Question 1
A cube has volume 64 cm³. Find its total surface area.
Steps only
a³ = 64, so a = 4 cm.
TSA = 6a² = 6 × 4² = 96 cm².
Answer
96 cm².
CLASS 9 · GANITA MANJARI PART II · CHAPTER 14
Math of Space: Surface Area and Volume
Use surface-area and volume formulas for cuboids, cylinders, cones, pyramids, spheres and real-life estimates.
69 worked answersAll 5 exercise setsAll 22 end-of-chapter questions
What you will learn
- Select the correct surface-area or volume formula.
- Explain changes caused by scaling and cutting.
- Make and communicate reasonable volume estimates.
69 worked answers · short steps first, explanations when you need them
Exercise Set 14.1
Textbook pages 125–126
Question 2
How many cubes of side 20 cm fit tightly in a cubical box of side 2 m?
Steps only
2 m = 200 cm, so 200 ÷ 20 = 10 cubes fit along each edge.
Total = 10³ = 1000.
Answer
1000 small cubes.
Question 3
A 40 m × 25 m × 10 m godown is filled with 2 m × 1.25 m × 1 m boxes. Find the number of boxes.
Steps only
Godown volume = 40 × 25 × 10 = 10,000 m³.
One box = 2 × 1.25 × 1 = 2.5 m³.
Number = 10,000 ÷ 2.5 = 4000.
Answer
4000 boxes.
Question 4
Two cubes of volume 125 cm³ are joined face-to-face. Find the surface area of the resulting cuboid.
Steps only
Cube edge = ∛125 = 5 cm.
The new cuboid is 10 cm × 5 cm × 5 cm.
TSA = 2(10×5 + 5×5 + 10×5) = 250 cm².
Answer
250 cm².
Question 5
A 4 cm cube is cut into 1 cm cubes. Find the ratio of the original surface area to the total surface area of all small cubes.
Steps only
Original area = 6 × 4² = 96 cm².
There are 4³ = 64 small cubes; their total area = 64 × 6 = 384 cm².
Ratio = 96:384 = 1:4.
Answer
1:4.
Question 6
Three adjacent face areas of a cuboid are 6 cm², 15 cm² and 10 cm². Find its volume.
Steps only
Let the edges be l, w, h: lw = 6, wh = 15 and lh = 10.
(lwh)² = (lw)(wh)(lh) = 6×15×10 = 900.
lwh = √900 = 30 cm³.
Answer
30 cm³.
Question 7(i)
A painted 5 cm cube is cut into 1 cm cubes. How many have exactly three painted faces?
Steps only
Only the eight corner cubes touch three outer faces.
Answer
8 cubes.
Question 7(ii)
How many small cubes have exactly two painted faces?
Steps only
Each of 12 edges has 5 − 2 = 3 non-corner cubes.
12 × 3 = 36.
Answer
36 cubes.
Question 7(iii)
How many small cubes have exactly one painted face?
Steps only
Each face has (5 − 2)² = 9 cubes away from its edges.
6 × 9 = 54.
Answer
54 cubes.
Question 7(iv)
How many small cubes have no painted face?
Steps only
The unpainted interior is (5 − 2) cubes long in each direction.
3³ = 27.
Answer
27 cubes.
Question 8(i)
Find an integer-edged cuboid with total surface area 100 cm².
Steps only
TSA = 2(lw + wh + lh), so lw + wh + lh = 50.
For (l,w,h) = (1,2,16): 2 + 32 + 16 = 50.
Answer
One answer is 1 cm × 2 cm × 16 cm.
Question 8(ii)
Is there another integer-edged cuboid with total surface area 100 cm²?
Steps only
For (2,4,7): lw + wh + lh = 8 + 28 + 14 = 50.
Thus TSA = 2 × 50 = 100 cm².
Answer
Yes: 2 cm × 4 cm × 7 cm.
Question 8(iii)
Show that all integer-edged cuboids of total surface area 100 cm² have been found.
Steps only
Order edges as l ≤ w ≤ h. Then lw + wh + lh = 50.
For fixed l,w, h = (50 − lw)/(l+w) must be an integer at least w.
Checking possible l gives only (1,2,16) and (2,4,7), up to rearrangement.
Answer
The complete list, apart from changing edge order, is 1×2×16 and 2×4×7 cm.
Exercise Set 14.2
Textbook pages 129–130
Question 1
Cylinder B has twice A’s radius and half A’s height. Compare curved areas and volumes.
Steps only
CSA ratio A:B = rh : (2r)(h/2) = 1:1.
Volume ratio A:B = r²h : (2r)²(h/2) = 1:2.
Answer
Curved surface areas = 1:1; volumes = 1:2.
Question 2(a)
Cylinder radii are 2:3 and heights are 3:2. Find their volume ratio.
Steps only
V₁:V₂ = r₁²h₁:r₂²h₂ = 2²×3 : 3²×2.
12:18 = 2:3.
Answer
2:3.
Question 2(b)
For the same cylinders, find the curved surface area ratio.
Steps only
CSA₁:CSA₂ = r₁h₁:r₂h₂ = 2×3 : 3×2 = 1:1.
Answer
1:1.
Question 3
A cube has edge r cm. Find the volume of the largest right circular cylinder cut from it.
Steps only
The circular base is inscribed in a square of side r, so cylinder radius = r/2.
Cylinder height = r.
V = π(r/2)²r = πr³/4.
Answer
πr³/4 cm³.
Question 4
A cylinder radius rises by 10% while height falls by x%; volume is unchanged. Find x.
Steps only
New volume factor = 1.1²(1 − x/100) = 1.
1 − x/100 = 1/1.21 = 100/121.
x = 2100/121 ≈ 17.36.
Answer
x ≈ 17.36%.
Question 5(i)
Find the volume of a metallic cube of side 12 cm.
Steps only
V = 12³ = 1728 cm³.
Answer
1728 cm³.
Question 5(ii)
Find the volume of one rod of radius 2 cm and height 12 cm.
Steps only
V = πr²h = π×2²×12 = 48π.
Using π ≈ 22/7 gives 1056/7 ≈ 150.86 cm³.
Answer
48π cm³, approximately 150.86 cm³.
Question 5(iii)
How many complete cylindrical rods can be made from the cube?
Steps only
1728 ÷ (1056/7) = 126/11 ≈ 11.45.
Only complete rods count, so take 11.
Answer
11 complete rods.
Exercise Set 14.3
Textbook pages 133–134
Question 1
Find the total surface area of a cone with slant height 21 m and base diameter 24 m.
Steps only
r = 12 m.
TSA = πr(l+r) = π×12×33 = 396π.
Using π = 22/7 gives about 1244.57 m².
Answer
396π m² ≈ 1244.57 m².
Question 2
Find the curved surface area for l = 10 cm and r = 7 cm.
Steps only
CSA = πrl = π×7×10 = 70π.
Using π = 22/7, CSA = 220 cm².
Answer
220 cm².
Question 3
A cone has h = 16 cm and r = 12 cm. Find curved and total surface areas.
Steps only
l = √(16²+12²) = √400 = 20 cm.
CSA = π×12×20 = 240π cm².
TSA = π×12×(20+12) = 384π cm².
Answer
CSA = 240π cm² ≈ 754.29 cm²; TSA = 384π cm² ≈ 1206.86 cm².
Question 4
A cone has height 15 cm and volume 1570 cm³. Find its base radius using π = 3.14.
Steps only
1570 = (1/3)×3.14×r²×15 = 15.7r².
r² = 100, so r = 10 cm.
Answer
10 cm.
Question 5(i)
A cone has curved area 308 cm² and slant height 14 cm. Find its radius.
Steps only
308 = πr×14.
With π = 22/7, 308 = 44r, so r = 7 cm.
Answer
7 cm.
Question 5(ii)
Find the total surface area of that cone.
Steps only
TSA = πr(l+r) = (22/7)×7×(14+7).
TSA = 462 cm².
Answer
462 cm².
Question 6
A conical cap has r = 7 cm and h = 24 cm. Find sheet area for 10 caps.
Steps only
l = √(7²+24²) = 25 cm.
One cap uses CSA = π×7×25 = 550 cm².
Ten caps use 5500 cm².
Answer
5500 cm².
Question 7
Find the length of 3 m wide tarpaulin for a tent with h = 8 m, r = 6 m, plus 20 cm wastage.
Steps only
l = √(8²+6²) = 10 m.
Tent area = πrl = 60π ≈ 188.57 m².
Length = 188.57÷3 + 0.20 ≈ 63.06 m.
Answer
Approximately 63.06 m of tarpaulin.
Question 8
A 6–8–10 right triangle rotates about its 8 cm side. Find the cone’s volume and curved area.
Steps only
The cone has r = 6 cm, h = 8 cm and l = 10 cm.
V = (1/3)π×6²×8 = 96π cm³.
CSA = π×6×10 = 60π cm².
Answer
Volume = 96π cm³ ≈ 301.71 cm³; curved area = 60π cm² ≈ 188.57 cm².
Question 9
A conical cup is filled to half its depth. What fraction of the cup’s volume contains water?
Steps only
At half the height, similarity gives water-surface radius R/2.
Volume scale factor = (1/2)³ = 1/8.
Answer
1/8 of the cup’s volume.
Exercise Set 14.4
Textbook pages 140–141
Question 1
Find the surface area of a ball bearing of radius 0.7 cm.
Steps only
A = 4πr² = 4×(22/7)×0.7².
A = 6.16 cm².
Answer
6.16 cm².
Question 2
Same-metal spheres weigh 5920 g and 740 g. The smaller diameter is 5 cm. Find the larger radius.
Steps only
Volume ratio = weight ratio = 5920:740 = 8:1.
Radius ratio = ∛8:∛1 = 2:1.
Smaller radius = 2.5 cm, so larger radius = 5 cm.
Answer
5 cm.
Question 3
The Moon’s diameter is about one fourth Earth’s. Find their surface-area ratio.
Steps only
Radius ratio Moon:Earth = 1:4.
Area ratio = 1²:4² = 1:16.
Answer
1:16.
Question 4(i)
Find the curved surface area of a hemisphere of radius 21 cm.
Steps only
CSA = 2πr² = 2×(22/7)×21² = 2772 cm².
Answer
2772 cm².
Question 4(ii)
Find the total surface area of the same hemisphere.
Steps only
TSA = 3πr² = 3×(22/7)×21² = 4158 cm².
Answer
4158 cm².
Question 5
Sixteen spheres of radius 2 cm are placed in a 16×8×8 cm box. Find the preservative-liquid volume.
Steps only
Box volume = 16×8×8 = 1024 cm³.
Sphere volume total = 16×(4/3)π×2³ = 512π/3 ≈ 536.17 cm³.
Liquid = 1024 − 536.17 ≈ 487.83 cm³.
Answer
Approximately 488 cm³.
Question 6
Show the volume change when a sphere’s radius increases by 10%.
Steps only
New radius factor = 1.10.
New volume factor = 1.10³ = 1.331.
Increase = (1.331−1)×100% = 33.1%.
Answer
The volume increases by 33.1%.
Question 7
A sphere’s volume increases by 72.8%. Find the percentage radius increase.
Steps only
New volume factor = 1.728.
Since 1.728 = 1.2³, the new radius factor is 1.2.
Radius increase = 20%.
Answer
x = 20%.
Question 8
A hemispherical dome has base circumference 35.2 m. Painting costs ₹10 per 100 cm². Find the cost.
Steps only
2πr = 35.2 gives r = 5.6 m.
Curved area = 2πr² = 197.12 m² = 1,971,200 cm².
Cost = 1,971,200÷100×₹10 = ₹197,120.
Answer
₹197,120.
End-of-Chapter Exercises
Textbook pages 144–147
Question 1
Estimate scoops from a 10 cm × 15 cm × 20 cm ice-cream container.
Steps only
Container volume = 3000 cm³.
Assume one scoop is a sphere of radius 2.5 cm: volume ≈ 65.45 cm³.
3000÷65.45 ≈ 45.8, so estimate about 46 scoops.
Answer
About 46 medium scoops under the stated assumption.
Question 2
How many unit cubes must be added to grow a cube of side a into side a+1?
Steps only
Added cubes = (a+1)³ − a³.
Expand: a³+3a²+3a+1−a³.
Answer
3a² + 3a + 1 unit cubes.
Question 3(i)
Could one person’s lifetime water intake fill a classroom?
Steps only
Assume 2 L/day for 75 years: 2×365×75 = 54,750 L = 54.75 m³.
A 8 m × 6 m × 3 m room holds 144 m³.
54.75 < 144.
Answer
Under these reasonable assumptions, no; it would fill about 38% of the room.
Question 3(ii)
Estimate the bricks in classroom walls.
Steps only
Assume room 8×6×3 m, wall thickness 0.23 m, openings 12 m².
Wall volume ≈ [2(8+6)×3−12]×0.23 = 16.56 m³.
Using about 0.002 m³ per brick including mortar gives 16.56÷0.002 ≈ 8280.
Answer
About 8,300 bricks for the stated model.
Question 4
Compare one 50 mm chocolate cube with one hundred 10 mm cubes.
Steps only
Option A volume = 50³ = 125,000 mm³.
Option B volume = 100×10³ = 100,000 mm³.
Answer
Option A gives 25,000 mm³, or 25%, more chocolate.
Question 5
Estimate the area occupied if the world population stands together.
Steps only
Assume 8 billion people and 4 people per m².
Area = 8,000,000,000÷4 = 2,000,000,000 m².
Since 1 km² = 1,000,000 m², area ≈ 2000 km².
Answer
About 2000 km² under these assumptions.
Question 6
Find milk needed for 1600 glasses of diameter 7 cm filled to height 12 cm.
Steps only
r = 3.5 cm. One glass = π×3.5²×12 = 147π = 462 cm³.
For 1600: 462×1600 = 739,200 cm³.
Divide by 1000 to convert to litres.
Answer
739.2 litres.
Question 7
A radius-5 sphere has five times the curved area of a radius-4 cone. Find the cone’s height and volume.
Steps only
Sphere area = 4π×5² = 100π, so cone CSA = 20π.
π×4×l = 20π gives l = 5 cm; h = √(5²−4²) = 3 cm.
V = (1/3)π×4²×3 = 16π cm³.
Answer
Height = 3 cm; volume = 16π cm³ ≈ 50.27 cm³.
Question 8(i)
Approximate the ratio of Jupiter’s volume to Earth’s using radii 69,900 km and 6370 km.
Steps only
Ratio = (69900/6370)³.
Calculator value ≈ 1321.34.
Answer
Approximately 1321:1.
Question 8(ii)
Approximate the ratio of the Sun’s volume to Earth’s using radii 695,700 km and 6370 km.
Steps only
Ratio = (695700/6370)³.
Calculator value ≈ 1,302,709.90.
Answer
Approximately 1,302,710:1.
Question 9
Show that a sphere’s volume is two thirds that of its smallest enclosing cylinder.
Steps only
The cylinder has radius r and height 2r, so Vc = πr²(2r) = 2πr³.
Sphere Vs = (4/3)πr³.
Vs/Vc = (4/3)/2 = 2/3.
Answer
Vsphere = (2/3)Vcylinder.
Question 10
A circular string is lengthened by 1 m and held uniformly above a sphere. Find the height for Earth, Moon, Jupiter and a volleyball.
Steps only
2π(R+h) − 2πR = 1.
2πh = 1, so h = 1/(2π) ≈ 0.159 m.
Answer
About 15.9 cm for every one of the four spheres.
Question 11
Find the volume ratio of a square pyramid fitted inside a radius-r, height-h cylinder as described.
Steps only
The square diagonal is 2r, so its side is r√2 and base area is 2r².
Pyramid volume = (1/3)×2r²h; cylinder volume = πr²h.
Ratio = 2/(3π).
Answer
Pyramid:cylinder = 2:3π, or 2/(3π):1.
Question 12(i)
A cuboid’s length l increases by 1 while w,h stay fixed. Find the volume change.
Steps only
Change = (l+1)wh − lwh = wh.
Answer
wh cubic units (option c).
Question 12(ii)
A cylinder’s radius r increases by 1 while h stays fixed. Find the volume change.
Steps only
Change = π(r+1)²h − πr²h.
= πh(2r+1) = 2πrh + πh.
Answer
2πrh + πh cubic units (option e).
Question 12(iii)
A sphere’s radius r decreases by 1. Find the change in volume.
Steps only
Old minus new = (4π/3)[r³−(r−1)³].
= (4π/3)(3r²−3r+1).
Answer
The volume decreases by (4π/3)(3r² − 3r + 1) cubic units.
Question 13
A cube has volume V. Express total surface area S in terms of V.
Steps only
a³ = V, so a = V^(1/3).
S = 6a² = 6V^(2/3).
Answer
S = 6V^(2/3).
Question 14
A cube has total surface area S. Express volume V in terms of S.
Steps only
a² = S/6, so a = √(S/6).
V = a³ = (S/6)^(3/2).
Answer
V = (S/6)^(3/2).
Question 15
A radius-4 cm glass must rise from 16 cm to 20 cm using radius-1 cm marbles. How many?
Steps only
Required displaced volume = π×4²×4 = 64π cm³.
One marble volume = (4/3)π×1³ = 4π/3 cm³.
Number = 64π ÷ (4π/3) = 48.
Answer
48 marbles, assuming they are fully submerged and the glass does not overflow.
Question 16
Estimate the ink volume in a new ballpoint refill.
Steps only
Example measurements: inner radius ≈ 1 mm = 0.1 cm; ink length ≈ 10 cm.
Model ink as a cylinder: V ≈ π×0.1²×10 = 0.1π cm³.
Answer
About 0.31 mL for the example measurements.
Question 17(i)
Estimate chapatis from a spherical dough ball of radius 6 cm.
Steps only
Dough volume = (4/3)π×6³ = 288π ≈ 905 cm³.
Assume one chapati is a cylinder of radius 7.5 cm and thickness 0.4 cm: volume ≈ π×7.5²×0.4 ≈ 70.7 cm³.
905÷70.7 ≈ 12.8.
Answer
About 13 chapatis under the stated size assumption.
Question 17(ii)
Describe how to estimate edible volume in half a coconut or muskmelon.
Steps only
Measure outer half as a hemisphere of radius R.
Measure shell or rind thickness t, so edible inner radius is about R−t.
Edible volume ≈ (2/3)π(R−t)³; subtract any hollow or seed-cavity volume.
Answer
Use (2/3)π(R−t)³ minus the measured cavity volume.
Question 18(i)
Would all pages of this textbook laid side-by-side cover a classroom floor?
Steps only
Assume 160 leaves, each 0.28 m × 0.22 m: total one-sided area ≈ 9.86 m².
Assume classroom floor 8 m × 6 m = 48 m².
9.86 < 48.
Answer
No under these assumptions; the pages cover about one fifth of the floor.
Question 18(ii)
Estimate the maximum number of textbooks in a 15 ft × 20 ft × 30 ft empty storeroom.
Steps only
Room volume = 9000 ft³ ≈ 254.85 m³.
Assume one book is 0.28×0.22×0.015 m, volume 0.000924 m³.
254.85÷0.000924 ≈ 275,800.
Answer
An ideal volume upper bound is about 276,000 books under these assumptions.
Question 19
Estimate the side of a cube that could contain the entire human population.
Steps only
Assume 8 billion people and 0.075 m³ per tightly packed person.
Total volume ≈ 600,000,000 m³.
Cube side ≈ ∛600,000,000 ≈ 843 m.
Answer
About 0.84 km per side under the stated assumptions.
Question 20(i)
Write an expression for the uniform depth of 1.38 billion km³ of water around Earth of radius 6371 km.
Steps only
Let depth be t km. Shell volume is (4π/3)[(6371+t)³−6371³].
Set this equal to 1.38×10⁹ km³.
t = [6371³ + 3(1.38×10⁹)/(4π)]^(1/3) − 6371.
Answer
t = [6371³ + 3(1.38×10⁹)/(4π)]^(1/3) − 6371 km.
Question 20(ii)
Calculate the water-layer depth from the expression.
Steps only
Calculator evaluation gives t ≈ 2.7044 km.
The thin-layer check V/(4πR²) gives about 2.7055 km, which is close.
Answer
Approximately 2.70 km, or 2704 m.
Question 21
Give cuboid dimensions illustrating a halved volume together with doubled surface area.
Steps only
Start with a 10×10×10 cuboid: V = 1000 and TSA = 600.
A cuboid about 23.615×23.615×0.896 has V ≈ 500.
Its TSA ≈ 1200.
Answer
One approximate pair is 10×10×10 and 23.615×23.615×0.896 units.
Question 22
Explain how to estimate the volume of a house.
Steps only
Split the usable interior into cuboidal rooms and any simple roof solids.
Measure each part; calculate cuboid volumes lwh and suitable roof volumes.
Add the parts and clearly list spaces or wall thicknesses omitted.
Answer
House volume ≈ the sum of the measured component volumes.