓πr²h + ⅔πr³ = π(36 + 18) = 54π ≈ 169.56 cm³.
7. Master Formula Table
| Combination | Surface area | Volume |
|---|---|---|
| Cone on hemisphere | πrl + 2πr² | ⅓πr²h + ⅔πr³ |
| Capsule (cyl + 2 hemis) | 2πrh + 4πr² | πr²h + (4/3)πr³ |
| Tent (cyl + cone) | 2πrh + πrl | πr²h + ⅓πr²H |
| Cube + hemisphere on top | 6a² + πr² | a³ + ⅔πr³ |
| Cylinder minus cone | 2πrh + πr² + πrl | πr²h − ⅓πr²h |
| Cylinder minus 2 hemis | 2πrh + 4πr² | πr²h − (4/3)πr³ |
8. Master Summary Table
| Step | Surface area | Volume |
|---|---|---|
| Identify parts | Which surfaces are visible? | Which parts are solid/hollow? |
| Combine | Add visible curved/flat surfaces | Add solids, subtract hollows |
| Common radius | Factor out πr | Factor out πr² |
| Check | Joined faces excluded | Units cm³ / m³ / litres |
Chart — volumes (r = 3 cm): cone (h = 12), hemisphere, ice-cream total:
9. Mind Map
- CSA, TSA, V
- l = √(r² + h²)
- Visible surfaces only
- Hidden faces out
- Add parts
- Subtract hollows
- Toy, capsule, tent
- Pen stand, bird-bath
10. Tips and Tricks
- Sketch the solid and shade the outer surface before writing the formula.
- If all parts share radius r, factor out πr (area) or πr² (volume) to simplify.
- Find hidden heights: total height minus the known parts (e.g. hemisphere height = r).
- Adding TSAs of both parts (double-counts the joined faces).
- Using cone height in place of slant height for curved surface.
- Forgetting the base circle when a cylinder stands on the ground and is "total".
11. Important One-liners
- Joined faces are not part of the surface area of a combined solid.
- Volumes of parts simply add up.
- A hemisphere of radius r has height r.
- The largest hemisphere on a cube face has diameter equal to the cube's side.
- Hollowing out a cone from a cylinder adds the cone's curved surface to the area.
12. Exam-pattern Questions with Detailed Answers
A solid made of a cylinder with a hemisphere on one end has total surface area:
- 2πrh + 2πr²
- 2πrh + 3πr²
- 2πrh + πr²
- 2πr(r + h)
Show answer
Cylinder curved (2πrh) + hemisphere curved (2πr²) + bottom circle (πr²) = 2πrh + 3πr².
Answer: (B)
Two cubes of 64 cm³ volume each are joined end to end. Find the surface area of the cuboid formed.
Show answer
Side 4; cuboid 8 × 4 × 4. SA = 2(32 + 16 + 32) = 160 cm².
Answer: 160 cm²
A cone and a hemisphere have equal bases and equal heights. Find the ratio of their volumes.
Show answer
h = r for both. Cone ⅓πr³ : hemisphere ⅔πr³ = 1 : 2.
Answer: 1 : 2
A vessel is a hollow hemisphere mounted by a hollow cylinder. The diameter is 14 cm and total height 13 cm. Find the inner surface area (π = 22/7).
Show answer
r = 7, cylinder h = 6. Area = 2πrh + 2πr² = 2π(7)(6 + 7) = 2 × 22 × 13 = 572 cm².
Answer: 572 cm²
A solid is a cone of radius 1 cm and height 1 cm on a hemisphere of radius 1 cm. Find its volume in terms of π.
Show answer
⅓π + ⅔π = π cm³.
Answer: π cm³
A medicine capsule is 14 mm long with diameter 5 mm. Find its surface area (π = 22/7).
Show answer
Area = 2πr(h + 2r) = 2πr × total length = 2 × (22/7) × 2.5 × 14 = 220 mm².
Answer: 220 mm²
A cylinder of radius 7 cm and height 10 cm has a hemisphere of radius 7 cm scooped out of one end. Find the volume remaining (π = 22/7).
Show answer
1540 − ⅔ × (22/7) × 343 = 1540 − 718.67 = 821.33 cm³.
Answer: ≈ 821.33 cm³
A tent is a cylinder (radius 7 m, height 3 m) with a conical top of slant height 25 m (same radius). Find the canvas area (π = 22/7).
Show answer
2πrh + πrl = (22/7)(7)(6 + 25) = 22 × 31 = 682 m².
Answer: 682 m²
A wooden article is made by scooping out a hemisphere from each end of a solid cylinder of height 10 cm and radius 3.5 cm. Find its total surface area (π = 22/7).
Show answer
2πrh + 4πr² = 2π(3.5)(10 + 7) = 2 × 22 × 0.5 × 17 = 374 cm².
Answer: 374 cm²
A solid toy is a hemisphere of radius 3.5 cm surmounted by a cone of the same radius; the total height is 15.5 cm. Find its volume (π = 22/7).
Show answer
Cone h = 12. V = ⅓π(12.25)(12) + ⅔π(42.875) = π(49 + 28.58) = (22/7)(77.58) ≈ 243.83 cm³.
Answer: ≈ 243.83 cm³
A hemispherical depression is cut from one face of a cube of edge l, with diameter l. Find the surface area of the remaining solid.
Show answer
6l² − π(l/2)² + 2π(l/2)² = 6l² + πl²/4 = (l²/4)(24 + π).
Answer: (l²/4)(π + 24)
A building has a cylindrical base (diameter 8 m, height 6 m) with a hemispherical dome. Find the volume of air inside, and the area to be painted inside (excluding the floor) (π = 3.14).
Show answer
r = 4. Volume = π(16)(6) + ⅔π(64) = π(96 + 42.67) = 3.14 × 138.67 ≈ 435.4 m³. Area = 2πrh + 2πr² = 2π(4)(6 + 4) = 80π ≈ 251.2 m².
Answer: ≈ 435.4 m³; ≈ 251.2 m²
13. Quick Revision Summary
- Know the six basic solids' CSA, TSA and volume.
- Combined surface area: only exposed surfaces count.
- Combined volume: add parts; subtract hollows.
- Find missing heights from total heights.
- Factor out common πr or πr² to simplify.