Surface Area and Volume teaches you to measure solid shapes: the cuboid and cube, the right circular cylinder, the right circular cone, the pyramid, the sphere and the hemisphere. For each one you find the area of the surface that covers it and the space it occupies, using formulas that come from nets and from comparisons such as three cones filling a cylinder. It also covers changing dimensions, recasting and combined solids, and it prepares the ground for the Class 10 board chapter on surface areas and volumes.
Key Concepts
1. Solids, Surface Area and Volume
A plane figure such as a rectangle lies flat and has only length and breadth. A solid has three dimensions: length, breadth and height. A brick, a tin can, a football and a tent are solids. Two different measurements describe any solid.
- Surface area: the amount of material needed to cover the outside. This is the total area of all its flat faces and curved surfaces. It is measured in square units: cm2, m2.
- Volume: the amount of space the solid occupies. This is measured in cubic units: cm3, m3. The volume a hollow container can hold is called its capacity, often given in litres.
Painting, wrapping and tin-plating are surface area problems. Filling a tank, digging a pit or melting metal are volume problems. Decide first which of the two a question asks for.
Nets: Seeing a Solid Flat
If you cut a cardboard box along some edges and open it out flat, you get its net: six rectangles joined together. The area of the net is the surface area of the box. A cylinder opens into two circles and a rectangle; a cone opens into a circle and a sector; a square pyramid opens into a square with four triangles. Each surface area formula below comes from adding the areas of the pieces of a net.
Units You Must Convert
| Quantity | Conversion | Reason |
|---|---|---|
| Length | 1 m = 100 cm | Basic |
| Area | 1 m2 = 100 × 100 = 10000 cm2 | Two lengths multiplied |
| Volume | 1 m3 = 100 × 100 × 100 = 1000000 cm3 | Three lengths multiplied |
| Capacity | 1 litre = 1000 cm3; 1 m3 = 1000 litres | Definition of the litre |
| Capacity | 1 cm3 = 1 millilitre (mL) | 1000 mL = 1 L |
Unless a question says otherwise, take π = 22/7. Choose it early: a radius of 7, 14, 3.5 or 10.5 is a signal that 22/7 will cancel neatly.
2. Cuboid and Cube
A cuboid is a solid bounded by six rectangular faces. A brick, a matchbox, a room and a book are cuboids. It has 6 faces, 12 edges and 8 vertices, and it is described by its length l, breadth b and height h.
Opposite faces of a cuboid are equal, so the six faces come in three pairs:
- top and bottom, each of area lb
- front and back, each of area lh
- two sides, each of area bh
Total surface area (TSA) = 2lb + 2bh + 2hl = 2(lb + bh + hl)
The four side faces (front, back and two sides) form the lateral surface area (LSA), the area of the four walls of a room. Opening them out gives one long rectangle whose length is the perimeter of the base, 2(l + b), and whose height is h.
LSA = 2h(l + b) = perimeter of base × height
The volume counts how many unit cubes fill the cuboid. A layer on the base holds l × b unit cubes, and there are h layers.
Volume = l × b × h = area of base × height
The longest rod that fits inside a cuboid runs from one corner to the opposite corner. This diagonal is √(l2 + b2 + h2), from the Baudhāyana-Pythagoras theorem used twice: once on the base, then on the upright right triangle with height h.
The Cube as a Special Case of the Cuboid
A cube is a cuboid whose length, breadth and height are all equal. Call the common edge a. Putting l = b = h = a into each cuboid formula gives the cube formulas directly, so a cube needs no new formulas at all:
| Measure | Cuboid | Put l = b = h = a |
|---|---|---|
| LSA | 2h(l + b) | 2a(2a) = 4a2 |
| TSA | 2(lb + bh + hl) | 2(3a2) = 6a2 |
| Volume | lbh | a3 |
| Diagonal | √(l2 + b2 + h2) | √(3a2) = a√3 |
This is what “special case” means: a general result (the cuboid) becomes a particular result (the cube) when an extra condition is added. Every cube is a cuboid; a cuboid is a cube only when all its edges are equal.
Illustration: A room is 5 m long, 4 m wide and 3 m high. The area of its four walls is 2 × 3 × (5 + 4) = 54 m2. Walls plus ceiling: 54 + (5 × 4) = 74 m2. The volume of air inside is 5 × 4 × 3 = 60 m3.
3. Right Circular Cylinder
A tin can, a water pipe, a pillar and a road roller are cylinders. A right circular cylinder has two equal, parallel circular ends (bases), and the line joining their centres (the axis) is perpendicular to the bases. It is fully described by two measurements: the radius r of the base and the height h, the distance between the two bases. In this chapter “cylinder” always means a right circular cylinder.
Curved Surface Area from the Net
Wrap a sheet of paper exactly once around a can, with no overlap, and cut it to the height of the can. Unroll it: the sheet is a rectangle. Its breadth is the height h of the can, and its length is the distance once round the base, the circumference 2πr.
CSA of cylinder = 2πr × h = 2πrh
Add the two circular ends, each of area πr2:
TSA of cylinder = 2πrh + 2πr2 = 2πr(r + h)
A cylinder open at one end (a glass, a tank without a lid) has surface 2πrh + πr2. A pipe open at both ends has only the curved surface 2πrh.
Volume
A cylinder is a stack of thin circular discs, each of area πr2, piled to height h, just as a cuboid is a stack of rectangles. So volume = area of base × height.
Volume of cylinder = πr2h
Hollow Cylinder
A pipe has an outer radius R and an inner radius r. The metal forms a ring-shaped cross-section of area πR2 − πr2, so:
- Volume of material = π(R2 − r2)h
- Outer CSA = 2πRh; inner CSA = 2πrh
- TSA = 2πRh + 2πrh + 2π(R2 − r2), the last term being the two ring-shaped ends
Illustration: A cylinder has r = 7 cm and h = 10 cm. CSA = 2 × (22/7) × 7 × 10 = 440 cm2. TSA = 2 × (22/7) × 7 × (7 + 10) = 44 × 17 = 748 cm2. Volume = (22/7) × 49 × 10 = 1540 cm3, which is 1.54 litres.
4. Right Circular Cone
An ice-cream cone, a birthday cap, a funnel and a heap of sand are cones. A right circular cone has a circular base and a vertex directly above the centre of the base. It is described by the radius r of the base and the height h, the perpendicular distance from the vertex to the centre of the base. A third length, the slant height l, is the distance from the vertex to any point on the edge of the base.
Slant Height
Cut a cone straight down through its vertex. The cut face is an isosceles triangle. Half of it is a right triangle with legs h (the height) and r (the radius), and hypotenuse l (the slant height). By the Baudhāyana-Pythagoras theorem:
l2 = r2 + h2, so l = √(r2 + h2)
Since l is the hypotenuse, the slant height is always longer than both the height and the radius.
Curved Surface Area from the Net
Cut a paper cone along one slant line and flatten it. The curved surface becomes a sector of a circle whose radius is the slant height l. The curved edge of the sector was the rim of the base, so its arc length is 2πr.
A full circle of radius l has circumference 2πl and area πl2. The sector is the fraction (2πr)/(2πl) = r/l of that circle, so its area is (r/l) × πl2 = πrl.
CSA of cone = πrl
TSA of cone = πrl + πr2 = πr(l + r)
A cap, a tent or a funnel has no base, so only the CSA is used. A solid cone standing on a table has the TSA.
Volume: One-Third of the Cylinder
Take a hollow cone and a hollow cylinder with the same base radius and the same height. Fill the cone with sand or water and pour it into the cylinder. It takes exactly three full cones to fill the cylinder. So:
Volume of cone = (1/3) × πr2h
Illustration: A cone has r = 7 cm and h = 24 cm. Slant height l = √(49 + 576) = √625 = 25 cm. CSA = (22/7) × 7 × 25 = 550 cm2. TSA = (22/7) × 7 × (25 + 7) = 22 × 32 = 704 cm2. Volume = (1/3) × (22/7) × 49 × 24 = 22 × 7 × 8 = 1232 cm3.
Two heights, two jobs: the vertical height h goes into the volume; the slant height l goes into the curved surface area. Mixing them up is the commonest cone error.
5. Pyramid: Base, Apex, Surface Area and Volume
The Great Pyramid of Giza and a tent with a square floor are pyramids. A pyramid has a polygon as its base and triangular faces, one on each side of the base, that meet at a single point called the apex. The apex does not lie on the base. A pyramid is named after its base: a triangular pyramid has a triangle as base, a square pyramid a square, a hexagonal pyramid a hexagon.
| Pyramid | Faces | Edges | Vertices |
|---|---|---|---|
| Triangular pyramid | 4 (all triangles) | 6 | 4 |
| Square pyramid | 5 (1 square, 4 triangles) | 8 | 5 |
| Base with n sides | n + 1 | 2n | n + 1 |
Two Different Heights
In a right pyramid the apex lies directly above the centre of the base. The height h is the perpendicular distance from the apex to the base. The slant height s is the height of a triangular face, measured from the apex to the midpoint of a base edge.
For a right square pyramid with base side a, the foot of the height is the centre of the square, which is a/2 from the midpoint of each side. The height, the half-side and the slant height form a right triangle:
s2 = h2 + (a/2)2, so s = √(h2 + a2/4)
Surface Area from the Net
A square pyramid unfolds into a square with a triangle attached to each side. Each triangle has base a and height s, so its area is (1/2)as. Four triangles:
LSA of square pyramid = 4 × (1/2) × a × s = 2as
TSA of square pyramid = a2 + 2as
For any right pyramid with a regular polygon base, LSA = (1/2) × perimeter of base × slant height; for a square, 4a gives 2as again.
Volume: One-Third Again
Joining the centre of a cube of edge a to its 8 corners splits it into 6 identical square pyramids, each with a face of the cube as base, height a/2 and volume a3/6. Now (1/3) × a2 × (a/2) = a3/6, which matches. The general rule, true for every pyramid whatever the shape of its base, is:
Volume of pyramid = (1/3) × base area × height
The cone follows the same pattern: each holds one-third of the prism or cylinder with the same base and height.
Illustration: A square pyramid has base side 12 cm and height 8 cm. Slant height s = √(64 + 36) = √100 = 10 cm. LSA = 2 × 12 × 10 = 240 cm2. TSA = 144 + 240 = 384 cm2. Volume = (1/3) × 144 × 8 = 384 cm3. (The two numbers happen to be equal here, but one is in cm2 and the other in cm3.)
6. Sphere and Hemisphere
A football, a marble, a ball bearing and (nearly) the Earth are spheres. A sphere is the set of all points in space at the same distance r from a fixed point, the centre. That distance is the radius, and the diameter is 2r. A sphere has one curved surface and no faces, edges or vertices.
Surface area of sphere = 4πr2
Volume of sphere = (4/3)πr3
Archimedes’ Sphere and Cylinder (Enrichment, Beyond the Syllabus List)
This result is for interest and is outside the listed syllabus competencies. Fit a sphere of radius r exactly inside a cylinder, so the cylinder has radius r and height 2r. Archimedes proved two facts about this pair:
- The surface area of the sphere, 4πr2, equals the curved surface area of the cylinder, 2πr × 2r = 4πr2.
- The volume of the sphere, (4/3)πr3, is two-thirds of the volume of the cylinder, πr2 × 2r = 2πr3.
Hemisphere
A plane through the centre cuts a sphere into two equal halves called hemispheres. A hemisphere has a curved surface (half the sphere) and a flat circular face of radius r.
- CSA of hemisphere = (1/2) × 4πr2 = 2πr2
- TSA of solid hemisphere = 2πr2 + πr2 = 3πr2
- Volume of hemisphere = (1/2) × (4/3)πr3 = (2/3)πr3
A bowl or a dome uses only the curved surface, 2πr2. A solid paperweight shaped like a hemisphere uses 3πr2.
Illustration: For r = 7 cm, the sphere has surface area 4 × (22/7) × 49 = 616 cm2 and volume (4/3) × (22/7) × 343 = 4312/3 ≈ 1437.33 cm3. The hemisphere of the same radius has CSA 308 cm2, TSA 462 cm2 and volume 2156/3 ≈ 718.67 cm3.
Hollow sphere (spherical shell): with outer radius R and inner radius r, the volume of material is (4/3)π(R3 − r3).
Cone, Hemisphere and Cylinder Side by Side
Take a cone, a hemisphere and a cylinder, all with base radius r and height r. Their volumes are (1/3)πr3, (2/3)πr3 and πr3, in the ratio 1 : 2 : 3.
7. Varying Dimensions, Recasting and Combined Solids
How Area and Volume Respond to Scaling
Surface area is built from a length times a length, so it depends on the square of the size. Volume is built from three lengths, so it depends on the cube. If every length of a solid is multiplied by k:
- surface area is multiplied by k2
- volume is multiplied by k3
A cube of edge 1 cm has surface area 6 cm2 and volume 1 cm3. A cube of edge 2 cm has surface area 24 cm2 (4 times) and volume 8 cm3 (8 times). Doubling the radius of a ball makes its surface 4 times and its volume 8 times.
When only some dimensions change, substitute into the formula instead of guessing. For a cylinder with the radius halved and height doubled: Vnew = π(r/2)2(2h) = (1/2)πr2h, so the volume halves. The halving and doubling do not cancel, because the radius is squared.
Enrichment (beyond the syllabus list): shapes of equal volume can need very different surfaces. Cuboids of volume 64 cm3 range from TSA 258 cm2 (64 × 1 × 1) down to 96 cm2 for the 4 × 4 × 4 cube: among all cuboids with a fixed volume, the cube has the least surface area (Competency Question 8 checks four whole-number designs).
Recasting
When a solid is melted and recast into another shape, no material is lost or gained, so the volume stays the same. The surface area usually changes.
Volume of the original solid = total volume of the new solid(s)
Example: a metal sphere of radius 6 cm is recast into a cylinder of radius 4 cm. Sphere volume = (4/3)π × 216 = 288π cm3. Cylinder volume = π × 16 × h. Setting 16πh = 288π gives h = 18 cm. To count small solids made from a big one, divide the big volume by one small volume.
Combined Solids
Many objects are two simple solids joined together: a toy is a cone on a hemisphere, a capsule is a cylinder with a hemisphere at each end, a tent is a cone on a cylinder.
- Volume of the combination = sum of the volumes of the parts.
- Surface area of the combination = sum of only the visible surfaces. Where two parts are joined, their touching faces are hidden and must be left out.
Example: a cone (radius 3 cm, height 4 cm, so slant height 5 cm) stands on a hemisphere of radius 3 cm. Volume = 12π + 18π = 30π cm3. Surface = 15π + 18π = 33π cm2; the two joined circles are not counted.
A step-by-step method works for every combined solid: name the parts, write the formula for each, find any missing length (often total height minus a radius), compute each part, then add only what the question asks for.
Formula and Theorem Sheet
| Solid | Lateral or Curved Surface Area | Total Surface Area | Volume |
|---|---|---|---|
| Cuboid | 2h(l + b) | 2(lb + bh + hl) | lbh |
| Cube | 4a2 | 6a2 | a3 |
| Cylinder | 2πrh | 2πr(r + h) | πr2h |
| Hollow cylinder | 2πRh (outer), 2πrh (inner) | 2π(R + r)h + 2π(R2 − r2) | π(R2 − r2)h |
| Cone | πrl | πr(l + r) | (1/3)πr2h |
| Square pyramid | 2as | a2 + 2as | (1/3)a2h |
| Any pyramid | Sum of triangular faces | Base area + LSA | (1/3) × base area × h |
| Sphere | 4πr2 | 4πr2 | (4/3)πr3 |
| Hemisphere | 2πr2 | 3πr2 | (2/3)πr3 |
| Hollow sphere | - | - | (4/3)π(R3 − r3) |
| Relation | Statement |
|---|---|
| Diagonal of cuboid | √(l2 + b2 + h2); for a cube a√3 |
| Slant height of cone | l = √(r2 + h2) |
| Slant height of square pyramid | s = √(h2 + (a/2)2) |
| Cone and cylinder, same r and h | Vcone = (1/3)Vcylinder |
| Sphere in cylinder (r, 2r), enrichment | Areas equal; Vsphere = (2/3)Vcylinder |
| Cone : hemisphere : cylinder (radius r, height r) | Volumes 1 : 2 : 3 |
| Scaling by k | Area × k2; volume × k3 |
| Recasting | Volume before = volume after |
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Important Definitions
- Solid: an object with three dimensions (length, breadth and height) that occupies space.
- Surface area: the total area of all the faces and curved surfaces of a solid, in square units.
- Lateral surface area: the area of the side faces only, leaving out the top and bottom (for a cuboid, the four walls).
- Curved surface area: the area of the curved part of a cylinder, cone, sphere or hemisphere, leaving out any flat faces.
- Total surface area: the curved or lateral surface area plus the area of all flat faces.
- Volume: the amount of space occupied by a solid, in cubic units.
- Capacity: the volume of the inside of a hollow container, often measured in litres.
- Cuboid: a solid bounded by six rectangular faces, with 12 edges and 8 vertices.
- Cube: a cuboid with all edges equal; its six faces are equal squares.
- Right circular cylinder: a solid with two equal parallel circular bases whose centres are joined by a line perpendicular to the bases.
- Right circular cone: a solid with a circular base and a vertex directly above the centre of the base.
- Slant height (cone): the distance from the vertex to any point on the circumference of the base, l = √(r2 + h2).
- Pyramid: a solid with a polygon base and triangular faces, one on each side of the base, meeting at a common point.
- Apex: the common vertex of all the triangular faces of a pyramid, not lying on the base.
- Slant height (pyramid): the height of a triangular face, from the apex to the midpoint of a base edge.
- Sphere: the set of all points in space at a fixed distance (the radius) from a fixed point (the centre).
- Hemisphere: one of the two equal halves of a sphere made by a plane through its centre.
Solved Examples
Take π = 22/7 unless stated otherwise.
Example 1: Painting a Tank
Question: An open cuboidal tank is 2.5 m long, 1.6 m wide and 1.2 m deep. Find its capacity in litres and the cost of painting its inner surface at ₹20 per m2.
Solution:
Capacity = l × b × h = 2.5 × 1.6 × 1.2 = 4.8 m3 = 4.8 × 1000 = 4800 litres.
The tank is open, so the inner surface is the base plus the four walls.
Base = 2.5 × 1.6 = 4 m2. Walls = 2h(l + b) = 2 × 1.2 × 4.1 = 9.84 m2.
Area to paint = 4 + 9.84 = 13.84 m2. Cost = 13.84 × 20 = ₹276.80.
Example 2: Cylinder from its Curved Surface
Question: The curved surface area of a cylindrical vessel is 4400 cm2 and the circumference of its base is 110 cm. Find its height and its capacity in litres.
Solution: CSA = circumference × height, so 110 × h = 4400 and h = 40 cm.
2πr = 110 gives r = 110 × 7/44 = 17.5 cm.
Volume = πr2h = (22/7) × 17.5 × 17.5 × 40 = 22 × 2.5 × 17.5 × 40 = 38500 cm3 = 38.5 litres.
Example 3: Metal Pipe
Question: A metal pipe is 77 cm long. Its inner diameter is 4 cm and outer diameter 4.4 cm. Find its inner curved surface area, outer curved surface area and the volume of metal.
Solution: r = 2 cm, R = 2.2 cm, h = 77 cm.
Inner CSA = 2πrh = 2 × (22/7) × 2 × 77 = 968 cm2.
Outer CSA = 2πRh = 2 × (22/7) × 2.2 × 77 = 1064.8 cm2.
Volume of metal = π(R2 − r2)h = (22/7) × (4.84 − 4) × 77 = 22 × 0.84 × 11 = 203.28 cm3.
Example 4: Conical Tent
Question: A conical tent is 24 m high and the radius of its base is 7 m. Find (a) its slant height, (b) the length of canvas 5 m wide needed to make it, ignoring wastage, and (c) the volume of air inside.
Solution: (a) l = √(72 + 242) = √(49 + 576) = √625 = 25 m.
(b) A tent has no floor, so canvas = CSA = πrl = (22/7) × 7 × 25 = 550 m2. Length = area ÷ width = 550 ÷ 5 = 110 m.
(c) Volume = (1/3)πr2h = (1/3) × (22/7) × 49 × 24 = 1232 m3.
Example 5: Cone from its Volume
Question: The volume of a right circular cone is 9856 cm3 and the diameter of its base is 28 cm. Find its height, slant height and curved surface area.
Solution: r = 14 cm. (1/3) × (22/7) × 196 × h = 9856.
(1/3) × 22 × 28 × h = 9856, so (616/3)h = 9856 and h = 9856 × 3/616 = 48 cm.
l = √(142 + 482) = √(196 + 2304) = √2500 = 50 cm.
CSA = πrl = (22/7) × 14 × 50 = 2200 cm2.
Example 6: Square Pyramid
Question: A right square pyramid has base side 10 m and slant height 13 m. Find its height, lateral surface area, total surface area and volume.
Solution: Half the base side is 5 m. From s2 = h2 + (a/2)2: h2 = 169 − 25 = 144, so h = 12 m.
LSA = 2as = 2 × 10 × 13 = 260 m2.
TSA = a2 + 2as = 100 + 260 = 360 m2.
Volume = (1/3) × a2 × h = (1/3) × 100 × 12 = 400 m3.
Example 7: Sphere from its Surface Area
Question: The surface area of a sphere is 154 cm2. Find its radius and volume.
Solution: 4 × (22/7) × r2 = 154, so r2 = 154 × 7/88 = 12.25 and r = 3.5 cm.
Volume = (4/3) × (22/7) × 3.5 × 3.5 × 3.5 = (4/3) × 22 × 0.5 × 12.25 = 539/3 ≈ 179.67 cm3.
Example 8: Hemispherical Bowl
Question: A hemispherical bowl has inner radius 10.5 cm. Find how much liquid it holds, in litres, and the cost of tin-plating its inner surface at ₹16 per 100 cm2.
Solution: Capacity = (2/3)πr3 = (2/3) × (22/7) × 10.5 × 10.5 × 10.5 = (2/3) × 22 × 1.5 × 110.25 = 2425.5 cm3 ≈ 2.43 litres.
Inner surface = CSA = 2πr2 = 2 × (22/7) × 110.25 = 693 cm2.
Cost = 693 × 16/100 = ₹110.88.
Example 9: Balls from a Cylinder
Question: A solid metal cylinder of diameter 6 cm and height 16 cm is melted and recast into spherical balls of diameter 2 cm. How many balls are made?
Solution: Volume of cylinder = π × 32 × 16 = 144π cm3.
Volume of one ball (r = 1 cm) = (4/3)π × 13 = (4/3)π cm3.
Number of balls = 144π ÷ (4/3)π = 144 × 3/4 = 108. The value of π cancels, so there is no need to substitute it.
Example 10: Toy Made of a Cone and a Hemisphere
Question: A toy is 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 and volume.
Solution: Height of cone = 15.5 − 3.5 = 12 cm. Slant height l = √(122 + 3.52) = √(144 + 12.25) = √156.25 = 12.5 cm.
Visible surface = CSA of cone + CSA of hemisphere = πrl + 2πr2.
πrl = (22/7) × 3.5 × 12.5 = 137.5 cm2. 2πr2 = 2 × (22/7) × 12.25 = 77 cm2.
TSA = 137.5 + 77 = 214.5 cm2.
Volume of cone = (1/3) × (22/7) × 12.25 × 12 = 154 cm3. Volume of hemisphere = (2/3) × (22/7) × 42.875 = 539/6 ≈ 89.83 cm3.
Volume of toy = 154 + 89.83 ≈ 243.83 cm3.
Example 11: Changing the Dimensions
Question: (a) The radius of a cylinder is doubled and its height is halved. Compare the new volume and curved surface area with the old. (b) The radius of a sphere is increased by 10%. By what percent do its surface area and volume increase?
Solution: (a) New volume = π(2r)2(h/2) = 2πr2h, so the volume is doubled. New CSA = 2π(2r)(h/2) = 2πrh, so the curved surface area is unchanged.
(b) New radius = 1.1r. Surface area is multiplied by 1.12 = 1.21, an increase of 21%. Volume is multiplied by 1.13 = 1.331, an increase of 33.1%.
Competency-Based Questions (with answers)
1. Case-based: The Rooftop Water Tank
A family installs a closed cylindrical water tank on its roof. The tank has internal diameter 1.4 m and height 2 m. The family uses about 440 litres of water a day.
(a) Find the capacity of the tank in litres. r = 0.7 m. V = (22/7) × 0.7 × 0.7 × 2 = 3.08 m3 = 3080 litres.
(b) For how many days will a full tank last? 3080 ÷ 440 = 7 days.
(c) The outside of the tank, including the lid and the base, is to be painted. Taking the outer radius as 0.7 m, find the area to be painted. TSA = 2πr(r + h) = 2 × (22/7) × 0.7 × 2.7 = 4.4 × 2.7 = 11.88 m2.
2. Case-based: The Ice-Cream Cone
An ice-cream shop fills cones of radius 3.5 cm and height 9 cm completely with ice cream and adds a hemispherical scoop of the same radius on top.
(a) Find the volume of ice cream inside the cone. (1/3) × (22/7) × 12.25 × 9 = 115.5 cm3.
(b) Find the total volume of ice cream in one serving. Hemisphere = (2/3) × (22/7) × 42.875 = 539/6 cm3. Total = 693/6 + 539/6 = 1232/6 ≈ 205.33 cm3.
(c) Ice cream arrives in cylindrical containers of radius 7 cm and height 20 cm. How many servings does one container fill? Container = (22/7) × 49 × 20 = 3080 cm3. Servings = 3080 ÷ (1232/6) = 18480/1232 = 15.
3. Case-based: A Pyramid Tent
A camping tent is a right square pyramid with a square floor of side 4 m and a height of 1.5 m at the centre. The four sloping sides are made of canvas; the floor is a separate sheet.
(a) Find the slant height of each sloping face. s = √(1.52 + 22) = √(2.25 + 4) = √6.25 = 2.5 m.
(b) Find the area of canvas for the sloping sides. LSA = 2as = 2 × 4 × 2.5 = 20 m2.
(c) Find the volume of air inside the tent. (1/3) × 16 × 1.5 = 8 m3.
4. Assertion-Reason
Assertion (A): If the radius of a sphere is doubled, its volume becomes 8 times. Reason (R): The volume of a sphere is proportional to the cube of its radius.
(a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true.
Answer: (a). V = (4/3)πr3, so replacing r by 2r multiplies V by 23 = 8. R gives the reason.
5. Assertion-Reason
Assertion (A): A cylinder and a cone have the same base radius and the same height, so the volume of the cylinder is three times the volume of the cone. Reason (R): The curved surface area of the cylinder is three times the curved surface area of the cone.
(a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true.
Answer: (c). πr2h is three times (1/3)πr2h, so A is true. The curved areas 2πrh and πrl have ratio 2h : l, which for r = 7, h = 24 is 48 : 25, and since l is always greater than h, 2h can never equal 3l. So R is false.
6. Assertion-Reason
Assertion (A): When a solid metal sphere is melted and recast into a cylinder, the surface area of the cylinder equals the surface area of the sphere. Reason (R): When a solid is recast, its volume does not change.
(a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true.
Answer: (d). Only the volume is conserved. In Concept 7, a sphere of radius 6 cm (surface area 144π cm2) became a cylinder of radius 4 cm and height 18 cm, whose TSA is 2π × 4 × 22 = 176π cm2. The areas differ.
7. Assertion-Reason
Assertion (A): The curved surface area of a hemisphere of radius 7 cm is 308 cm2. Reason (R): The total surface area of a solid hemisphere of radius r is 3πr2.
(a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true.
Answer: (b). CSA = 2πr2 = 2 × (22/7) × 49 = 308 cm2, so A is true. R is also true, but it describes the total surface area, so it does not explain the value of the curved surface area.
8. Data-based: Same Volume, Different Boxes
A company must pack 64 cm3 of sweets in a closed cuboidal cardboard box with whole-number edges. It compares four designs: 64 × 1 × 1, 16 × 2 × 2, 8 × 4 × 2 and 4 × 4 × 4 (all in cm).
(a) Find the cardboard needed for each design. Using 2(lb + bh + hl): 258, 136, 112 and 96 cm2 respectively.
(b) Which design uses the least cardboard, and what pattern do you see? The cube, 4 × 4 × 4, uses the least (96 cm2). As the three edges get closer to one another, the surface area falls.
(c) What percent of cardboard does the cube save compared with the 8 × 4 × 2 box? Saving = 112 − 96 = 16 cm2. Percent = 16/112 × 100 ≈ 14.3%.
Important Questions for Exams
1 Mark Questions
Q1. The volume of a cube is 343 cm3. Find its edge.
a3 = 343 = 73, so a = 7 cm.
Q2. Write the total surface area of a solid hemisphere of radius r.
2πr2 + πr2 = 3πr2.
Q3. Find the slant height of a cone of radius 5 cm and height 12 cm.
l = √(25 + 144) = √169 = 13 cm.
Q4. How many faces, edges and vertices does a square pyramid have?
5 faces, 8 edges, 5 vertices.
Q5. The radius of a cylinder is halved and its height is unchanged. What happens to its volume?
V = π(r/2)2h = (1/4)πr2h. It becomes one-fourth.
2 Marks Questions
Q6. The diameter of a sphere is decreased by 25%. By what percent does its surface area decrease?
New radius = 0.75r. New area = 4π(0.75r)2 = 0.5625 × 4πr2. Decrease = 1 − 0.5625 = 0.4375, that is 43.75%.
Q7. How many cubes of edge 2 cm can be cut from a cuboid 12 cm × 8 cm × 6 cm?
Along the edges: 12/2 = 6, 8/2 = 4, 6/2 = 3. Number = 6 × 4 × 3 = 72. (Check: 576 ÷ 8 = 72.)
Q8. A cone has volume 1570 cm3 and base radius 10 cm. Find its height. (Use π = 3.14.)
(1/3) × 3.14 × 100 × h = 1570, so h = 1570 × 3/314 = 15 cm.
3 Marks Questions
Q9. The inside of a hemispherical dome of radius 3.5 m is to be painted at ₹40 per m2. Find the cost.
Area = CSA = 2πr2 = 2 × (22/7) × 3.5 × 3.5 = 77 m2. Cost = 77 × 40 = ₹3080.
Q10. A conical heap of grain has base diameter 12 m and height 8 m. Find its volume and the area of canvas needed to just cover it.
r = 6 m, l = √(36 + 64) = 10 m. Volume = (1/3) × (22/7) × 36 × 8 = 2112/7 ≈ 301.71 m3. Canvas = πrl = (22/7) × 6 × 10 = 1320/7 ≈ 188.57 m2.
Q11. A right square pyramid has base side 16 cm and height 6 cm. Find its total surface area and volume.
s = √(62 + 82) = √100 = 10 cm. TSA = a2 + 2as = 256 + 320 = 576 cm2. Volume = (1/3) × 256 × 6 = 512 cm3.
5 Marks Questions
Q12. A medicine capsule is a cylinder with a hemisphere at each end. Its total length is 20 mm and its diameter is 7 mm. Find its surface area and volume.
r = 3.5 mm. Length of cylindrical part = 20 − 2 × 3.5 = 13 mm.
Surface area = 2πrh + 2 × 2πr2 = 2 × (22/7) × 3.5 × 13 + 4 × (22/7) × 12.25 = 286 + 154 = 440 mm2.
Volume = πr2h + (4/3)πr3 = (22/7) × 12.25 × 13 + (4/3) × (22/7) × 42.875 = 500.5 + 539/3 ≈ 500.5 + 179.67 = 680.17 mm3.
The two hemispheres together form one full sphere, which is why 4πr2 and (4/3)πr3 appear.
Q13. A tent is a cylinder of radius 6 m and height 3 m, topped by a cone of the same radius and slant height 10 m. Find (a) the canvas needed, (b) its cost at ₹70 per m2, and (c) the volume of air inside.
(a) Canvas = CSA of cylinder + CSA of cone = 2π × 6 × 3 + π × 6 × 10 = 36π + 60π = 96π = 2112/7 ≈ 301.71 m2. The floor and the joining circle are not covered with canvas.
(b) Cost = (2112/7) × 70 = ₹21120.
(c) Height of cone = √(100 − 36) = 8 m. Volume = π × 36 × 3 + (1/3) × π × 36 × 8 = 108π + 96π = 204π = 4488/7 ≈ 641.14 m3.
Q14. A solid hemisphere of radius 6 cm is melted and recast into a cone of the same radius. (a) Find the height of the cone. (b) Find its slant height. (c) Find the ratio of the curved surface area of the hemisphere to that of the cone.
(a) (1/3)πr2h = (2/3)πr3 gives h = 2r = 12 cm.
(b) l = √(36 + 144) = √180 = 6√5 cm ≈ 13.42 cm.
(c) CSA of hemisphere = 2π × 36 = 72π. CSA of cone = π × 6 × 6√5 = 36√5π. Ratio = 72 : 36√5 = 2 : √5.
Common Mistakes and Examiner Tips
- Using the diameter as the radius. Fix: the moment you read “diameter”, write r = d/2 on the first line of your answer.
- Using the height instead of the slant height in πrl or 2as. Fix: vertical height goes into volume; slant height goes into curved or lateral area. Find the slant height first with the Baudhāyana-Pythagoras theorem.
- Forgetting the 1/3 for cones and pyramids. Fix: remember that three cones fill the matching cylinder, and six pyramids fill a cube.
- Taking TSA when the object is open. Fix: ask what is really covered. A tent, a cap, a pipe, a bowl or an open tank has no top or base in the answer.
- Writing TSA of a hemisphere as 2πr2. Fix: 2πr2 is only the curved part. A solid hemisphere adds the flat circle, giving 3πr2.
- Mixing units. Fix: convert every length to one unit before multiplying. Remember 1 m2 = 10000 cm2 and 1 m3 = 1000000 cm3, not 100.
- Adding hidden faces in a combined solid. Fix: add only the surfaces you could touch from outside; the circles where two parts meet disappear.
- Using the total height of a toy as the height of one part. Fix: subtract the radius of the hemisphere (or the height of the other part) first.
- Expecting surface area to stay the same after recasting. Fix: only the volume is conserved. Equate volumes, never areas.
- Assuming a doubled length doubles the volume. Fix: area scales by k2 and volume by k3. When only some dimensions change, substitute into the formula.
Quick Revision Points
- Surface area is in square units; volume is in cubic units; capacity is volume inside a container.
- 1 L = 1000 cm3; 1 m3 = 1000 L; 1 m2 = 10000 cm2.
- Every surface area formula comes from the area of the solid’s net.
- Cuboid: LSA 2h(l + b), TSA 2(lb + bh + hl), V = lbh, diagonal √(l2 + b2 + h2).
- A cube is a cuboid with l = b = h = a: LSA 4a2, TSA 6a2, V = a3, diagonal a√3.
- Cylinder: CSA 2πrh (rectangle 2πr by h), TSA 2πr(r + h), V = πr2h.
- Hollow cylinder: V = π(R2 − r2)h.
- Cone: l = √(r2 + h2), CSA πrl (a sector of radius l), TSA πr(l + r), V = (1/3)πr2h.
- Slant height is always longer than both the height and the radius.
- Pyramid: polygon base, triangular faces meeting at the apex; named by its base.
- Square pyramid: s2 = h2 + (a/2)2; LSA 2as; TSA a2 + 2as.
- Any pyramid: V = (1/3) × base area × height.
- Sphere: surface area 4πr2, V = (4/3)πr3.
- Hemisphere: CSA 2πr2, TSA 3πr2, V = (2/3)πr3.
- Enrichment, beyond the syllabus list (Archimedes): a sphere in its cylinder has equal curved area and 2/3 of the volume.
- Cone, hemisphere, cylinder with radius r and height r: volumes 1 : 2 : 3.
- Scale every length by k: area × k2, volume × k3.
- Enrichment, beyond the syllabus list: for a fixed volume, the cube has the least surface area of all cuboids.
- Recasting: volume before = volume after; divide volumes to count pieces.
- Combined solid: volumes add; surface = visible surfaces only.
Weightage and Exam Pattern
Class 9 has no board exam; this chapter is assessed in school tests and the annual exam. In the CBSE Class IX Mathematics course structure for 2026-27, the chapter is part of Unit V, Mensuration, which carries 14 marks out of the 80-mark theory paper and is allotted 27 teaching periods. The unit has two chapters, Area and Perimeter and Surface Area and Volume, and the course structure does not split the 14 marks between them. The other 20 marks are internal assessment.
| Unit | Marks (out of 80) |
|---|---|
| I. Number System | 07 |
| II. Algebra | 20 |
| III. Coordinate Geometry | 04 |
| IV. Geometry | 25 |
| V. Mensuration | 14 |
| VI. Statistics and Probability | 10 |
The CBSE question paper design weights questions by typology: about 54% remembering and understanding, 24% applying, and 22% analysing, evaluating and creating. In this chapter that means recalling and choosing the right formula, applying it to real objects such as tanks, tents and toys, and analysing questions on changing dimensions, recasting and combined solids.
These formulas return in the Class 10 board chapter on surface areas and volumes.
Class 9 Maths – swipe through all 10 cards to understand the whole chapter.
Cuboid: three pairs of faces
A cuboid has 6 rectangular faces in three equal pairs: lb, bh and hl.
LSA is the four walls of a room.
- LSA = perimeter of base × height
- Volume = area of base × height
- Diagonal = √(l2 + b2 + h2)
Cube as a special case
Put l = b = h = a into the cuboid formulas.
Every cube is a cuboid, not every cuboid is a cube.
- Six equal square faces
- Diagonal = a√3
- TSA 486 cm2 gives a = 9 cm
Cylinder: unroll the label
The curved surface opens into a rectangle 2πr long and h high.
Halve the diameter first.
- Open at one end: 2πrh + πr2
- Hollow pipe: V = π(R2 − r2)h
- Roller area per turn = CSA
Cone: slant height first
Radius, height and slant height form a right triangle.
l goes into area, h goes into volume.
- Net of curved surface is a sector of radius l
- Tent or cap: CSA only
- r = 7, h = 24 gives l = 25
Three cones fill a cylinder
Same radius and same height: the cone holds one-third.
Use the vertical height h here, never the slant height l.
- Cylinder = 3 × cone
- r = 7, h = 24 gives 1232 cm3
- Forgetting 1/3 triples the answer
Pyramid: base, apex, slant height
A polygon base with triangular faces meeting at the apex.
Square pyramid: 5 faces, 8 edges, 5 vertices.
- TSA = a2 + 2as
- Six pyramids fill a cube
- Named after its base shape
Sphere: 4πr2 and (4/3)πr3
All points at distance r from the centre.
Halve the diameter before cubing the radius.
- r = 7 cm: area 616 cm2
- r = 3.5 cm: area 154 cm2
- Hollow sphere: (4/3)π(R3 − r3)
Hemisphere: curved or total
Half a sphere, with one flat circular face.
Bowl or dome: use 2πr2.
- Solid hemisphere adds πr2
- r = 7 cm: CSA 308 cm2, TSA 462 cm2
- Cone : hemisphere : cylinder = 1 : 2 : 3
Scale by k
Area depends on length squared, volume on length cubed.
Substitute when only some dimensions change.
- Double the radius: area × 4, volume × 8
- Radius up 10%: area up 21%
- Cylinder: r halved, h doubled gives volume halved
Recasting and combined solids
Melting keeps volume; joining hides faces.
Combined surface = visible surfaces only.
- Sphere r 6 into cylinder r 4 gives h = 18
- Volumes of parts simply add
- Toy height minus radius gives cone height
📝 Practice Surface Area and Volume - 10 questions
Concept-check questions · with answers & explanations
Start →Close ✕
Why A: Step 1 (joined cubes form a cuboid): dimensions 12 cm × 4 cm × 4 cm. Step 2 (TSA of cuboid, 2(lb + bh + hl)): 2(12 × 4 + 4 × 4 + 4 × 12) = 2(48 + 16 + 48) = 224 cm². Check: three cubes have 3 × 96 = 288 cm², and the two joints hide 4 faces of 16 cm², giving 288 – 64 = 224 cm².
Why not B: 288 cm² adds the surface areas of the three separate cubes and ignores the 4 faces hidden at the two joints
Why not C: 128 cm² is the lateral area 2h(l + b) = 2 × 4 × 16, which leaves out the two square end faces
Why not D: 192 is 3 × 4³, the volume of the cuboid in cm³, not its surface area
Remember: Joining solids hides faces: surface area falls, volume just adds.
Why A: Step 1 (diameter to radius): r = 1.4 / 2 = 0.7 m, h = 2 m. Step 2 (one revolution covers the CSA, 2πrh): 2 × 22/7 × 0.7 × 2 = 8.8 m². Step 3 (5 revolutions): 5 × 8.8 = 44 m².
Why not B: 8.8 m² is the area levelled in one revolution only; the roller turns 5 times
Why not C: 88 m² uses the diameter 1.4 m as the radius, doubling the answer
Why not D: 22 m² uses πrh per revolution, half the curved surface area
Remember: One roll of a roller paints its curved surface once.
Why C: Step 1 (right triangle r, h, l): h = √(5² – 3²) = √16 = 4 cm. Step 2 (volume of cone, (1/3)πr²h): (1/3) × π × 9 × 4 = 12π cm³.
Why not A: 15π uses the slant height 5 as the vertical height in (1/3)πr²h
Why not B: 24π is πr(l + r) = 3π × 8, the total surface area, not the volume
Why not D: 36π is πr²h without the 1/3, the volume of the matching cylinder
Remember: Volume needs vertical height h; curved area needs slant height l.
Why B: Step 1 (volume of pyramid, (1/3) × a² × h): (1/3) × a² × 6 = 200, so 2a² = 200. Step 2 (solve for a): a² = 100, so a = 10 cm.
Why not A: 100 is a², the base area in cm²; the side is its square root
Why not C: About 5.8 cm comes from a² × 6 = 200, dropping the 1/3
Why not D: About 8.2 cm comes from using 1/2 instead of 1/3, giving 3a² = 200
Remember: Undo the formula step by step: multiply by 3, divide by h, then take the square root.
Why A: Step 1 (surface area, 4πr²): 4 × 22/7 × r² = 616, so r² = 616 × 7 / 88 = 49. Step 2 (square root): r = √49 = 7 cm.
Why not B: 49 is r²; the square root still has to be taken
Why not C: 14 cm is the diameter, twice the radius
Why not D: About 9.9 cm comes from 2πr² = 616, the hemisphere formula, giving r² = 98
Remember: From area to radius: divide by 4π, then square root.
Why A: Step 1 (slant height of cone): l = √(9 + 16) = 5 cm. Step 2 (cone CSA, πrl): π × 3 × 5 = 15π cm². Step 3 (hemisphere CSA, 2πr²): 2π × 9 = 18π cm². Step 4 (joined circular faces are hidden, add visible parts only): 15π + 18π = 33π cm².
Why not B: 42π adds the circle of area 9π where the parts are joined, which is hidden inside the toy
Why not C: 51π uses the full sphere’s surface 36π instead of the hemisphere’s curved surface 18π
Why not D: 30π uses the vertical height 4 cm in πrl, giving 12π for the cone
Remember: Combined solid area = visible curved surfaces only; joints hide faces.
Why D: Step 1 (base face): 10 × 8 = 80 cm². Step 2 (four walls, lateral area 2h(l + b)): 2 × 5 × (10 + 8) = 180 cm². Step 3 (no lid, so add base and walls only): 80 + 180 = 260 cm².
Why not A: 340 cm² is the full total surface area 2(80 + 50 + 40), which includes the missing lid
Why not B: 180 cm² is only the four walls; the box still needs a base
Why not C: 400 is 10 × 8 × 5, the volume of the box in cm³, not an area
Remember: Open box = total surface area minus the missing lid, lb.
Why D: Step 1 (volume = base area × height): V = πr²h. Step 2 (substitute): 22/7 × 7 × 7 × 10 = 22 × 7 × 10 = 1540 cm³.
Why not A: 440 is 2πrh, the curved surface area in cm², not the volume
Why not B: 748 is 2πr(r + h), the total surface area in cm²
Why not C: 3080 doubles the correct answer, as if the volume were 2πr²h
Remember: Cylinder volume = area of the circular base × height.
Why B: Step 1 (slant height from the right triangle): l = √(6² + 8²) = √100 = 10 m. Step 2 (tent = curved surface only, πrl): π × 6 × 10 = 60π m², about 188.6 m².
Why not A: 48π uses the vertical height 8 m in place of the slant height in πrl
Why not C: 96π adds the base circle 36π; the tent has no floor, so only the curved surface is canvas
Why not D: 120π uses 2πrl, the cylinder-style formula, doubling the correct area
Remember: Tents, caps and funnels: curved surface πrl only.
Why D: Step 1 (right triangle with height and half the base side): s = √(8² + 6²) = √100 = 10 cm. Step 2 (lateral area, 2as): 2 × 12 × 10 = 240 cm². Step 3 (add the square base): 240 + 144 = 384 cm².
Why not A: 336 uses the vertical height 8 cm as the height of each triangular face: 144 + 2 × 12 × 8
Why not B: 240 is only the four triangular faces; the square base is part of the total surface
Why not C: 624 drops the 1/2 in each triangle: 144 + 4 × 12 × 10
Remember: Face triangles use slant height s = √(h² + (a/2)²).
Chapter Navigation
Previous: Sequences and Progressions Class 9 Notes
Related Chapters in Class 9 Maths
- Coordinate Geometry Class 9 Notes
- Introduction to Polynomials Class 9 Notes
- Number System Class 9 Notes
- Exploring Algebraic Identities Class 9 Notes
- Circles Class 9 Notes
- Area and Perimeter Class 9 Notes
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Frequently Asked Questions
Surface area is the total area of all the faces and curved surfaces of a solid, the material needed to cover it, measured in square units such as cm² or m². Volume is the space the solid occupies, measured in cubic units such as cm³ or m³. Painting or wrapping is a surface area problem; filling or melting is a volume problem.
A cube is a cuboid whose length, breadth and height are all equal. Putting l = b = h = a into the cuboid formulas gives the cube formulas directly: LSA 2a(2a) = 4a², TSA 2(3a²) = 6a² and volume a³. Every cube is a cuboid, but a cuboid is a cube only when all its edges are equal.
Cut the cone through its vertex. Half of the cut face is a right triangle with legs r (the radius) and h (the height), and hypotenuse l (the slant height). By the Baudhāyana-Pythagoras theorem, l = √(r² + h²). Use l in the curved surface area πrl and h in the volume (1/3)πr²h.
If a hollow cone and a hollow cylinder have the same base radius and the same height, it takes exactly three cones full of sand or water to fill the cylinder. So the cone’s volume is (1/3)πr²h. A pyramid follows the same rule: its volume is one-third of base area times height.
The height h is the perpendicular distance from the apex to the base. The slant height s is the height of a triangular face, from the apex to the midpoint of a base edge. For a square pyramid of base side a, s² = h² + (a/2)². The slant height gives the lateral area 2as; the height gives the volume (1/3)a²h.
A solid hemisphere has a curved surface, half of the sphere’s 4πr², which is 2πr², and a flat circular face of area πr². So its total surface area is 3πr². A bowl or dome has no flat face, so only the curved surface area 2πr² is used for it.
Only the volume stays the same, because no material is lost or gained. Equate the volume of the original solid to the total volume of the new solid or solids. The surface area usually changes. For example, a sphere of radius 6 cm recast into a cylinder of radius 4 cm gives a cylinder of height 18 cm.
If every length of a solid is multiplied by k, its surface area is multiplied by k² and its volume by k³. Doubling the edge of a cube makes its surface area 4 times and its volume 8 times. When only some dimensions change, substitute the new values into the formula.
