Class 10 Maths · Original EZT practice
Surface Areas and Volumes (Class 10 Chapter 12): summary and practice
Surface Areas and Volumes is about solids built from simpler solids: a cylinder with a hemisphere on top, a cone on a cylinder, a cuboid with a hole. Volumes just add, but surface areas need care - the faces where two solids join are hidden and must not be counted. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 22 October 2026:
1. The basic formulas
Cuboid: 2(lb + bh + hl) and lbh. Cube: 6a^2 and a^3. Cylinder: CSA 2 pi r h, total 2 pi r (r + h), volume pi r^2 h. Cone: CSA pi r l with l^2 = r^2 + h^2, volume (1/3) pi r^2 h. Sphere: 4 pi r^2 and (4/3) pi r^3. Hemisphere: CSA 2 pi r^2, total 3 pi r^2, volume (2/3) pi r^3.
2. Surface area of a combination of solids
Add the exposed surfaces only. Where two solids join, the shared faces disappear from the outside - a hemisphere on a cylinder contributes its curved 2 pi r^2, not 3 pi r^2, because its flat face is glued down.
3. Volume of a combination of solids
Volume is the easy one: just add the volumes of the pieces. Joining hides surfaces but never hides space.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Volume of a cylinder
Choose the method: V = pi r^2 h - base area times height.
EZT practice question
A cylinder has radius 3 cm and height 7 cm. Find its volume.
Show the worked solution
- r^2 = 9.
- V = pi x 9 x 7.
- Simplify: 63 pi.
Answer: 63 pi cubic cm, about 197.9 cubic cm.
Check: Base area 9 pi x height 7 = 63 pi - the formula is exactly base times height, so the check is the method itself.
Watch for this: Do not use 2 pi r h - that is curved surface area, not volume.
Now try it yourself
Volume of a cylinder with r = 2 cm, h = 5 cm?
Check your answer
20 pi cubic cm.
pi x 4 x 5.
Volume and slant height of a cone
Choose the method: V = (1/3) pi r^2 h; l^2 = r^2 + h^2.
EZT practice question
A cone has radius 3 cm and height 4 cm. Find its slant height and volume.
Show the worked solution
- l^2 = 9 + 16 = 25, so l = 5 cm.
- V = (1/3) pi x 9 x 4.
- Simplify: 12 pi.
Answer: l = 5 cm; V = 12 pi cubic cm, about 37.7 cubic cm.
Check: A cylinder on the same base and height holds 36 pi - the cone is exactly a third of it, which is the definition of the cone formula.
Watch for this: Do not drop the 1/3 - a cone is not a cylinder.
Now try it yourself
Cone with r = 6 cm, h = 8 cm - slant height?
Check your answer
10 cm.
6-8-10 triangle.
Volume of a combination of solids
Choose the method: Add the volumes of the pieces - nothing is hidden.
EZT practice question
A toy is a cylinder of radius 3 cm and height 4 cm with a cone of the same base radius and height 3 cm on top. Find its volume.
Show the worked solution
- Cylinder: pi x 9 x 4 = 36 pi.
- Cone: (1/3) pi x 9 x 3 = 9 pi.
- Add: 36 pi + 9 pi.
Answer: 45 pi cubic cm, about 141.4 cubic cm.
Check: Volumes always add - the join hides surfaces, never space, so no subtraction appears in volume questions.
Watch for this: Do not subtract the shared circle - volume has no shared SURFACE problem; both solids keep their full volume.
Now try it yourself
A hemisphere of radius 3 cm on a cube of side 6 cm - total volume?
Check your answer
216 + 18 pi cubic cm.
Cube 6^3 plus hemisphere (2/3) pi x 27.
Surface area of a combination: the join is hidden
Choose the method: Add exposed surfaces only; the glued faces count zero times.
EZT practice question
A hemisphere of radius 3 cm sits on top of a cylinder of radius 3 cm and height 4 cm. Find the total surface area.
Show the worked solution
- Hemisphere contributes only its curved part: 2 pi r^2 = 18 pi (its flat circle is glued to the cylinder).
- Cylinder contributes CSA 2 pi r h = 24 pi, plus its bottom circle pi r^2 = 9 pi (the top circle is glued).
- Add: 18 pi + 24 pi + 9 pi.
Answer: 51 pi sq cm, about 160.2 sq cm.
Check: Count the hidden faces: the hemisphere's flat face and the cylinder's top both vanish - two faces removed, and 2 pi r^2 + 24 pi + 9 pi uses neither.
Watch for this: Do not use the hemisphere's total 3 pi r^2 - its flat face is not on the outside of the toy.
Now try it yourself
A cone (r = 3, l = 5) on a cylinder (r = 3, h = 4) - total surface area?
Check your answer
15 pi + 24 pi + 9 pi = 48 pi sq cm.
Cone CSA + cylinder CSA + bottom circle.
Volume and surface of a sphere
Choose the method: V = (4/3) pi r^3; SA = 4 pi r^2.
EZT practice question
Find the volume and surface area of a sphere of radius 3 cm.
Show the worked solution
- r^3 = 27.
- V = (4/3) pi x 27 = 36 pi.
- SA = 4 pi x 9 = 36 pi - the numbers coincide here, the units do not.
Answer: V = 36 pi cubic cm, SA = 36 pi sq cm.
Check: At r = 3 the numerical coincidence (4/3 r^3 = 4 r^2) happens only because r = 3 - a useful memory hook, not a rule.
Watch for this: Do not mix the units - volume is cubic, surface is square, even when the numbers match.
Now try it yourself
Volume of a sphere of radius 6 cm?
Check your answer
288 pi cubic cm.
(4/3) pi x 216.
Hemisphere: curved versus total surface
Choose the method: Curved = 2 pi r^2; total = 3 pi r^2 (curved plus the flat circle).
EZT practice question
A hemispherical bowl has radius 6 cm. Find its curved surface area and total surface area.
Show the worked solution
- Curved: 2 pi x 36 = 72 pi sq cm.
- Flat circle: pi x 36 = 36 pi sq cm.
- Total: 72 pi + 36 pi.
Answer: CSA = 72 pi, total = 108 pi sq cm.
Check: Total is exactly 1.5 times the curved area - the flat circle adds half of 2 pi r^2.
Watch for this: Which one does the question want? A bowl's OUTER surface usually means curved only; painting the bowl including its rim face means total.
Now try it yourself
Total surface area of a hemisphere of radius 7 cm?
Check your answer
147 pi sq cm.
3 pi x 49.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Cylinder volume?
Show answer and reason
pi r^2 h.
Base area times height.
2. Cone volume?
Show answer and reason
(1/3) pi r^2 h.
A third of the matching cylinder.
3. Hemisphere total surface?
Show answer and reason
3 pi r^2.
2 pi r^2 curved plus pi r^2 flat.
4. In a combination solid, what happens at the join?
Show answer and reason
The shared faces vanish from the surface area; volumes just add.
Surfaces get hidden, space does not.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- Volume of a cylinder with r = 2 cm, h = 5 cm.
- Slant height of a cone with r = 6 cm, h = 8 cm.
- Total surface area of a hemisphere of radius 7 cm.
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Questions about this practice page
- Is the frustum of a cone in the current Class 10 book?
- No. The current NCERT Class 10 Maths book (Reprint 2026-27) covers only surface area of a combination of solids and volume of a combination of solids - the frustum section from older editions is not in the current book.
- Is melting and recasting (converting one solid into another) in the current book?
- No. That section too was in older editions; the current chapter has no solid-conversion section - only combinations of solids, their surface areas and their volumes.
- How do I find the surface area of a combination of solids?
- Add only the exposed surfaces. Where two solids join, both glued faces disappear: a hemisphere on a cylinder contributes its curved 2 pi r^2, not its total 3 pi r^2, and the cylinder loses its top circle.
- Does volume work the same way for combinations?
- Simpler, actually - volumes always just add. Joining hides surfaces but never space, so a cylinder-plus-cone toy has volume pi r^2 h + (1/3) pi r^2 h with nothing subtracted.
- Are these the NCERT exercise questions for Surface Areas and Volumes (Class 10)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.