Math of Space: Surface Area and Volume: Class 9 Ganita Manjari · Chapter 14

Wrapping and filling boxes, tins, cones, pyramids and balls

Ganita Manjari Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 13 Oct 2026 · Updated 13 Oct 2026

80 min
52 practice questions
21 MCQs

What you will learn in Math of Space: Surface Area and Volume

Derive and use the surface area and volume of a cuboid and a cube, including cube roots of volumes
Explain why solids with equal volume can have different surface areas, and why a cube needs the least
Describe right and oblique cylinders and their cross-sections, and find the curved and total surface areas and volume of open and closed cylinders
Find a cone's slant height, derive its curved surface area by unrolling it into a sector, and use V = ⅓πr²h
Find the volume of any pyramid as one third of base area times height, and its surface area by adding faces
Use the surface area and volume formulas for spheres and hemispheres and connect them to Archimedes' cylinder and to thin cones
Make guesstimates for real objects, stating assumptions and recognising upper bounds

Original companion text aligned to NCERT Mathematics Class 9, Chapter 14, pp. 123–148. Not affiliated with or endorsed by NCERT.

Wrapping and filling

Pick up any box at home: a carton of milk, a steel tiffin, a packet of tea. Two questions about it matter in real life. How much material does it take to make or cover it? How much can it hold? The first question is about its surface area (पृष्ठीय क्षेत्रफल); the second is about its volume (आयतन) (Fig. 14.1).

2/2Filling the inside measures volume, in cubic units.

Fig. 14.1Surface area tells you how much it takes to wrap a solid; volume tells you how much it takes to fill it.

Step 1 shows a box covered in orange wrapping paper, with the note wrap it and the label surface area in square centimetres; a faint second box is on the right. In step 2 the wrapped box fades and the box on the right comes forward, filled with small unit cubes in rows, columns and layers, with the note fill it and the label volume in cubic centimetres.

You already know a lot that helps. You have folded nets into cubes, cylinders and cones. In Chapter 6 of Part 1 you found areas of rectangles, triangles, circles and sectors, and used the Baudhāyana–Pythagoras theorem. We now put all of that to work on five families of solids (cuboids, cylinders, cones, pyramids and spheres) and finish by making sensible estimates about real objects.

Definition

Surface area

The total area of all the faces and curved parts that make up the outside of a solid. It is measured in square units such as $c m^{2}$ or $m^{2}$ .

ExampleThe cardboard needed to make a closed box.

Definition

Volume

How much space a solid fills, or how much a container can hold, counted in unit cubes: $c m^{3}$ , $m^{3}$ and so on.

ExampleThe milk a carton can hold.

14.1 Cuboids and Cubes

A brick, a matchbox, a steel almirah and a shoebox share one shape. Each is a cuboid (घनाभ): a box whose faces are all rectangles, meeting at right angles along 12 edges and 8 corners. Three measurements pin down any cuboid: its length $l$ , width $w$ and height $h$ . A rectangle needs only two; the cuboid adds a third direction.

To find its surface area, notice that the faces come in pairs. The top and bottom are equal, the front and back are equal, and the two sides are equal (Fig. 14.2). So we need only three areas, each counted twice. If you have the printed net of a cuboid from the textbook, fold it and check which faces pair up.

4/4Add the three areas and double the result.

Fig. 14.2Six faces, three equal pairs: that is why every term in the formula is doubled.NCERT Fig. 14.2 · p. 123

A cuboid is drawn with its length l along the front, width w going back and height h up the side. In step 2 the visible top, front and side faces are coloured blue, orange and green, and a note says each has a hidden twin. In step 3 the six faces are laid out on the right as three pairs: two blue l by w rectangles, two orange l by h rectangles and two green w by h rectangles. In step 4 the formula TSA = 2(lw + lh + wh) appears below the pairs.

Adding the three pairs gives the total surface area (TSA) of a cuboid. Writing $S$ for the TSA:

Total surface area (TSA) of a cuboid

S = 2 (l w + l h + w h)

(14.1)

To measure volume we need a measuring block, just as we used unit squares for area. That block is the unit cube, with every edge 1 unit long. Its volume is called 1 cubic unit: $1 c m^{3}$ for a 1 cm cube, $1 m^{3}$ for a 1 m cube. Counting how many unit cubes fill a cuboid, with no gaps, gives its volume.

Fill a cuboid layer by layer (Fig. 14.3). The bottom layer is a rectangle of $l \times w$ cubes. There are $h$ such layers, one on top of another. So the cuboid holds $l \times w \times h$ unit cubes. Notice that $l \times w$ is just the area of the base, the same product that gives the area of a rectangle.

3/4Three equal layers hold $12 \times 3 = 36$ cubes.

Fig. 14.3Volume counts unit cubes: one layer of

l \times w

, repeated

h

times.NCERT Fig. 14.3 · p. 124

On the left is a single unit cube with edge 1 cm. In step 2 one layer of 4 by 3 cubes appears, labelled l = 4 and w = 3, with the count 4 × 3 = 12. In step 3 two more layers slide down onto it, so the stack is 3 layers high, labelled h = 3, with the count 12 × 3 = 36. In step 4 the rule V = (l × w) × h is written beside the stack.

Volume of a cuboid

\begin{aligned} V & = l w h \\ = base area \times h \end{aligned}

(14.2)

Make all three measurements equal and the cuboid becomes a cube (घन), like a dice or a sugar cube. With every edge equal to $a$ , formula (14.1) shrinks to $2 (a^{2} + a^{2} + a^{2}) = 6 a^{2}$ and formula (14.2) to $a \times a \times a = a^{3}$ :

Total surface area of a cube

TSA = 6 a^{2}

(14.3)

Volume of a cube

V = a^{3}

(14.4)

Now the name cube number makes sense: $5^{3} = 125$ is exactly how many unit cubes build a cube of edge 5. Reading a table of cubes backwards gives cube roots, and the edge of a cube is the cube root of its volume. A cube that holds $343 c m^{3}$ has edge $\sqrt[3]{343} = 7 c m .$

Example 14.1NCERT Example 1 · p. 125

Two blocks of jaggery

A shop moulds jaggery (gur) into blocks. Block A is a cube of edge 4 cm. Block B is a cuboid 8 cm × 4 cm × 2 cm. Compare their volumes and surface areas. Which block needs more wrapping paper, and which will harden faster as it cools?

Try it yourself first. Then reveal the solution one step at a time.

The example shows a big idea. Boxes with the same volume can have different surface areas, and the cube needs the least material. Packaging designers use this to save card. Engineers use the opposite: a car radiator or the fins on a motorbike engine spread metal thin to expose more surface and lose heat quickly.

InteractiveBox lab

Fig. 14.4Build a box from unit cubes, or compare every whole-number box with the same volume.

Quick check

1 of 3

Four closed boxes each hold exactly

64 c m^{3}

. Which one needs the least card to make?

4 cm × 4 cm × 4 cm8 cm × 4 cm × 2 cm16 cm × 2 cm × 2 cm8 cm × 8 cm × 1 cm

Show answer

Answer (A) 4 cm × 4 cm × 4 cm

The cube

4 \times 4 \times 4

needs

96 c m^{2}

. The others need 112, 136 and

160 c m^{2}

. Equal volume does not mean equal surface; the shape closest to a cube needs the least.

Practice · 8 questions

Practice 14.1: questions with answers

NumericalEasyNCERT Exercise Set 14.1, Q1
A cube-shaped water tank holds $512 m^{3}$ of water. Find the area of steel sheet needed for all six of its faces.
m²
Show solution
The edge is $a = \sqrt[3]{512} = 8 m$ , since $8^{3} = 512$ .
TSA $= 6 a^{2} = 6 \times 64 = 384 m^{2} .$
NumericalEasyNCERT Exercise Set 14.1, Q2
How many wooden cubes of edge 25 cm can be packed, with no gaps, into a cubical crate of inner edge 1.5 m?
cubes
Show solution
1.5 m = 150 cm, and $150 \div 25 = 6$ cubes fit along each edge.
So the crate holds $6 \times 6 \times 6 = 216$ cubes. (Check: $150^{3} \div 25^{3} = 3 375 000 \div 15 625 = 216$ .)
NumericalMediumNCERT Exercise Set 14.1, Q3
A warehouse is 30 m long, 18 m wide and 8 m high inside. It is to be filled with cartons that each measure 1.5 m × 1.2 m × 0.8 m. How many cartons fit?
cartons
Show solution
Along the length: $30 \div 1.5 = 20$ . Along the width: $18 \div 1.2 = 15$ . Up the height: $8 \div 0.8 = 10$ .
Number of cartons $= 20 \times 15 \times 10 = 3000$ .
Check with volumes: $(30 \times 18 \times 8) \div (1.5 \times 1.2 \times 0.8) = 4320 \div 1.44 = 3000.$ The counts divide exactly, so no space is wasted.
NumericalMediumNCERT Exercise Set 14.1, Q4
Three cubes, each of volume $64 c m^{3}$ , are glued face to face in a straight row. Find the total surface area of the new solid.
cm²
Show solution
Each cube has edge $\sqrt[3]{64} = 4 c m$ , so the row is a cuboid $12 c m \times 4 c m \times 4 c m$ .
TSA $= 2 (12 \times 4 + 12 \times 4 + 4 \times 4) = 2 (48 + 48 + 16) = 224 c m^{2} .$
Check: three loose cubes have $3 \times 96 = 288 c m^{2}$ ; gluing hides 4 faces of $16 c m^{2}$ , and $288 - 64 = 224$ .
NumericalMediumNCERT Exercise Set 14.1, Q5
A cube of edge 3 cm is cut into cubes of edge 1 cm. Find the ratio of the surface area of the big cube to the total surface area of all the small cubes.
Show solution
Answer1 : 3
Big cube: $6 \times 3^{2} = 54 c m^{2}$ .
There are $3^{3} = 27$ small cubes, each with $6 c m^{2}$ , so $162 c m^{2}$ in all.
Ratio $= 54 : 162 = 1 : 3$ . The volume is unchanged, but cutting triples the surface. This is why a spoonful of sugar dissolves faster than a sugar cube: more surface touches the water.
NumericalHardNCERT Exercise Set 14.1, Q6
Three faces of a cuboid share one corner. Their areas are $12 c m^{2}$ , $15 c m^{2}$ and $20 c m^{2}$ . How much does the cuboid hold, and what are its edges?
cm³
Show solution
Say $l w = 12$ , $w h = 15$ and $l h = 20$ . Multiplying: $(l w) (w h) (l h) = l^{2} w^{2} h^{2} = (l w h)^{2}$ .
So $(l w h)^{2} = 12 \times 15 \times 20 = 3600$ , and $V = l w h = 60 c m^{3} .$
Then $h = 60 \div 12 = 5$ , $l = 60 \div 15 = 4$ and $w = 60 \div 20 = 3$ (in cm). Check: $4 \times 3 = 12$ , $3 \times 5 = 15$ , $4 \times 5 = 20$ .
7
Short answerHardNCERT Exercise Set 14.1, Q7
A wooden cube of edge 4 cm is painted red on all six faces and then cut into $1 c m^{3}$ cubes. How many of the small cubes have paint on
1. exactly three faces?
  cubes
  Show solution
  Only the 8 corner cubes have three painted faces.
2. exactly two faces?
  cubes
  Show solution
  Each of the 12 edges has $4 - 2 = 2$ cubes between its corners: $12 \times 2 = 24$ .
3. exactly one face?
  cubes
  Show solution
  Each of the 6 faces has a $2 \times 2$ block of cubes away from its edges: $6 \times 4 = 24$ .
4. no face at all?
  cubes
  Show solution
  The unpainted core is a $2 \times 2 \times 2$ cube: 8 cubes. Check: $8 + 24 + 24 + 8 = 64 = 4^{3}$ .
8
ProofHardNCERT Exercise Set 14.1, Q8
Find cuboids whose edges are whole numbers of centimetres and whose total surface area is exactly $64 c m^{2}$ .
1. Is there more than one such cuboid?
  YesNo
  Show solution
  Yes. For example $4 \times 4 \times 2$ gives $2 (16 + 8 + 8) = 64$ and $7 \times 2 \times 2$ gives $2 (14 + 4 + 14) = 64$ .
2. Find them all.
  Show solution
  Answer10 × 2 × 1, 7 × 2 × 2, 4 × 4 × 2
  There are exactly three: $10 \times 2 \times 1$ , $7 \times 2 \times 2$ and $4 \times 4 \times 2$ (in cm).
3. Show that there are no others.
  Show solution
  We need $l w + w h + h l = 32$ with $h \leq w \leq l$ . Since each product is at least $h^{2}$ , $3 h^{2} \leq 32$ , so $h \leq 3$ .
  If $h = 1$ : $l w + l + w = 32$ , so $(l + 1) (w + 1) = 33 = 3 \times 11$ . This gives $w = 2$ , $l = 10$ .
  If $h = 2$ : $l w + 2 l + 2 w = 32$ , so $(l + 2) (w + 2) = 36$ with $w + 2 \geq 4$ . The pairs are $4 \times 9$ ( $w = 2, l = 7$ ) and $6 \times 6$ ( $w = l = 4$ ).
  If $h = 3$ : $(l + 3) (w + 3) = 41$ , a prime, which has no factor pair with both factors at least 6. So there are no more cuboids.

14.2 Right Circular Cylinder

A water pipe, a gas cylinder, an unsharpened pencil and a tin of ghee are all cylinders. A cylinder has two equal, parallel circular ends joined by a curved surface. The line joining the centres of the two circles is its axis.

In a right circular cylinder (लंब वृत्तीय बेलन) the ends are circles and the axis is perpendicular to them, like a neat stack of coins. Push the stack sideways and the axis leans: that is an oblique cylinder. Its ends are still circles, but the axis is no longer at right angles to them. In this chapter we study only right circular cylinders, and simply call them cylinders.

Definition

Cross-section

The flat shape you see when you cut a solid with one very thin, straight slice.

ExampleSlicing a cucumber across gives circles.

The cross-section (अनुप्रस्थ काट) depends on how you cut (Fig. 14.5). Slice a cylinder parallel to its base and every slice is the same circle. Slice it along the axis and you get a rectangle. When we say 'the cross-section of a cylinder', we usually mean the circle. A cucumber sliced for salad gives near-circles, and every slice of a bread loaf is a cross-section of the loaf. A CT or MRI scan works the same way: it shows doctors the body one thin slice at a time.

1/3Right: the axis is perpendicular to the base. Oblique: the axis leans.

Fig. 14.5A right cylinder's axis is perpendicular to its base. Slices across it are circles; slices along it are rectangles.NCERT Cross-sections · p. 127

Step 1 shows two cylinders side by side: on the left a right cylinder whose dashed axis meets the base at a marked right angle, and on the right an oblique cylinder whose dashed axis leans. In step 2 the oblique cylinder fades, a horizontal slice is marked across the right cylinder and an arrow leads to the slice drawn flat: a circle. In step 3 a vertical slice through the axis is marked and the flat slice beside it is a rectangle.

Definition

Right circular cylinder

A solid with two equal, parallel circular ends whose axis is perpendicular to the ends.

ExampleA new pencil, a gas cylinder without its cap.

Curved surface. Imagine peeling the paper label off a tin (Fig. 14.6). Cut straight down and unroll it: the label becomes a rectangle. Its height is the height $h$ of the tin. Its length went once round the tin, so it equals the circumference, $2 π r$ . This rectangle's area is the cylinder's curved surface area (CSA).

4/4A closed tin also has two circular lids, each $π r^{2}$ .

Fig. 14.6Unrolled, the curved surface is a rectangle

2 π r

long and

h

high.NCERT Fig. 14.5A · p. 128

Step 1 shows a tin of radius r and height h whose curved side is covered by an orange label. Step 2 marks a dashed cut line down the front of the label. In step 3 the label unrolls to the right into a long rectangle of height h and length 2πr, labelled CSA = 2πrh. In step 4 two blue circles, the top and bottom lids, appear touching the rectangle, and the label changes to TSA = 2πrh + 2πr².

Curved surface area of a cylinder

CSA = 2 π r h

(14.5)

A closed cylinder, such as a steel dabba with its lid on, also has two circular ends of area $π r^{2}$ each. Its total surface area is:

Cylinder closed at both ends

\begin{aligned} TSA & = 2 π r h + 2 π r^{2} \\ = 2 π r (h + r) \end{aligned}

(14.6)

A cylinder open at one end, such as a glass or a pencil stand, has only one circular end:

Cylinder open at one end

\begin{aligned} TSA & = 2 π r h + π r^{2} \\ = π r (2 h + r) \end{aligned}

(14.7)

Volume. Every slice across a cylinder is the same circle of area $π r^{2}$ . Just as with a cuboid, the volume is the area of the cross-section times the height:

Volume of a cylinder

V = π r^{2} h

(14.8)

14.2.1 Another way of understanding why the above formulas are true

Stack identical coins into a tall tower (Fig. 14.7). Each coin is a very short cylinder: a circle of area $π r^{2}$ with a tiny thickness $t$ . If the tower has $n$ coins, its height is $h = n t$ and its volume is $n \times π r^{2} t = π r^{2} (n t) = π r^{2} h$ . Now look at the edges. Each coin's rim is a thin strip about $2 π r$ long and $t$ wide. Together the rims cover the side of the tower, with area $n \times 2 π r t = 2 π r h$ . Both formulas fall out of one picture. A stack of rotis or papads tells the same story.

3/3Faces give $V = π r^{2} h$ ; rims give $2 π r h$ .

Fig. 14.7A cylinder is a stack of thin discs: the faces build the volume, the rims build the curved surface.NCERT 14.2.1 · p. 128

Step 1 shows a single coin seen from slightly above, with radius r and thickness t marked; its top face is labelled area πr². In step 2 a tower of twelve coins appears on the right with its height h marked beside it. In step 3 the rims of all the coins are highlighted together and two formulas appear on the left: V = πr² × h and rims = 2πr × h.

Example 14.2NCERT Example 2 · p. 129

Covering a pencil stand

Ritu covers the curved outside of a cylindrical pencil stand with coloured paper. The stand has radius 4.2 cm and height 10 cm. How much paper does she need, ignoring overlaps? Use $π \approx \frac{22}{7}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example

A water tank on the roof

A closed cylindrical water tank has radius 0.7 m and height 2 m. (a) How many litres does it hold? (b) How much steel sheet is needed to make it? Use $π \approx \frac{22}{7}$ and $1 m^{3} = 1000$ litres.

Try it yourself first. Then reveal the solution one step at a time.

Practice · 5 questions

Practice 14.2: questions with answers

1
NumericalMediumNCERT Exercise Set 14.2, Q1
Cylinder Q has 3 times the radius of cylinder P but only one third of its height. Find the ratio of their curved surface areas (P : Q) and of their volumes (P : Q).
1. CSA of P : CSA of Q
  Show solution
  Answer1 : 1
  $2 π r h : 2 π (3 r) (\frac{h}{3}) = 2 π r h : 2 π r h = 1 : 1$ . The curved surfaces are equal.
2. Volume of P : volume of Q
  Show solution
  Answer1 : 3
  $π r^{2} h : π (3 r)^{2} (\frac{h}{3}) = π r^{2} h : 3 π r^{2} h = 1 : 3$ . Q holds three times as much.
2
NumericalMediumNCERT Exercise Set 14.2, Q2
Two cylinders have radii in the ratio $3 : 4$ and heights in the ratio $2 : 5$ . Compare (a) how much they hold and (b) their curved surfaces, as ratios.
1. volumes
  Show solution
  Answer9 : 40
  $π (3 k)^{2} (2 m) : π (4 k)^{2} (5 m) = 18 : 80 = 9 : 40.$
2. curved surface areas
  Show solution
  Answer3 : 10
  $2 π (3 k) (2 m) : 2 π (4 k) (5 m) = 6 : 20 = 3 : 10.$
NumericalMediumNCERT Exercise Set 14.2, Q3
The largest possible right circular cylinder is carved out of a wooden cube of edge 14 cm. Find the volume of the cylinder, and the fraction of the cube that is used. Use $π \approx \frac{22}{7}$ .
cm³
Show solution
The cylinder's diameter and height both equal the edge, so $r = 7 c m$ and $h = 14 c m$ .
$V = \frac{22}{7} \times 7^{2} \times 14 = 22 \times 7 \times 14 = 2156 c m^{3} .$
The cube's volume is $14^{3} = 2744 c m^{3}$ , and $2156 \div 2744 = \frac{11}{14} \approx 0.79$ . In general the fraction is $\frac{π}{4}$ , about 79 %, for any cube.
NumericalHardNCERT Exercise Set 14.2, Q4
The radius of a cylinder is increased by 25 %. By what percentage must its height be decreased so that the volume stays the same?
%
Show solution
$π (1.25 r)^{2} (k h) = π r^{2} h$ gives $1.5625 k = 1$ , so $k = 0.64$ .
The height becomes 64 % of what it was, a decrease of 36 %.
5
NumericalMediumNCERT Exercise Set 14.2, Q5
A solid copper cube of edge 14 cm is melted down. All the copper is then cast into solid rods shaped as cylinders, each 14 cm long with radius 1 cm. Use $π \approx \frac{22}{7}$ .
1. Find the volume of the cube.
  cm³
  Show solution
  $14^{3} = 2744 c m^{3}$ .
2. Find the volume of one rod.
  cm³
  Show solution
  $\frac{22}{7} \times 1^{2} \times 14 = 44 c m^{3}$ .
3. How many complete rods can be made?
  rods
  Show solution
  $2744 \div 44 \approx 62.4$ , so 62 complete rods (with a little copper left over).

14.3 Cones

An ice-cream cone, a birthday cap, a funnel and a heap of grain on a threshing floor are all roughly cones (शंकु). Take a right-angled triangle and spin it once around one of its two shorter sides (Fig. 14.8). The other short side sweeps out a circle, the base. The long side sweeps out the slanting surface. The solid you get is a right circular cone.

Three lengths describe it. The radius $r$ belongs to the base. The height $h$ runs from the apex straight down to the centre of the base. The slant height $l$ (तिर्यक ऊँचाई) runs along the surface from the apex to the rim. These are the three sides of the spinning triangle, so the Baudhāyana–Pythagoras theorem links them:

Slant height of a cone

l^{2} = h^{2} + r^{2}

(14.9)

5/5Thin triangles of height $l$ with bases adding to $2 π r$ .

Fig. 14.8Spin a right triangle to get a cone; unroll the cone to get a sector of radius

l

and arc

2 π r

.NCERT Fig. 14.8, 14.9 · p. 130

Step 1 shows a right triangle with vertical side h, horizontal side r and slanting side l, standing on a dashed axis. Step 2 shows its mirror image dashed and a curved arrow around the base: the triangle is turning. Step 3 shows the finished cone with h, r and l labelled and the formula l² = h² + r². In step 4 a dashed seam is drawn down the cone and, on the right, the curved surface lies flat as a sector of a circle of radius l whose arc is 2πr long. In step 5 the sector is cut into many thin triangles, one of them shaded, and the formula CSA = ½ × 2πr × l = πrl appears.

Definition

Slant height

The distance from the apex of a cone to any point on the rim of its base, measured along the surface.

ExampleA cone with $r = 5$ and $h = 12$ has $l = 13$ .

Slit a paper party cap along one slant line and press it flat: you get a sector of a circle. The sector's radius is the slant height $l$ . Its curved edge was the rim of the base, so the arc is $2 π r$ long. Roll the sector up until its two straight edges meet, and the cone is back. The cut-out nets at the end of the textbook let you try this.

DerivationCurved surface area of a cone

First way: thin triangles. Slice the flattened sector, like a pizza, into $n$ narrow wedges (the last step of Fig. 14.8). When $n$ is huge, each wedge is practically a tall, thin triangle of height $l$ . Its short base is a tiny bit of the arc: call these bits $b_{1}, b_{2}, \dots, b_{n}$ . Adding the triangle areas gives the curved surface area: $A = \frac{1}{2} b_{1} l + \frac{1}{2} b_{2} l + \dots + \frac{1}{2} b_{n} l .$ Taking out the common factor, $A = \frac{1}{2} (b_{1} + b_{2} + \dots + b_{n}) l .$

The pieces $b_{1}, \dots, b_{n}$ make up the whole arc, which is $2 π r$ . So:

\begin{aligned} A & = \frac{1}{2} \times 2 π r \times l \\ = π r l \end{aligned}

Second way: the sector angle. Let the sector have angle $θ$ at its centre. From Part 1, a sector of radius $l$ has area $π l^{2} \times \frac{θ}{360}$ and arc length $2 π l \times \frac{θ}{360}$ . The arc must equal the rim of the base, $2 π r$ :

2 π l \times \frac{θ}{360} = 2 π r

so \frac{θ}{360} = \frac{r}{l}

Then the area of the sector is $π l^{2} \times \frac{r}{l} = π r l$ , the same answer. Adding the circular base gives the total surface area.

Curved surface area of a cone

CSA = π r l

(14.10)

Total surface area of a cone

\begin{aligned} TSA & = π r l + π r^{2} \\ = π r (l + r) \end{aligned}

(14.11)

Volume. Here is the surprise of the section: a cone holds exactly one third of the cylinder that fits around it (same base, same height).

Volume of a cone

\begin{aligned} V & = \frac{1}{3} π r^{2} h \\ = \frac{1}{3} \times A \times h \end{aligned}

(14.12)

where

$A$: the area of the base, $π r^{2}$

Why a third? A complete proof needs a limiting argument of the kind you will meet in calculus in Class 12. But you can check the rule with your hands in the activity below, and later in this chapter you will see the same $\frac{1}{3}$ appear for every pyramid.

ActivityNCERT Experiment (Fig. 14.10) · p. 132

Three cones fill one tin

You’ll need

an open tin
a sheet of thin card
tape and scissors
dry rice or sand
a ruler

Predict first

How many conefuls of rice will it take to fill the tin?

234about 1.5

Find a tin with an open top (a small powder or coffee tin works). Measure its radius and height.
Roll a sheet of card into a cone that matches the tin: its rim exactly as wide as the tin's mouth, and its tip exactly as deep as the tin is tall. Tape it and trim the rim level.
Fill the cone to the brim with dry rice or sand. Level the top with a ruler.
Pour it into the tin. Repeat, counting the conefuls, until the tin is full to the brim. Record your results.

Table 14.1

Conefuls poured	How full is the tin?
1
2
3

Your entries are saved automatically.

A second way. Roll a lump of modelling clay into a solid cylinder and note its radius and height. Squash it, then remould all of it into cones that each match the cylinder's radius and height. Count the cones you get.

3/4After two conefuls it is two thirds full.

Fig. 14.9Three conefuls fill a tin with the same radius and height.NCERT Fig. 14.10 · p. 132

On the left is a cone filled with yellow rice, point down, with radius r and height h marked. On the right is an empty tin with the same r and h and two dashed lines at one third and two thirds of its height. In step 2 an arrow shows the cone being poured and the rice in the tin rises to the first dashed line, labelled one third full. In step 3 it rises to the second line, two thirds full. In step 4 the tin is full to the brim after three cones.

Example 14.3NCERT Example 3 · p. 133

One triangle, two cones

A right triangle has sides 20 cm, 21 cm and 29 cm. It is spun through 360° about the side of 21 cm. Find the volume of the cone formed. What happens if it is spun about the 20 cm side instead? Use $π \approx \frac{22}{7}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example

A heap of wheat

Freshly harvested wheat lies in a conical heap 4.2 m across and 2.8 m high. Find (a) its slant height, (b) the area of tarpaulin needed to cover it, and (c) the volume of wheat. Use $π \approx \frac{22}{7}$ .

Try it yourself first. Then reveal the solution one step at a time.

Practice · 9 questions

Practice 14.3: questions with answers

NumericalEasyNCERT Exercise Set 14.3, Q1
A conical tent has slant height 25 m and base diameter 14 m. Find its total surface area, including the floor. Use $π \approx \frac{22}{7}$ .
m²
Show solution
$r = 7 m$ , so $TSA = \frac{22}{7} \times 7 \times (25 + 7) = 22 \times 32 = 704 m^{2} .$
NumericalEasyNCERT Exercise Set 14.3, Q2
Find the curved surface area of a cone with base radius 3.5 cm and slant height 12 cm. Use $π \approx \frac{22}{7}$ .
cm²
Show solution
$\frac{22}{7} \times 3.5 \times 12 = 11 \times 12 = 132 c m^{2} .$
3
NumericalMediumNCERT Exercise Set 14.3, Q3
A cone is 15 cm high and has base radius 8 cm. Find its curved and total surface areas. Use $π \approx 3.14$ .
1. curved surface area
  cm²
  Show solution
  $l = \sqrt{15^{2} + 8^{2}} = \sqrt{289} = 17 c m$ .
  CSA $= 3.14 \times 8 \times 17 = 427.04 c m^{2} .$
2. total surface area
  cm²
  Show solution
  TSA $= π r (l + r) = 3.14 \times 8 \times 25 = 628 c m^{2} .$
NumericalMediumNCERT Exercise Set 14.3, Q4
A cone of height 12 cm has volume $314 c m^{3}$ . Find the radius of its base. Use $π \approx 3.14$ .
cm
Show solution
$r^{2} = \frac{3 V}{π h} = \frac{942}{3.14 \times 12} = \frac{942}{37.68} = 25$ , so $r = 5 c m .$
5
NumericalMediumNCERT Exercise Set 14.3, Q5
A cone has slant height 25 cm, and its curved surface covers $550 c m^{2}$ . Use $π \approx \frac{22}{7}$ .
1. Find the radius of the base.
  cm
  Show solution
  $\frac{22}{7} \times r \times 25 = 550$ gives $r = \frac{550 \times 7}{550} = 7 c m .$
2. Find the total surface area.
  cm²
  Show solution
  TSA $= 550 + \frac{22}{7} \times 7^{2} = 550 + 154 = 704 c m^{2} .$
NumericalMediumNCERT Exercise Set 14.3, Q6
A party cap is a cone of base radius 3.5 cm and height 12 cm. How much card is needed for 20 such caps (curved surface only)? Use $π \approx \frac{22}{7}$ .
cm²
Show solution
$l = \sqrt{12^{2} + {3.5}^{2}} = \sqrt{144 + 12.25} = \sqrt{156.25} = 12.5 c m$ .
One cap: $\frac{22}{7} \times 3.5 \times 12.5 = 11 \times 12.5 = 137.5 c m^{2}$ .
Twenty caps: $20 \times 137.5 = 2750 c m^{2} .$
NumericalHardNCERT Exercise Set 14.3, Q7
A conical tent is 12 m high with base radius 5 m. It is made from canvas 2.5 m wide. Allowing an extra 0.36 m of canvas for seams and cutting, what length of canvas is needed? Use $π \approx 3.14$ .
m
Show solution
$l = \sqrt{12^{2} + 5^{2}} = 13 m$ , so CSA $= 3.14 \times 5 \times 13 = 204.1 m^{2}$ .
Length of canvas $= 204.1 \div 2.5 = 81.64 m$ .
With the allowance: $81.64 + 0.36 = 82 m .$
8
NumericalMediumNCERT Exercise Set 14.3, Q8
A right triangle with sides 9 cm, 12 cm and 15 cm is spun through 360° about its 12 cm side. How much space does the solid fill, and how much area does its curved surface have? Use $π \approx 3.14$ .
1. volume
  cm³
  Show solution
  The solid is a cone with $h = 12$ , $r = 9$ , $l = 15$ (cm).
  $V = \frac{1}{3} \times 3.14 \times 81 \times 12 = 3.14 \times 324 = 1017.36 c m^{3} .$
2. curved surface area
  cm²
  Show solution
  CSA $= 3.14 \times 9 \times 15 = 423.9 c m^{2} .$
NumericalHardNCERT Exercise Set 14.3, Q9
A drinking cup is a cone, point down. Water is poured in until it reaches two thirds of the cup's depth. What fraction of the cup is full?
$\frac{2}{3}$ $\frac{4}{9}$ $\frac{8}{27}$ $\frac{1}{3}$
Show solution
The water is a cone with the same shape, scaled down: height $\frac{2}{3} h$ and radius $\frac{2}{3} r$ (similar triangles).
Its volume is $\frac{1}{3} π (\frac{2}{3} r)^{2} (\frac{2}{3} h) = (\frac{2}{3})^{3} \times \frac{1}{3} π r^{2} h$ .
So the cup is $\frac{8}{27}$ full, under one third, even though the water is two thirds of the way up. Most of a cone's volume is near its wide end.

14.4 Pyramidal Shapes

To build a pyramid, start with a flat shape as the base: a triangle, a square, a hexagon or any other closed outline. Choose a point, the apex, outside the plane of that base, and fill in every straight line from the base to the apex. When the base is a polygon, the slanting sides are triangles that meet at the apex (Fig. 14.10).

The great pyramids of Egypt have square bases. The tall tower (vimana) over the sanctum of the Brihadeeswara temple in Thanjavur rises as a stepped pyramid. Some tea bags are folded as triangular pyramids. Even mountains can take this shape: where glaciers grind away a peak from several sides, they leave a sharp pyramid-like horn, such as Shivling in Uttarakhand.

3/3Unfold the square pyramid: one square and four triangles.

Fig. 14.10Any base, one apex: the volume uses base area and height; the surface is the sum of the faces.NCERT Fig. 14.11 · p. 134

Step 1 shows a pyramid on a square base, drawn in perspective, with its hidden edge dashed and its height h drawn as a dashed line from the apex straight down to the centre of the base. In step 2 a pyramid on a triangular base appears on the right, also with its height h dashed. In step 3 the triangular pyramid is replaced by the net of the square pyramid: a square labelled base with a triangle on each of its four sides, labelled base plus 4 triangles.

Definition

Pyramid

A solid made by joining every point of a flat base (a polygon or other closed shape) to one apex outside the plane of the base.

The volume of every pyramid follows the same rule as the cone, whatever the shape of its base:

Volume of a pyramid

V = \frac{1}{3} \times A \times h

(14.13)

where

$A$: the area of the base
$h$: the perpendicular height

Here $h$ is the perpendicular height, from the apex straight down to the plane of the base, not the length of a sloping edge. Compare (14.13) with (14.12): a cone is simply a pyramid whose base is a circle. Surface area needs nothing new: list the faces (the base and the triangles), find the area of each, and add.

Example

A pyramid tent

A tent is a square pyramid. Its floor is a square of side 10 m and its apex is 12 m above the centre of the floor. Find (a) the volume of air inside and (b) the canvas needed for the four sloping sides.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

A triangular pyramid has base area

30 c m^{2}

and height 10 cm. What is its volume?

cm³

Show answer

Answer 100 cm³

V = \frac{1}{3} \times 30 \times 10 = 100 c m^{3} .

14.5 Spheres and Hemispheres

Every point on the skin of a well-pumped football is the same distance from its centre. That single property defines a sphere (गोला): the set of all points in space at a fixed distance $r$ (the radius) from a fixed point $O$ (the centre). The same rule in a flat plane gives a circle. You can also get a sphere by spinning a circle about a diameter, just as spinning a right triangle gave a cone. Marbles, laddoos and ripe oranges are all close to spheres.

Definition

Sphere

The set of all points in space that are at the same distance $r$ (the radius) from a fixed point $O$ (the centre).

ExampleThe surface of a football.

Surface area. Archimedes found a beautiful way to see the surface of a sphere. Slide the sphere into the snuggest possible cylinder, one with radius $r$ and height $2 r$ (Fig. 14.11). His discovery: the ball's surface and the cylinder's curved side have equal areas. Any thin horizontal band of the sphere matches the band of the cylinder at the same height. The sphere's band is closer to the axis, so it is shorter around, but it slopes, so it is wider; the two effects cancel exactly. The cylinder's curved surface is $2 π r \times 2 r$ , so:

Surface area of a sphere

S = 4 π r^{2}

(14.14)

4/4So the sphere's area is $2 π r \times 2 r = 4 π r^{2}$ .

Fig. 14.11Archimedes: the sphere's surface equals the curved surface of the cylinder that just encloses it.NCERT Fig. 14.12 · p. 136

Step 1 shows a sphere of radius r with its centre and a dashed equator. In step 2 a cylinder is drawn around it, just touching it, with its height marked 2r; the text says a cylinder that just fits the ball. In step 3 a thin band of the sphere and the band of the cylinder at the same height are shaded orange, with the note that the two bands have the same area. In step 4 the shading goes and the result appears on the right: 2πr × 2r = 4πr².

Volume. Archimedes also found the volume of a sphere:

Volume of a sphere

V = \frac{4}{3} π r^{3}

(14.15)

You will derive it properly with calculus in later classes. But notice something: $\frac{4}{3} π r^{3} = \frac{1}{3} \times 4 π r^{2} \times r$ . That is one third of (surface area) × (radius), which looks very like the cone formula $\frac{1}{3} \times$ (base area) × (height).

DerivationThe volume of a sphere from thin cones

Mark out the ball's skin into thousands of small patches, as fine as you like, so that each one is practically flat. Join every patch to the centre $O$ with straight lines (Fig. 14.12). Each patch and its lines form a long, thin cone (strictly, a thin pyramid) whose apex is $O$ and whose height is the radius $r$ .

Together these cones make up the whole ball. If the patch areas are $a_{1}, a_{2}, \dots, a_{n}$ , then $V = \frac{1}{3} a_{1} r + \frac{1}{3} a_{2} r + \dots + \frac{1}{3} a_{n} r,$ which is $V = \frac{r}{3} (a_{1} + a_{2} + \dots + a_{n}) .$

The patches make up the whole surface, $4 π r^{2}$ , so:

\begin{aligned} V & = \frac{r}{3} \times 4 π r^{2} \\ = \frac{4}{3} π r^{3} \end{aligned}

3/3Their volume is $\frac{1}{3} \times$ (total base area) $\times r$ .

Fig. 14.12Every thin cone has height

r

, and their bases add up to the whole surface.NCERT Visualising Volume Formula · p. 138

Step 1 shows a sphere with centre O and one tiny patch on its surface shaded orange, joined to O to form a thin cone of height r; the text says a tiny patch plus a thin cone to O. In step 2 the whole ball is split into many thin wedges of alternating colour, all meeting at O, with the text many thin cones fill the whole ball. In step 3 the formula V = ⅓ × 4πr² × r = 4⁄3 πr³ appears on the right.

Example

Leather and air in a football

A football has radius 11 cm. Find (a) the area of leather on its surface and (b) the volume of air inside, ignoring the thickness of the leather. Use $π \approx 3.14$ .

Try it yourself first. Then reveal the solution one step at a time.

14.5.1 Hands-on activity to verify the formula for surface area of a sphere

ActivityNCERT 14.5.1 Activity · p. 138

Peel an orange

You’ll need

a round orange (or a rubber ball and thick string)
two rulers
paper and a compass

Predict first

How many circles of the orange's radius will its peel fill?

about 2about 3about 4about 6

Choose an orange that is as round as possible. Measure its diameter with two rulers held upright on either side, and halve it to get $r$ .
On a sheet of paper, draw several circles of radius $r$ .
Peel the orange carefully in small pieces. Lay the pieces flat inside the circles, filling one circle before starting the next, with as few gaps as you can.
Count how many circles the peel fills.

Table 14.2

Radius r (cm)	Circles filled by the peel

Your entries are saved automatically.

Another way. Push a pin into the top of a rubber ball. Starting at the pin, wind thick string round and round the ball until it is covered, holding it in place with pins. Unwind it, and coil the string inside circles of the ball's radius.

3/3The peel fills about four circles: area $\approx 4 π r^{2}$ .

Fig. 14.13The peel of an orange fills about four circles of the orange's radius.NCERT Fig. 14.14 · p. 139

Step 1 shows an orange with its radius r marked. In step 2 four empty circles of the same radius r appear in a two by two grid, labelled 4 circles, radius r. In step 3 the orange is shown peeled, as a dashed outline, and the four circles are filled with flat pieces of orange peel.

14.5.2 Hemisphere

Slice a ball exactly through its centre and you get two identical halves, each a hemisphere (अर्धगोला). A solid half-ball has two surfaces (Fig. 14.14). One is the curved half of the ball's skin. The other is the flat circle where the knife went through, called its base, with the same radius $r$ . A katori, a half coconut shell, the dome of an observatory and the great dome of Gol Gumbaz in Vijayapura are all close to hemispheres. Geography uses the word too: the equator divides the Earth into the Northern and Southern Hemispheres.

Definition

Hemisphere

Half of a sphere, cut off by a plane through the centre. It has a curved surface and a flat circular base.

3/3Curved part $2 π r^{2}$ , flat base $π r^{2}$ .

Fig. 14.14A hemisphere has half the sphere's curved surface plus one flat circle.NCERT Fig. 14.15 · p. 139

Step 1 shows a sphere with a dashed circle around its middle where it will be cut. In step 2 the top half lifts away, revealing the flat circular face of the bottom half, coloured blue. In step 3 leader lines and labels name the parts: the curved surface of a hemisphere is 2πr², the flat base is πr², and the total surface area is 3πr².

The curved part is half of the sphere's surface, and the solid hemisphere is half of the ball:

Curved surface of a hemisphere

CSA = 2 π r^{2}

(14.16)

Solid hemisphere

\begin{aligned} TSA & = 2 π r^{2} + π r^{2} \\ = 3 π r^{2} \end{aligned}

(14.17)

Volume of a hemisphere

V = \frac{2}{3} π r^{3}

(14.18)

Be careful with a hollow bowl: its inside or outside is just the curved surface $2 π r^{2}$ , because a bowl has no flat lid. Use $3 π r^{2}$ only for a solid hemisphere.

Example 14.4NCERT Example 4 · p. 140

A katori of dal

A steel katori is a hemisphere with inner radius 7 cm. How much dal can it hold when full to the brim? Use $π \approx \frac{22}{7}$ and give the answer in millilitres ( $1 c m^{3} = 1$ mL).

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

A ball has radius 7 cm. How much area does its surface cover? Use

π \approx \frac{22}{7}

cm²

Show answer

Answer 616 cm²

4 π r^{2} = 4 \times \frac{22}{7} \times 49 = 616 c m^{2} .

Practice · 8 questions

Practice 14.4: questions with answers

NumericalEasyNCERT Exercise Set 14.4, Q1
A steel marble has radius 1.4 cm. Find its surface area. Use $π \approx \frac{22}{7}$ .
cm²
Show solution
$4 \times \frac{22}{7} \times {1.4}^{2} = 4 \times \frac{22}{7} \times 1.96 = 4 \times 6.16 = 24.64 c m^{2} .$
NumericalHardNCERT Exercise Set 14.4, Q2
Two solid spheres are made of the same brass. One weighs 200 g and the other 5400 g. The smaller one has diameter 4 cm. Find the radius of the larger one.
cm
Show solution
The mass ratio is $5400 \div 200 = 27$ , so the volume ratio is 27.
Since $V \propto r^{3}$ , the radius ratio is $\sqrt[3]{27} = 3$ .
The smaller radius is 2 cm, so the larger radius is $3 \times 2 = 6 c m .$
NumericalMediumNCERT Exercise Set 14.4, Q3
The diameter of Mars is roughly half the diameter of the Earth. Treating both as spheres, find the ratio of the surface area of Mars to that of the Earth.
Show solution
Answer1 : 4
If Mars has radius $r$ , the Earth has radius $2 r$ .
Ratio $= 4 π r^{2} : 4 π (2 r)^{2} = 1 : 4$ . (Their volumes are in the ratio $1 : 8$ .)
4
NumericalEasyNCERT Exercise Set 14.4, Q4
A solid hemisphere has radius 14 cm. Use $π \approx \frac{22}{7}$ .
1. Find its curved surface area.
  cm²
  Show solution
  $2 \times \frac{22}{7} \times 196 = 1232 c m^{2} .$
2. Find its total surface area.
  cm²
  Show solution
  $3 \times \frac{22}{7} \times 196 = 1848 c m^{2} .$
NumericalMediumNCERT Exercise Set 14.4, Q5
Six solid balls, each of radius 3 cm, are packed in one layer in a box whose inside measures 18 cm × 12 cm × 6 cm. The box is then filled up with oil. Find the volume of oil, to the nearest $c m^{3}$ . Use $π \approx 3.14$ .
cm³
Show solution
Box: $18 \times 12 \times 6 = 1296 c m^{3}$ . (The balls fit as $3 \times 2 \times 1$ , each in a 6 cm cube.)
Balls: $6 \times \frac{4}{3} \times 3.14 \times 27 = 678.24 c m^{3}$ .
Oil $= 1296 - 678.24 = 617.76 \approx 618 c m^{3} .$
6
NumericalMediumNCERT Exercise Set 14.4, Q6
A sphere's radius grows by 30 %.
1. By what percentage does its surface area increase?
  %
  Show solution
  ${1.3}^{2} = 1.69$ , an increase of 69 %.
2. By what percentage does its volume increase?
  %
  Show solution
  ${1.3}^{3} = 2.197$ , an increase of 119.7 %: the volume more than doubles.
NumericalHardNCERT Exercise Set 14.4, Q7
A spherical balloon slowly loses air and its volume falls by 48.8 %. Assuming it stays a sphere, by what percentage has its radius fallen?
%
Show solution
$(new r \div old r)^{3} = 1 - 0.488 = 0.512$ .
$\sqrt[3]{0.512} = 0.8$ , so the radius is 80 % of before: it has fallen by 20 %.
NumericalHardNCERT Exercise Set 14.4, Q8
The outside of a hemispherical dome is to be painted. The circumference of its base is 44 m, and painting costs ₹5 per $100 c m^{2}$ . Find the cost. Use $π \approx \frac{22}{7}$ .
₹
Show solution
$2 \times \frac{22}{7} \times r = 44$ , so $r = 7 m$ .
Curved surface $= 2 \times \frac{22}{7} \times 49 = 308 m^{2}$ .
$1 m^{2} = 10 000 c m^{2}$ , so the rate is ₹5 × 100 = ₹500 per $m^{2}$ .
Cost $= 308 \times 500 =$ ₹1,54,000.

14.6 Areas and Volumes Around Us

Look around your home and school with the five families of solids in mind. A water tank on the roof may be a cuboid or a cylinder. An LPG cylinder is a cylinder with rounded ends. A clay matka is close to a sphere, a funnel is a cone, and a steel almirah is a cuboid. Real objects are rarely perfect, but a good model gets you a useful answer quickly.

ActivityNCERT Activity · p. 141

How much does it hold?

You’ll need

a measuring tape
a ruler
a notebook

Pick four containers at home, for example a pressure cooker, a bucket, a matka and the inside of a fridge (or an overhead tank, with an adult's help).
Decide which solid each is closest to, and measure what you need (radius, height, length and so on) in centimetres.
Work out each volume in $c m^{3}$ , then change it to litres using $1000 c m^{3} = 1$ litre.
If the label or bill gives the capacity, compare it with your estimate.

Table 14.3

Object	Model solid	Measurements (cm)	Volume (litres)
Pressure cooker
Bucket
Matka
Fridge (inside)

Your entries are saved automatically.

14.6.1 Guesstimates: did you ever wonder?

How many litres of water does your school use in a day? How many grains are in a kilo of rice? Questions like these have no exact data, but you can still reason your way to a sensible number. This is a guesstimate: an estimate built from simple models, reasonable assumptions and rounded numbers. Two careful people can end with different numbers, and both can be sensible. A guesstimate is judged by how clearly its steps can be checked, not by matching an answer key.

Definition

Guesstimate

A sensible value for something we cannot measure exactly, reasoned out from a simple model, stated assumptions and rounded numbers.

A reliable routine has five moves:

Guess before any working, and write the guess down.
Model: replace the real object with shapes you can measure.
Assume: list every size or rate you had to make up.
Calculate with round numbers, keeping the units straight.
Look back: compare with your guess, and ask which assumption would change the answer most.

Example 14.5NCERT Example 5 · p. 141

Laddoos from a vat

A sweet shop has a cylindrical vat of laddoo mixture with radius 20 cm, filled to a depth of 30 cm. About how many laddoos can be made from it? Guess first: tens, hundreds or thousands?

Try it yourself first. Then reveal the solution one step at a time.

Example 14.6NCERT Example 6 · p. 142

Footballs in a storeroom

A school storeroom is 4 m long, 3 m wide and 2.5 m high. About how many footballs of radius 11 cm could be packed into it? Guess first.

Try it yourself first. Then reveal the solution one step at a time.

2/2Nestling rows into the hollows fits 48 balls in the same box.

Fig. 14.15Seen from the side: nestling each row into the hollows below fits more balls in the same box.NCERT Fig. 14.17 · p. 143

A rectangular box seen from the side. In step 1 it holds circles in straight rows and columns, 10 across and 4 high, labelled square rows: 40 balls, with clear gaps between the balls. In step 2 every second row is shifted by half a ball so each ball sits in the hollow between two below; the rows sit closer together and 5 rows fit, labelled nestled rows: 48 balls.

The gaps also explain why we measure volume with cubes, not balls. Unit cubes fit together with no space left over, so counting them gives the volume exactly. Spheres always leave gaps, however cleverly you stack them: at best they fill about 74 % of the space.

Quick check

1 of 3

How many litres are there in

1 m^{3}

litres

Show answer

Answer 1000 litres

1 m^{3} = 100 \times 100 \times 100 = 10^{6} c m^{3}

, and

1000 c m^{3} = 1

litre, so

1 m^{3} = 1000

litres.

End-of-chapter exercises

For every guesstimate below, write down a quick guess before you calculate, and list your assumptions.

Table 14.4All the formulas of this chapterNCERT Chapter Summary · p. 147

Solid	Surface area	Volume
Cuboid	$2 (l w + l h + w h)$	$l w h$
Cube	$6 a^{2}$	$a^{3}$
Cylinder	CSA $2 π r h$ ; TSA $2 π r (h + r)$ ; one end open $π r (2 h + r)$	$π r^{2} h$
Cone	CSA $π r l$ ; TSA $π r (l + r)$	$\frac{1}{3} π r^{2} h$
Pyramid	sum of the face areas	$\frac{1}{3} \times$ base area $\times h$
Sphere	$4 π r^{2}$	$\frac{4}{3} π r^{3}$
Hemisphere	CSA $2 π r^{2}$ ; TSA $3 π r^{2}$	$\frac{2}{3} π r^{3}$

Recap

Math of Space: Surface Area and Volume: key points and summary

Surface area measures the outside of a solid (square units); volume measures the space it fills (cubic units); see Wrapping and filling.
Cuboid: TSA $= 2 (l w + l h + w h)$ , $V = l w h$ ; cube: $6 a^{2}$ and $a^{3}$ ; see Section 14.1.
Equal volumes can have different surface areas; the cube needs the least; see Example 14.1.
A right circular cylinder's axis is perpendicular to its circular ends; slices across it are circles; see Section 14.2.
Cylinder: CSA $= 2 π r h$ , TSA $= 2 π r (h + r)$ , one end open $π r (2 h + r)$ , $V = π r^{2} h$ ; see Section 14.2.1.
Cone: $l^{2} = h^{2} + r^{2}$ , CSA $= π r l$ , TSA $= π r (l + r)$ , $V = \frac{1}{3} π r^{2} h$ ; see Section 14.3.
Any pyramid: $V = \frac{1}{3} \times$ base area $\times$ height; surface area = sum of the faces; see Section 14.4.
Sphere: $S = 4 π r^{2}$ (Archimedes' cylinder) and $V = \frac{4}{3} π r^{3} = \frac{1}{3} \times S \times r$ ; see Section 14.5.
An orange's peel fills about four circles of its radius; see Section 14.5.1.
Hemisphere: CSA $2 π r^{2}$ , solid TSA $3 π r^{2}$ , $V = \frac{2}{3} π r^{3}$ ; see Section 14.5.2.
Scaling every length by $k$ multiplies areas by $k^{2}$ and volumes by $k^{3}$ ; see Practice 14.4.
Guesstimates: guess, model, assume, calculate, look back; volume ÷ object volume gives only an upper bound; see Section 14.6.1.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise Set 14.1, Q1	Practice 14.1 · Q1
NCERT Exercise Set 14.1, Q2	Practice 14.1 · Q2
NCERT Exercise Set 14.1, Q3	Practice 14.1 · Q3
NCERT Exercise Set 14.1, Q4	Practice 14.1 · Q4
NCERT Exercise Set 14.1, Q5	Practice 14.1 · Q5
NCERT Exercise Set 14.1, Q6	Practice 14.1 · Q6
NCERT Exercise Set 14.1, Q7	Practice 14.1 · Q7
NCERT Exercise Set 14.1, Q8	Practice 14.1 · Q8
NCERT Exercise Set 14.2, Q1	Practice 14.2 · Q1
NCERT Exercise Set 14.2, Q2	Practice 14.2 · Q2
NCERT Exercise Set 14.2, Q3	Practice 14.2 · Q3
NCERT Exercise Set 14.2, Q4	Practice 14.2 · Q4
NCERT Exercise Set 14.2, Q5	Practice 14.2 · Q5
NCERT Exercise Set 14.3, Q1	Practice 14.3 · Q1
NCERT Exercise Set 14.3, Q2	Practice 14.3 · Q2
NCERT Exercise Set 14.3, Q3	Practice 14.3 · Q3
NCERT Exercise Set 14.3, Q4	Practice 14.3 · Q4
NCERT Exercise Set 14.3, Q5	Practice 14.3 · Q5
NCERT Exercise Set 14.3, Q6	Practice 14.3 · Q6
NCERT Exercise Set 14.3, Q7	Practice 14.3 · Q7
NCERT Exercise Set 14.3, Q8	Practice 14.3 · Q8
NCERT Exercise Set 14.3, Q9	Practice 14.3 · Q9
NCERT Exercise Set 14.4, Q1	Practice 14.4 · Q1
NCERT Exercise Set 14.4, Q2	Practice 14.4 · Q2
NCERT Exercise Set 14.4, Q3	Practice 14.4 · Q3
NCERT Exercise Set 14.4, Q4	Practice 14.4 · Q4
NCERT Exercise Set 14.4, Q5	Practice 14.4 · Q5
NCERT Exercise Set 14.4, Q6	Practice 14.4 · Q6
NCERT Exercise Set 14.4, Q7	Practice 14.4 · Q7
NCERT Exercise Set 14.4, Q8	Practice 14.4 · Q8
NCERT End-of-chapter Exercises Q1	End-of-chapter exercises · Q1
NCERT End-of-chapter Exercises Q2	End-of-chapter exercises · Q2
NCERT End-of-chapter Exercises Q3	End-of-chapter exercises · Q3
NCERT End-of-chapter Exercises Q4	End-of-chapter exercises · Q4
NCERT End-of-chapter Exercises Q5	End-of-chapter exercises · Q5
NCERT End-of-chapter Exercises Q6	End-of-chapter exercises · Q6
NCERT End-of-chapter Exercises Q7	End-of-chapter exercises · Q7
NCERT End-of-chapter Exercises Q8	End-of-chapter exercises · Q8
NCERT End-of-chapter Exercises Q9	End-of-chapter exercises · Q9
NCERT End-of-chapter Exercises Q10	End-of-chapter exercises · Q10
NCERT End-of-chapter Exercises Q11	End-of-chapter exercises · Q11
NCERT End-of-chapter Exercises Q12	End-of-chapter exercises · Q12
NCERT End-of-chapter Exercises Q13	End-of-chapter exercises · Q13
NCERT End-of-chapter Exercises Q14	End-of-chapter exercises · Q14
NCERT End-of-chapter Exercises Q15	End-of-chapter exercises · Q15
NCERT End-of-chapter Exercises Q16	End-of-chapter exercises · Q16
NCERT End-of-chapter Exercises Q17	End-of-chapter exercises · Q17
NCERT End-of-chapter Exercises Q18	End-of-chapter exercises · Q18
NCERT End-of-chapter Exercises Q19	End-of-chapter exercises · Q19
NCERT End-of-chapter Exercises Q20	End-of-chapter exercises · Q20
NCERT End-of-chapter Exercises Q21	End-of-chapter exercises · Q21
NCERT End-of-chapter Exercises Q22	End-of-chapter exercises · Q22

Quick check: MCQs on Math of Space: Surface Area and Volume

14.1 Cuboids and Cubes3 questions
14.2.1 Another way of understanding why the above formulas are true4 questions
14.3 Cones4 questions
14.4 Pyramidal Shapes3 questions
14.5.2 Hemisphere4 questions
14.6.1 Guesstimates: did you ever wonder?3 questions

Glossary · 14

Key terms (English–Hindi)

coneशंकुshanku: A solid with a circular base joined to one apex; a right circular cone has its apex directly above the centre.First seen in the chapter →
cross-sectionअनुप्रस्थ काटanuprasth kaat: The flat shape made by one thin, straight slice through a solid.First seen in the chapter →
cubeघनghan: A cuboid whose length, width and height are equal.First seen in the chapter →
cuboidघनाभghanaabh: A box-shaped solid with six rectangular faces meeting at right angles.First seen in the chapter →
curved surface areaवक्र पृष्ठीय क्षेत्रफलvakra prishthiya kshetrafal: The area of only the curved part of a cylinder, cone or sphere.First seen in the chapter →

guesstimateअनुमानanumaan: A reasoned estimate made with a simple model and stated assumptions when exact data are missing.First seen in the chapter →

hemisphereअर्धगोलाardhgola: Half of a sphere cut by a plane through its centre.First seen in the chapter →

pyramidपिरामिडpiraamid: A solid with a flat base whose every point is joined to one apex outside the base.First seen in the chapter →

right circular cylinderलंब वृत्तीय बेलनlamb vrittiya belan: A solid with two equal parallel circular ends whose axis is perpendicular to them.First seen in the chapter →

slant heightतिर्यक ऊँचाईtiryak oonchaai: The distance from a cone's apex to its rim along the surface: $l = \sqrt{h^{2} + r^{2}}$ .First seen in the chapter →
sphereगोलाgola: All points in space at a fixed distance $r$ from a centre.First seen in the chapter →
surface areaपृष्ठीय क्षेत्रफलprishthiya kshetrafal: The total area of the outside of a solid, in square units.First seen in the chapter →

total surface areaसंपूर्ण पृष्ठीय क्षेत्रफलsampoorn prishthiya kshetrafal: The area of every face of a solid, flat and curved, added together.First seen in the chapter →

volumeआयतनaayatan: The amount of space a solid occupies or a container holds, in cubic units.First seen in the chapter →

Math of Space: Surface Area and Volume: Class 9 Ganita Manjari · Chapter 14

What you will learn in Math of Space: Surface Area and Volume

Wrapping and filling

14.1 Cuboids and Cubes

Practice 14.1: questions with answers

14.2 Right Circular Cylinder

14.2.1 Another way of understanding why the above formulas are true

Practice 14.2: questions with answers

14.3 Cones

Practice 14.3: questions with answers

14.4 Pyramidal Shapes

14.5 Spheres and Hemispheres

14.5.1 Hands-on activity to verify the formula for surface area of a sphere

14.5.2 Hemisphere

Practice 14.4: questions with answers

14.6 Areas and Volumes Around Us

14.6.1 Guesstimates: did you ever wonder?

End-of-chapter exercises

Practice questions with answers (NCERT “End-of-chapter exercises”)

Math of Space: Surface Area and Volume: key points and summary

Find your NCERT exercise question

Quick check: MCQs on Math of Space: Surface Area and Volume

Key terms (English–Hindi)

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