NCERT Solutions for Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume – step-by-step answers to every exercise and in-text question of the chapter, as per the latest NCERT syllabus for 2027. Read every answer online, free.
Chapter 14 moves from flat shapes to solids. It finds how much surface covers a solid and how much space it fills, for cuboids and cubes, right circular cylinders, cones, pyramids, spheres and hemispheres. Each formula is explained by a model — a stack of cards, unrolled paper, cones filled with salt, string wound on a ball. By the end you can use these formulas to solve problems, compare shapes when lengths are scaled, and make sensible guesstimates of everyday volumes.
NCERT Solutions for Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume – Questions and Answers
All 61 questions of the chapter (textbook pages Part II, 123 – 148) are answered below, in the order of the book – in-text questions, activities and the end-of-chapter exercises. Each answer is written the way you would write it in the exam.
In-text Questions — Page 124Section 14.1 Cuboids and Cubes
The two formulas have the same pattern. The volume formula just has one more factor, the height.
Volume of a cuboid = l × w × h = (area of the base rectangle) × h (cubic units)
Area counts the unit squares that cover a flat shape. Volume counts the unit cubes that fill a solid. A cuboid is a pile of h layers, and each layer is a rectangle of l × w unit cubes. For example, a 4 cm × 3 cm rectangle has area 12 cm², and a 4 cm × 3 cm × 5 cm cuboid has volume 12 × 5 = 60 cm³.
Think and Reflect — Page 124Section 14.1 Cuboids and Cubes
A photograph of a pack of playing cards: one neat stack of cards lies on a table, and a few loose cards are spread out beside it. The stack has the shape of a cuboid made of many thin, identical rectangular cards lying flat on top of one another.
Think of the cuboid as a pack of n identical thin cards, each a rectangle of length l and width w, and each of very small thickness t.
Volume of the pack = n × lw × t = lw × (n × t)
But n × t = height of the pack = h
So volume = lw × h = area of base × height
Every card has exactly the same area as the base, so stacking cards only multiplies the base area by the height. A taller pack has more cards and so more volume, in exact proportion to its height.
For a cube of side a:
Volume = a × a × a = a³
A cube has 6 faces, and each one is a square of area a². For example, a cube of side 3 cm has surface area 6 × 9 = 54 cm² and volume 27 cm³.
In-text Questions — Page 124Section 14.1 Cuboids and Cubes — the cube as a special cuboid
A cube is a cuboid whose length, width and height are all equal. Put l = w = h = a in the cuboid formulas:
V = whl = a · a · a = a³
So the cube formulas are not new formulas. They are the cuboid formulas with all three edges equal: the three different pairs of faces become three equal pairs of squares.
Exercise Set 14.1 — Pages 125–126Section 14.1 Cuboids and Cubes
TSA = 6a² = 6 × 4² = 6 × 16 = 96 cm²
The total surface area is 96 cm².
First use the same unit: 2 m = 200 cm.
Number of cubes = 10 × 10 × 10 = 1000
1000 small cubes can be packed.
Line the boxes up along each edge of the godown:
Along 25 m: 25 ÷ 1.25 = 20 boxes
Along 10 m: 10 ÷ 1 = 10 boxes
Number of boxes = 20 × 20 × 10 = 4000
The godown holds 4000 boxes.
Resulting cuboid: l = 5 + 5 = 10 cm, w = 5 cm, h = 5 cm
TSA = 2(lw + wh + hl) = 2(10×5 + 5×5 + 5×10)
= 2(50 + 25 + 50) = 2 × 125 = 250 cm²
The surface area of the cuboid is 250 cm².
Number of small cubes = 4 × 4 × 4 = 64
Surface area of one small cube = 6 × 1² = 6 cm²
Total surface area of the small cubes = 64 × 6 = 384 cm²
Ratio = 96 : 384 = 1 : 4
The ratio is 1 : 4. The volume stays 64 cm³, but cutting makes 4 times as much surface.
The three faces at a corner are l × w, w × h and h × l. Let lw = 6, wh = 15, hl = 10.
l²w²h² = 900
(lwh)² = 900 ⇒ lwh = 30
The volume of the cuboid is 30 cm³.
(i) exactly three faces painted?
(ii) exactly two faces painted?
(iii) exactly one face painted?
(iv) no face painted?
There are 5 × 5 × 5 = 125 small cubes in all.
(ii) Two faces painted: cubes on an edge but not at a corner
each edge has 5 − 2 = 3 such cubes, 12 edges → 12 × 3 = 36
(iii) One face painted: inner cubes of each face
each face has 3 × 3 = 9, 6 faces → 6 × 9 = 54
(iv) No face painted: the hidden inner cube 3 × 3 × 3 → 27
Check: 8 + 36 + 54 + 27 = 125.
(i) Is there more than one such cuboid?
(ii) Can you find them all?
(iii) Show that you have found them all.
2(lw + wh + hl) = 100, so lw + wh + hl = 50. Take l ≥ w ≥ h ≥ 1.
(i) Yes. (ii) Exactly two: 16 × 2 × 1 and 7 × 4 × 2 (in cm).
(iii) Each product is at least h², so 3h² ≤ 50 and h ≤ 4. Adding h² to both sides of lw + h(l + w) = 50 gives
h = 1: 51 = 3 × 17 → w = 2, l = 16
h = 2: 54 = 6 × 9 → w = 4, l = 7 (2 × 27 and 3 × 18 have a factor below 4)
h = 3: 59 is prime → none
h = 4: 66 = 6 × 11, 3 × 22, 2 × 33, 1 × 66 → no factor reaches 8 → none
All cases are checked, so there are no others.
Think and Reflect — Page 126Section 14.1 Cuboids and Cubes
| Pool A | Pool B | |
|---|---|---|
| Depth | 8 ft ≈ 2.4 m | 6 ft ≈ 1.8 m |
| Water (volume) | 8 × 50 × 20 = 8000 ft³ ≈ 2,26,400 L | 6 × 100 × 15 = 9000 ft³ ≈ 2,54,700 L |
| Water surface (land needed) | 50 × 20 = 1000 ft² | 100 × 15 = 1500 ft² |
| Area to tile (floor + 4 walls) | 1000 + 2(50 + 20) × 8 = 2120 ft² | 1500 + 2(100 + 15) × 6 = 2880 ft² |
Option B suits a school better. 8 ft is far above the head of a Class 9 student (about 5 ft tall), which is unsafe for learners; B is 2 ft shallower. B is also 100 ft (about 30 m) long, good for swimming laps and races, and its larger surface lets more students swim at once. The price is about 28,300 L more water, 50% more land and more tiling.
Parameters considered: safety (depth), purpose (lessons, laps), water needed, land available, and cost of digging and tiling.
Exercise Set 14.2 — Pages 129–130Section 14.2 Right Circular Cylinder
Let cylinder A have radius r and height h. Then B has radius 2r and height h/2.
CSA of B = 2π(2r)(h/2) = 2πrh
CSA of A : CSA of B = 1 : 1
Volume of A = πr²h
Volume of B = π(2r)²(h/2) = π × 4r² × h/2 = 2πr²h
Volume of A : Volume of B = 1 : 2
Let the radii be 2k and 3k, and the heights 3m and 2m.
= 12k²m : 18k²m = 2 : 3
(b) CSA₁ : CSA₂ = 2π(2k)(3m) : 2π(3k)(2m)
= 12km : 12km = 1 : 1
The largest cylinder stands on one face of the cube. Its circular base just touches the four sides of that square face, so its diameter is r and its radius is r/2. Its height is the edge, r.
That is about 0.785 r³, roughly 78.5% of the cube.
Let the original radius be r and height h. New radius = 1.1r, new height = h(1 − x/100).
1.21 × (1 − x/100) = 1
1 − x/100 = 1/1.21 = 100/121
x/100 = 21/121
x = 2100/121 = 1743⁄121 ≈ 17.36
The height must decrease by about 17.36% (exactly 2100/121 %).
(i) the volume of the cube,
(ii) the volume of one cylindrical rod,
(iii) the approximate number of complete cylindrical rods that can be formed. (π ≈ 22/7)
(ii) Volume of one rod = πr²h = 22/7 × 2² × 12 = 1056/7 ≈ 150.86 cm³
(iii) Number of rods = 1728 ÷ (1056/7) = 1728 × 7/1056 ≈ 11.45
So 11 complete rods can be formed. The metal left over (about 68.6 cm³) is not enough for a twelfth rod.
Exercise Set 14.3 — Pages 133–134Section 14.3 Cones (use π ≈ 22/7 or π ≈ 3.14)
TSA = πr(l + r) = 22/7 × 12 × (21 + 12)
= 22/7 × 12 × 33 = 8712/7 ≈ 1244.57 m²
The total surface area is about 1244.57 m².
The curved surface area is 220 cm².
CSA = πrl = 3.14 × 12 × 20 = 753.6 cm²
TSA = πr(l + r) = 3.14 × 12 × (20 + 12) = 3.14 × 384 = 1205.76 cm²
CSA = 753.6 cm² and TSA = 1205.76 cm².
1570 = (1/3) × 3.14 × r² × 15 = 15.7 r²
r² = 1570 ÷ 15.7 = 100
r = 10 cm
The radius of the base is 10 cm.
22/7 × r × 14 = 308 ⇒ 44r = 308 ⇒ r = 7 cm
(ii) TSA = CSA + area of base = 308 + 22/7 × 7² = 308 + 154 = 462 cm²
A cap has no base, so only its curved surface is made of sheet.
CSA of one cap = πrl = 22/7 × 7 × 25 = 550 cm²
Sheet for 10 caps = 10 × 550 = 5500 cm²
The sheet required is 5500 cm² (0.55 m²).
The tent has no floor, so the tarpaulin covers only the curved surface.
CSA = πrl = 3.14 × 6 × 10 = 188.4 m²
Length of tarpaulin = area ÷ width = 188.4 ÷ 3 = 62.8 m
Add 20 cm = 0.2 m for margins and wastage: 62.8 + 0.2 = 63 m
The length of tarpaulin required is 63 m.
6² + 8² = 36 + 64 = 100 = 10², so the right angle is between the 6 cm and 8 cm sides. Turning about the 8 cm side gives a cone with h = 8 cm, r = 6 cm, l = 10 cm.
CSA = πrl = 3.14 × 6 × 10 = 60π ≈ 188.4 cm²
Volume ≈ 301.44 cm³, curved surface area ≈ 188.4 cm².
Let the cup have radius r and depth h. The water forms a smaller cone of the same shape with depth h/2. By similar triangles its radius is r/2.
= (1/8) × volume of the cup
The water fills only 1/8 of the cup.
In-text Question — Page 137Section 14.5 Spheres and Hemispheres
Compare with the cone formula, V = (1/3) × area of base × height. The two match if the sphere is built out of many thin cones:
Divide the surface of the sphere into very tiny pieces. Join every piece to the centre O. Each piece becomes the base of a thin cone whose apex is O and whose height is the radius r.
= (1/3) × b₁ × r + (1/3) × b₂ × r + … + (1/3) × bₙ × r
= (1/3) × (b₁ + b₂ + … + bₙ) × r
= (1/3) × (surface area of sphere) × r
= (1/3) × 4πr² × r = (4/3)πr³
Exercise Set 14.4 — Pages 140–141Section 14.5 Spheres and Hemispheres
The surface area is 6.16 cm².
Both spheres are of the same metal, so weight is proportional to volume.
(4/3)πR³ : (4/3)πr³ = 8 : 1 ⇒ R³ / r³ = 8 ⇒ R/r = 2
r = 5 ÷ 2 = 2.5 cm ⇒ R = 2 × 2.5 = 5 cm
The radius of the larger sphere is 5 cm.
If the moon’s diameter is 1/4 of the earth’s, its radius is also 1/4 of the earth’s: r(moon) = R/4.
= 1/16 : 1 = 1 : 16
(ii) TSA = 3πr² = 3 × 22/7 × 21 × 21 = 3 × 22 × 3 × 21 = 4158 cm²
Volume of one sphere = (4/3)π × 2³ = 32π/3 cm³
Volume of 16 spheres = 16 × 32π/3 = 512π/3 ≈ 512 × 3.14 ÷ 3 ≈ 535.9 cm³ (π ≈ 3.14)
Volume of liquid = 1024 − 535.9 ≈ 488.1 cm³
The volume of the liquid is about 488 cm³.
Let the radius be r. The new radius is 1.1r.
Increase = 1.331V − V = 0.331V
Percentage increase = 0.331 × 100 = 33.1%
So the volume increases by 33.1%, as required.
(1 + x/100)³ = 1.728
1 + x/100 = ∛1.728 = 1.2 (since 12³ = 1728)
x/100 = 0.2 ⇒ x = 20
The radius was increased by 20%.
Area to paint = CSA = 2πr² = 2 × 22/7 × 5.6 × 5.6 = 197.12 m²
= 197.12 × 10,000 cm² = 19,71,200 cm²
Cost = 19,71,200 ÷ 100 × ₹10 = ₹1,97,120
The cost of painting the dome is ₹1,97,120.
In-text Questions — Page 141Section 14.6 Areas and Volumes Around Us
| Shape | Objects and structures around us |
|---|---|
| Cube | dice, Rubik’s cube, ice cubes, sugar cubes, some cardboard boxes |
| Cuboid | bricks, books, matchbox, almirah, classroom, shoe box, packet of milk or juice, water tank on a roof (some) |
| Cylinder | water pipes, gas cylinder, drums, candles, pillars, tumblers, tube light, coins (very short cylinders) |
| Cone | ice-cream cone, birthday cap, funnel, traffic cone, the roof of a round hut |
| Sphere / hemisphere | balls, marbles, globe, laddoo, oranges; bowls, domes and igloos are hemispheres |
| Pyramid | the pyramids of Egypt, a tetrahedral puzzle, some roofs and temple towers, tents with a square base, peaks such as Shivling |
Method: measure each object with a measuring tape, model it as a cylinder or a cuboid, use V = πr²h (π ≈ 3.14) or V = lwh, and divide cm³ by 1000 to get litres.
| Object (sample measurements) | Model and working | Volume |
|---|---|---|
| Gas cylinder: diameter 30 cm, height of the straight part 50 cm | 3.14 × 15² × 50 = 35,325 cm³ | ≈ 35 L |
| Bucket: top diameter 30 cm, bottom 24 cm, height 30 cm | cylinder of average diameter 27 cm: 3.14 × 13.5² × 30 ≈ 17,168 cm³ | ≈ 17 L |
| Steel vessel: diameter 20 cm, height 15 cm | 3.14 × 10² × 15 = 4710 cm³ | ≈ 4.7 L |
| Cupboard: 90 cm × 50 cm × 180 cm | 90 × 50 × 180 = 8,10,000 cm³ | 810 L |
| Cylindrical water tank: diameter 110 cm, height 120 cm | 3.14 × 55² × 120 = 11,39,820 cm³ | ≈ 1140 L |
These are sample values: a tape and a formula are enough to estimate any container.
In-text Question — Page 143Section 14.6.1 Guesstimates: Did You Ever Wonder? (Example 6, tennis balls in a classroom)
Three pictures, labelled A, B and C. Each shows identical blue balls in two layers on a flat square base; in each layer the balls touch in straight rows, 3 balls by 2. In A every ball of the upper layer rests directly on top of a ball below. In B the whole upper layer is shifted sideways by half a ball, so each upper ball rests in the groove between two balls below. In C the balls look the most tightly packed: the upper layer nestles into the hollows of the layer below.
Count as the book does for the first configuration (room 900 × 900 × 450 cm, ball diameter 6 cm), but the layers now sit lower.
Layers = (450 − 6) ÷ 5.2 + 1 ≈ 86, alternately 150 × 150 and 149 × 150 balls
Total ≈ 43 × 22,500 + 43 × 22,350 ≈ 19,28,550
Third (C): balls rest in the hollows of four; layer gap = √(6² − (3√2)²) ≈ 4.24 cm
Layers = (450 − 6) ÷ 4.24 + 1 ≈ 105, alternately 150 × 150 and 149 × 149 balls
Total ≈ 53 × 22,500 + 52 × 22,201 ≈ 23,46,950
About 19 lakh balls (second) and 23 lakh (third): more than 16.9 lakh, less than the 33.5 lakh bound.
End-of-Chapter Exercises — Pages 144–147Whole chapter (for guesstimate problems, make a guess first before solving)
Guess first: perhaps about 50.
Assume a scoop is a ball of diameter 5 cm (r = 2.5 cm)
Volume of a scoop = (4/3) × 3.14 × 2.5³ ≈ 65.4 cm³
Number of scoops ≈ 3000 ÷ 65.4 ≈ 46
About 45 scoops can be obtained (a little is lost in the corners).
= (a³ + 3a² + 3a + 1) − a³
= 3a² + 3a + 1
Picture it: add three square slabs a × a × 1 on three faces (3a²), three rods a × 1 × 1 along the edges where the slabs meet (3a), and one corner cube (1).
(i) Could a person drink enough water in a lifetime to fill an entire room the size of your classroom?
(ii) Estimate the number of bricks used to build the walls of your classroom.
Sample classroom: 8 m × 6 m × 3 m.
Lifetime drinking ≈ 2 L a day × 365 × 70 years ≈ 51,100 L
No. A lifetime of drinking fills only about one-third of the room. It would take about 200 years.
Minus 1 door (1 × 2.1) and 3 windows (1.5 × 1.2 each): 84 − 7.5 = 76.5 m²
Wall volume (thickness 20 cm) = 76.5 × 0.2 = 15.3 m³ = 1,53,00,000 cm³
One brick with mortar ≈ 20 × 10 × 10 = 2000 cm³
Bricks ≈ 1,53,00,000 ÷ 2000 ≈ 7650
About 7,500 to 8,000 bricks.
(B) 100 × 10³ = 100 × 1000 = 1,00,000 mm³ = 100 cm³
Option A gives more chocolate, 25 cm³ (25%) more, even though B “sounds” bigger.
Guess first, then model it.
Space for one standing person in a tight crowd ≈ 50 cm × 50 cm = 0.25 m²
Area ≈ 8 × 10⁹ × 0.25 = 2 × 10⁹ m²
1 km² = 10⁶ m² ⇒ area ≈ 2000 km²
All of humanity would fit, standing, in about 2000 km²: a square about 45 km on each side (√2000 ≈ 45), about one and a third times the area of the National Capital Territory of Delhi (about 1,480 km²).
Milk in one glass = πr²h = 22/7 × 3.5 × 3.5 × 12 = 462 cm³
For 1600 students = 1600 × 462 = 7,39,200 cm³
= 7,39,200 ÷ 1000 = 739.2 L
739.2 litres of milk are needed.
CSA of cone = 100π ÷ 5 = 20π cm²
π × 4 × l = 20π ⇒ l = 5 cm
h = √(l² − r²) = √(25 − 16) = 3 cm
V = (1/3)πr²h = (1/3) × π × 16 × 3 = 16π ≈ 50.29 cm³
The height of the cone is 3 cm and its volume is 16π ≈ 50.29 cm³ (π ≈ 22/7).
(i) the ratio of the volume of Jupiter to the volume of the Earth;
(ii) the ratio of the volume of the Sun to the volume of the Earth.
(Calculator can be used.)
The ratio of volumes of two spheres is the cube of the ratio of their radii, since (4/3)π cancels.
(ii) 6,95,700 ÷ 6370 ≈ 109.2 ⇒ ratio ≈ 109.2³ ≈ 13,03,000 : 1
About 1300 Earths would fit inside Jupiter, and about 13 lakh Earths inside the Sun (by volume).
The smallest cylinder that encloses a sphere of radius r fits it tightly: its radius is r and its height is the diameter, 2r.
(2/3) × 2πr³ = (4/3)πr³ = volume of sphere
Hence volume of sphere = (2/3) × volume of the enclosing cylinder.
Repeat this for: (i) the Moon (ii) Jupiter (iii) a volleyball. Are you surprised by the three answers? Why or why not?
Let the radius be R (in metres) and the gap be x.
Second string = 2π(R + x) = 2πR + 1
2πx = 1 ⇒ x = 1/(2π) ≈ 0.159 m ≈ 16 cm
The second string is about 16 cm above the ground. R cancelled out, so the answer is the same 16 cm for (i) the Moon, (ii) Jupiter and (iii) a volleyball.
Most people are surprised. 1 m is a tiny addition to the Earth’s 40,000 km equator, so we expect a microscopic gap; for a volleyball it more than doubles the string, so we expect a big gap. But circumference grows by exactly 2π for every unit of radius, whatever the size of the circle.
The square has its corners on the circle, so its diagonal is a diameter, 2r. If its side is a, then a² + a² = (2r)².
Volume of pyramid = (1/3) × 2r² × h = (2/3)r²h
Volume of cylinder = πr²h
Ratio = (2/3)r²h : πr²h = 2 : 3π (≈ 0.21)
The pyramid fills about 21% of the cylinder.
(i) the length of a cuboid with dimensions l, w, h is increased by 1 unit?
(a) 1 cubic unit (b) (lwh + 1) cubic unit (c) wh cubic units (d) lw cubic units (e) lh cubic units
(ii) the radius of a cylinder with dimensions r, h is increased by 1 unit?
(a) 1 cubic unit (b) πr²h cubic units (c) πr² cubic units (d) 2πrh + 2πh cubic units (e) 2πrh + πh cubic units
*(iii) the radius of a sphere is decreased by 1 unit?
(ii) π(r + 1)²h − πr²h = πh(r² + 2r + 1 − r²) = πh(2r + 1)
= 2πrh + πh → (e)
(iii) (4/3)πr³ − (4/3)π(r − 1)³
= (4/3)π[r³ − (r³ − 3r² + 3r − 1)]
= (4/3)π(3r² − 3r + 1)
= 4πr² − 4πr + (4/3)π cubic units (a decrease)
S = 6a² = 6(∛V)²
S = 6(∛V)² = 6V2/3
V = a³ = (√(S/6))³ = (S/6)√(S/6)
V = (S/6)√(S/6) = (S/6)3/2
The marbles sink and push the water up. Their total volume must equal the volume of water needed to raise the level by 20 − 16 = 4 cm.
Volume of one marble = (4/3)π × 1³ = (4/3)π cm³
Number of marbles = 64π ÷ (4/3)π = 64 × 3/4 = 48
The crow should drop 48 marbles.
Method: take out the refill. The ink fills a long thin cylinder inside it. Measure the length of the ink column with a ruler and estimate the inner diameter of the tube (look at the open end, or subtract the wall from the outer diameter).
V = πr²h = 3.14 × 0.1 × 0.1 × 10 = 0.314 cm³
A new pen holds about 0.3 cm³ of ink, i.e. about 0.3 millilitre (1 cm³ = 1 mL): only a few drops.
(i) A ball of chapati dough of radius 6 cm is prepared. Estimate how many chapatis can be made from it?
*(ii) Cut a coconut/muskmelon in half and find out the approximate volume of edible coconut flesh/fruit by taking the necessary measurements.
Assume one chapati is a disc of diameter 18 cm (r = 9 cm) and thickness 2 mm (0.2 cm)
One chapati = π × 9² × 0.2 = 16.2π cm³ ≈ 51 cm³
Number ≈ 288π ÷ 16.2π ≈ 17.8
(i) About 17 or 18 chapatis.
(ii) Half a coconut’s flesh is a hemispherical shell: a big hemisphere minus the hollow inside. Sample: outer radius of the flesh R = 5.5 cm, thickness 1 cm, so inner radius r = 4.5 cm.
= (2/3) × 3.14 × 75.25 ≈ 157.5 cm³
About 160 cm³ of flesh in half a coconut (sample values).
(i) If all the pages of this textbook were laid out side-by-side on the floor would they cover the entire classroom floor?
(ii) What is the maximum number of textbooks that can fit in an empty storeroom of dimensions 15 ft × 20 ft × 30 ft?
Sample measurements: a page is about 20 cm × 27 cm; the book has about 150 printed pages (this chapter ends on page 148), i.e. about 75 sheets; the book is about 1 cm thick.
Even counting all 150 pages separately: 150 × 0.054 ≈ 8.1 m²
Classroom floor (8 m × 6 m) = 48 m²
(i) No. The pages would cover only about one-sixth of the floor.
One book = 20 × 27 × 1 = 540 cm³
Upper bound = 25,48,50,000 ÷ 540 ≈ 4,72,000
Packing books upright (457 cm ÷ 1, 610 cm ÷ 20, 914 cm ÷ 27): 457 × 30 × 33 = 4,52,430
(ii) About 4.5 lakh textbooks.
Guess first. Then model each standing person as needing a box 0.5 m × 0.5 m × 2 m = 0.5 m³ (people stand on many floors, as in the picture).
Volume needed ≈ 8 × 10⁹ × 0.5 = 4 × 10⁹ m³
Side = ∛(4 × 10⁹) ≈ 1587 m
The cube would need a side of about 1.6 km.
(i) Write an expression that gives the thickness of this water layer.
(ii) Simplify the expression in (i) using a calculator.
(i) The layer is very thin compared with the Earth, so its volume ≈ surface area × thickness.
(ii)
t ≈ 1.38 × 10⁹ ÷ 5.10 × 10⁸ ≈ 2.7 km
The water layer would be about 2.7 km thick.
We need a cuboid that can be reshaped so that its surface area doubles while its volume halves. One answer: a cuboid 14 cm × 14 cm × 8 cm, reshaped to 28 cm × 28 cm × 1 cm.
S = 2(14×14 + 14×8 + 8×14) = 2(196 + 112 + 112) = 840 cm²
New: V = 28 × 28 × 1 = 784 cm³ = 1568 ÷ 2 (halved)
S = 2(28×28 + 28×1 + 1×28) = 2(784 + 28 + 28) = 1680 cm² = 2 × 840 (doubled)
In general, an a × a × (4a/7) cuboid works: doubling the length and width and making the height one-eighth halves the volume and doubles the surface area.
Plan: treat each room as a cuboid, measure its length and width (pace it out or use a tape), take the ceiling height, find each volume, and add.
| Room (sample flat, ceiling 3 m) | Floor area | Volume (m³) |
|---|---|---|
| Living room | 5 × 4 = 20 m² | 20 × 3 = 60 |
| Bedroom 1 | 4 × 3.5 = 14 m² | 42 |
| Bedroom 2 | 3.5 × 3 = 10.5 m² | 31.5 |
| Kitchen | 3 × 2.5 = 7.5 m² | 22.5 |
| Two bathrooms | 2 × (2 × 1.5) = 6 m² | 18 |
| Total | 58 m² | 174 m³ |
Sample result: the inside of the house is about 174 m³ (1,74,000 litres).
Approximations: rooms taken as perfect cuboids; wall thickness, cupboards and the balcony ignored; one ceiling height for all rooms.
Chapter 14 of 14 – Maths
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Also for This Chapter
NCERT Solutions Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume – Overview
| Class | Class 9 |
| Subject | Maths |
| Chapter | 14 – Math of Space: Surface Area and Volume |
| Textbook pages | Part II, 123 – 148 |
| Questions solved | 61 |
| Session | 2027 |
| Format | Full answers on this page |
NCERT Solutions Class 9 Maths Chapter 14 – FAQs
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Is this for the 2027 session?
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All 61 questions of the chapter (textbook pages Part II, 123 – 148) – the in-text questions, activities and the end-of-chapter exercises.
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