The total surface area of a solid hemisphere of radius \(r\) is
- (a)\(2\pi r^2\)
- (b)\(4\pi r^2\)
- (c)\(3\pi r^2\)
- (d)\(\pi r^2\)
Surface Areas and Volumes Class 10: notes and important questions
Revision notes, 37 board-style questions with the step marks shown, and a four-step route from the basics to 95+, each step ending in a short checkpoint.
- 37 questions
- 10 multiple choice, 3 assertion–reason, 8 very short answer, 8 short answer, 5 long answer, 3 case study
- About 11 hours to master
Mensuration unit: 10 of 80 theory marks (Areas Related to Circles and Surface Areas and Volumes).
Revision notes
Formulae (radius \(r\), height \(h\), slant height \(l\))
| Solid | Curved/lateral SA | Total SA | Volume |
|---|---|---|---|
| Cuboid \(l\times b\times h\) | \(2h(l+b)\) | \(2(lb+bh+hl)\) | \(lbh\) |
| Cube (edge \(a\)) | \(4a^2\) | \(6a^2\) | \(a^3\) |
| Cylinder | \(2\pi rh\) | \(2\pi r(r+h)\) | \(\pi r^2h\) |
| Cone (\(l=\sqrt{r^2+h^2}\)) | \(\pi rl\) | \(\pi r(l+r)\) | \(\frac13\pi r^2h\) |
| Sphere | \(4\pi r^2\) | \(4\pi r^2\) | \(\frac43\pi r^3\) |
| Hemisphere | \(2\pi r^2\) | \(3\pi r^2\) | \(\frac23\pi r^3\) |
Combinations of solids
- Volume of a combined solid \(=\) sum of the volumes of the parts. If a part is scooped out, subtract its volume.
- Surface area is not simply the sum of the parts' surface areas. Add only the surfaces that are actually exposed. Where two solids are joined, the common face disappears from both.
- Scooping a hemisphere (or cone) out of a face: remove the circular base area \(\pi r^2\) from the face, and add the curved surface of the cavity.
- Placing a hemisphere on a face: face area loses \(\pi r^2\), and \(2\pi r^2\) is gained. Net change \(=+\pi r^2\).
Worked example 1
A toy is a cone of radius 4.2 cm on a hemisphere of the same radius. The total height is 9.8 cm. Find its total surface area \(\left(\pi=\frac{22}{7}\right)\).
Cone height \(=9.8-4.2=5.6\), \(l=\sqrt{5.6^2+4.2^2}=7\). TSA \(=\pi rl+2\pi r^2=92.4+110.88=203.28\ \text{cm}^2\).
Worked example 2
A solid is a cylinder of radius 7 cm and height 10 cm with a hemisphere on top. Find its volume.
\(\pi r^2h+\frac23\pi r^3=1540+718.67=2258.67\ \text{cm}^3\).
Worked example 3
Two cubes of edge 6 cm are joined face to face. Find the surface area of the cuboid formed.
Cuboid \(12\times6\times6\): \(2(72+36+72)=360\ \text{cm}^2\). Check: \(2\times6\times36-2\times36=360\).
Common errors
- Adding the total surface areas of the parts and so counting the hidden joined faces.
- Using the vertical height \(h\) in \(\pi rl\) instead of the slant height.
- Using the diameter as the radius, or mixing units (cm and m, mm and cm).
- For a vessel, "inner surface area" excludes the open top.
Board-exam tips
- Syllabus limit: combinations of any two of cubes, cuboids, spheres, hemispheres, right circular cylinders and cones.
- List which surfaces are exposed before calculating. This is where most marks are lost.
- Take out \(\pi\) as a common factor to simplify arithmetic: e.g. \(\pi r(2h+l)\).
- \(1\ \text{m}^3=1000\) litres and \(1000\ \text{cm}^3=1\) litre.
Topics in this chapter: Volume of basic solids · Surface area of basic solids · Surface area of combined solids · Volume of combined solids.
Route to 95: four steps
Work through the steps in order. Take each checkpoint closed book, about 1.5 minutes per mark; pass at 80% to move on. Two misses in a row means going back one step.
Step 1
Secure the basics
You can use the surface area and volume formulae of the cube, cuboid, cylinder, cone, sphere and hemisphere.
Step 2
Board standard
You can find the surface area and volume of solids made from two shapes, deciding which faces are exposed.
Step 3
Full marks on long answers
You can write complete 5-mark answers on combined solids and real objects (tents, toys, vessels) with units and conversions.
Step 4
95+ stretch (HOTS)
You can handle drilled holes, hollow objects and the 'largest cone in a cube' questions that decide 95+.
Practice questions
Original questions in the board's styles. Multiple-choice answers are checked as you go; for written answers, compare your working with the step mark scheme and record your marks. 9 of the 37 are competency-based (case studies and questions set in a real-life situation): the "Competency-based" button shows just those.
The slant height of a cone with base radius 5 cm and height 12 cm is
- (a)13 cm
- (b)17 cm
- (c)\(\sqrt{119}\) cm
- (d)7 cm
The volume of a hemisphere of radius 21 cm is \(\left(\pi=\frac{22}{7}\right)\)
- (a)\(38808\ \text{cm}^3\)
- (b)\(19404\ \text{cm}^3\)
- (c)\(2772\ \text{cm}^3\)
- (d)\(9702\ \text{cm}^3\)
Where marks are lost in Surface Areas and Volumes
- Adding total surface areas of the parts and counting hidden joined faces. Fix: list exposed surfaces in words (curved surface of cone + curved surface of hemisphere) before any numbers.
- Using the vertical height h in πrl. Fix: find l = √(r² + h²) as a separate line.
- Mixing units (cm with m, cm³ with litres). Fix: convert at the start; 1000 cm³ = 1 litre and 1 m³ = 1000 litres.
- Using the diameter as the radius. Fix: write r = d/2 on the first line.
- Inner surface area of a vessel including the open top. Fix: read 'open', 'hollow', 'inner' carefully and drop that face.
- Long arithmetic slips with π. Fix: factor out π (e.g. πr(2h + l)) and simplify before multiplying.
Surface Areas and Volumes in our sample papers
Once step 3 is passed, test the chapter inside a full timed paper on the 2026-27 pattern (original papers by us, not official CBSE papers).
- Sample paper 1 (Board standard): questions 14, 31
- Sample paper 2 (Board standard): questions 15, 16, 34
- Sample paper 3 (Above board): questions 15, 20, 35
- Sample paper 4 (Above board): questions 15, 16, 25, 36
- Sample paper 5 (95+ challenge): questions 15, 16, 34