CBSE Class 10 · Boards 2027 · NCERT 2026-27 syllabus
NCERT Solutions Class 10 Maths Chapter 12 Surface Areas and Volumes
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Boards begin February 2027
Pre-boards are in December
Surface Areas and Volumes is in Mensuration, which carries 10 of the 80 marks in the 2026-27 CBSE Maths paper
8 NCERT textbook questions from Chapter 12
Exercise 12.1
NCERT Ex 12.1 Q11 mark
Two cubes each of volume 64 cm³ are joined end to end. The surface area of the resulting cuboid is
Step 1Side of each cube = ∛64 = 4 cm.
Step-by-step solution
Side of each cube = ∛64 = 4 cm.
When joined end to end, the resulting cuboid has dimensions l × b × h = 8 × 4 × 4 cm.
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. (Take π = 22/7.)
Step 1Greatest possible diameter = side of the cube = 7 cm, so radius r = 3.5 cm.
Step-by-step solution
Greatest possible diameter = side of the cube = 7 cm, so radius r = 3.5 cm.
SA of solid = TSA of cube − base area of hemisphere + CSA of hemisphere = 6a² − πr² + 2πr² = 6a² + πr².
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area. (Take π = 22/7.)
Step 1Radius r = 5/2 = 2.5 mm. Cylindrical length = 14 − 2(2.5) = 9 mm.
Step-by-step solution
Radius r = 5/2 = 2.5 mm. Cylindrical length = 14 − 2(2.5) = 9 mm.
SA = CSA of cylinder + CSA of two hemispheres = 2πrh + 4πr² = 2πr(h + 2r).
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel. (Take π = 22/7.)
Step 1Radius r = 14/2 = 7 cm. Height of cylindrical part h = 13 − 7 = 6 cm.
Step-by-step solution
Radius r = 14/2 = 7 cm. Height of cylindrical part h = 13 − 7 = 6 cm.
Inner SA = CSA of hemisphere + CSA of cylinder = 2πr² + 2πrh = 2πr(r + h).
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 the total surface area of the toy. (Take π = 22/7.)
Step 1Radius r = 3.5 cm. Height of cone h = 15.5 − 3.5 = 12 cm.
Step-by-step solution
Radius 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 = CSA of cone + CSA of hemisphere = πrl + 2πr² = πr(l + 2r).
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model. (Take π = 22/7.)
Step 1Radius r = 3/2 = 1.5 cm. Length of cylinder h = 12 − 2(2) = 8 cm. Cone height hc = 2 cm.
Step-by-step solution
Radius r = 3/2 = 1.5 cm. Length of cylinder h = 12 − 2(2) = 8 cm. Cone height hc = 2 cm.
A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends, with length 5 cm and diameter 2.8 cm. (Take π = 22/7.)
Step 1Radius r = 2.8/2 = 1.4 cm. Cylinder length = 5 − 2(1.4) = 2.2 cm.
Step-by-step solution
Radius r = 2.8/2 = 1.4 cm. Cylinder length = 5 − 2(1.4) = 2.2 cm.
Volume of one gulab jamun = πr²h + (4/3)πr³ = πr²[h + (4/3)r] = (22/7)(1.4)²[2.2 + (4/3)(1.4)] cm³.
Rasheed got a playing top (lattu) which is shaped like a cone surmounted by a hemisphere. The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he has to colour. (Take π = 22/7.)
Step 1Radius r = 3.5/2 = 1.75 cm. Height of cone h = 5 − 1.75 = 3.25 cm.
Step-by-step solution
Radius r = 3.5/2 = 1.75 cm. Height of cone h = 5 − 1.75 = 3.25 cm.
Slant height l = √(r² + h²) = √(1.75² + 3.25²) = √(3.0625 + 10.5625) = √13.625 ≈ 3.7 cm.
Area to colour = CSA of hemisphere + CSA of cone = 2πr² + πrl = πr(2r + l).
CBSE board pattern: Assertion-Reason and case-based questions
Written in the CBSE board format. These are not numbered NCERT/Exemplar questions; some case studies are built on textbook problems. Neither NCERT book has Assertion-Reason questions: they, and case-based questions, are a CBSE board-paper format, here on this chapter's 2026-27 syllabus, with solutions by LazyTopper.
CBSE board pattern · Case-based4 marks
A tent is in the shape of a cylinder surmounted by a conical top. The height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the conical top is 2.8 m. (i) Find the area of the canvas required for making the tent (the base of the tent is NOT covered with canvas). (ii) Find the cost of the canvas at ₹500 per m². (Take π = 22/7.)
Step 1Radius r = 4/2 = 2 m. Cylinder height h = 2.1 m. Cone slant l = 2.8 m.
Step-by-step solution
Radius r = 4/2 = 2 m. Cylinder height h = 2.1 m. Cone slant l = 2.8 m.
Canvas area = CSA of cylinder + CSA of cone = 2πrh + πrl = πr(2h + l).
= (22/7)(2)(2 × 2.1 + 2.8) m² = (22/7)(2)(4.2 + 2.8) m² = (22/7)(2)(7) m² = 44 m².
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Questions from the NCERT Class 10 Mathematics textbook (the CBSE board-pattern section is in the board format, not numbered book questions); the solutions are written by LazyTopper. LazyTopper is an independent platform and is not affiliated with CBSE or NCERT.