Chapter 14: Math of Space: Surface Area and Volume

Class 9 Ganita Manjari Part 2 · notes, exercises and mini test.

Learning map

What this chapter is teaching

Three-dimensional solids occupy volume in space and are bounded by surfaces. This chapter derives the total surface area and volume of cuboids, cylinders, cones, pyramids, spheres, and hemispheres, and applies them to real-world geometric problems and scientific guesstimates.

Prisms & CuboidsBase area times height: V = lwh and V = π r² h. Unrolling gives net surface areas.
Cones & PyramidsOne-third of the enclosing prism: V = ⅓ × base area × h. Slant height l = √(h² + r²).
Spheres & HemispheresArchimedes' circumscribing cylinder theorem: SA = 4π r² and V = ⁴⁄₃π r³.
Guesstimates & ScalingModeling real scenarios with reasonable assumptions, packing ratios, and scale factors.
Study tip: Pay close attention to unit consistency — convert all dimensions (cm, m, ft) to the same unit before calculating.
Coverage check: Sections 1–10 cover all textbook definitions, Examples 1–6, Archimedes' theorems, cone decompositions, and real-life modeling.

1Cuboids and Cubes

A cuboid is a three-dimensional rectangular solid bounded by six rectangular faces. It is characterized by three independent perpendicular dimensions: length (l), width (w), and height (h).

  • Total Surface Area (TSA): Summing the areas of the 3 pairs of identical opposite faces:
    TSA = 2(lw + wh + hl)
  • Volume (V): The measure of 3D space occupied. Thinking of the cuboid as a stack of thin rectangular playing cards (base area lw) piled to height h:
    V = Base Area × Height = lwh
  • The Cube (Special Case l = w = h = a):
    TSA = 6a² and V = a³
Solved Example 1 (Page 125) — Same Volume, Different Surface Area:

Cube A has side a = 6cm. Cuboid B has dimensions 9cm × 6cm × 4cm. Compare their volumes and surface areas:

  1. Cube A: V_A = 6³ = 216cm³; TSA_A = 6(6²) = 216cm².
  2. Cuboid B: V_B = 9 × 6 × 4 = 216cm³; TSA_B = 2(9 × 6 + 6 × 4 + 4 × 9) = 2(54 + 24 + 36) = 2(114) = 228cm².
  3. Conclusion: Both shapes have identical volume (216cm³), but the cuboid has a greater surface area (228 > 216cm²).
  4. Practical Significance: A cube maximizes storage while minimizing surface packaging. In biology, smaller animals lose heat faster due to higher surface-area-to-volume ratio!
PracticeExercise 14.1 Q1–Q8

2Right Circular Cylinders

A right circular cylinder is generated by rotating a rectangle about one of its sides. Its axis is strictly perpendicular to its circular bases of radius r. (If the axis is inclined, it is an oblique cylinder).

  • Cross-sections: Slicing parallel to the base gives a circle ( ext{area} = π r²). Slicing parallel to the axis gives a rectangle. (This slicing principle is identical to medical CT and MRI scans).
  • Curved Surface Area (CSA): Unrolling the curved wall yields a flat rectangle of length 2π r (base circumference) and height h:
    CSA = 2π rh
  • Total Surface Area (TSA):
    • Closed at both ends (e.g. tin can): TSA = 2π rh + 2π r² = 2π r(h + r)
    • Open at one end (e.g. drinking tumbler): TSA = 2π rh + π r² = π r(2h + r)
  • Volume (V): Stacking thin circular discs (like circular khakhras) of base area π r² to height h:
    V = π r² h
Solved Example 2 (Page 129) — Savitri's Cylindrical Kaleidoscope:

Savitri makes a cylindrical kaleidoscope of length h = 25cm and radius r = 3.5cm. Find the area of chart paper needed for the curved surface (π = 22/7):

Area = CSA = 2π rh = 2 × ²²⁄₇ × 3.5 × 25 = 2 × 22 × 0.5 × 25 = 550cm²
PracticeExercise 14.2 Q1–Q5

3Right Circular Cones

A right circular cone is formed by rotating a right-angled triangle about one of its perpendicular legs. The vertex is called the apex.

l² = h² + r² ⇒ l = √(h² + r²) (Baudh\=ayana– Pythagoras Theorem)

where r is the base radius, h is the vertical height, and l is the slant height.

Curved Surface Area Derivations:

  1. Method 1 (Triangular Strips): Cutting along the slant height unrolls the cone into a circular sector of radius l. Dividing this sector into n narrow isosceles triangles of height l and bases b_1, b_2, …, b_n:
    CSA = ½(b_1 + b_2 + … + b_n)l = ½(circumference)l = ½(2π r)l = π rl
  2. Method 2 (Sector Ratio): Arc length s = 2π r = (θ / 360^°) × 2π l ⇒ (θ / 360^°) = (r / l). Sector area = π l² × (θ / 360^°) = π l² ((r / l)) = π rl.
  3. Total Surface Area: TSA = CSA + Base Area = π rl + π r² = π r(l + r).

Volume of a Cone & The 1/3 Ratio:

V = ⅓π r² h = ⅓ × Base Area × Height
  • Experimental Verification (Salt Activity): A conical container with radius r and height h fills exactly ⅓ of a cylinder of the same radius r and height h. It takes exactly 3 full cones of salt to fill the cylinder to the brim!
  • Modelling Clay Activity: Reshaping a solid cylinder of clay produces exactly 3 identical cones of equal radius and height.
  • Historical Milestone: This fundamental ⅓ relationship was first discovered by the great Greek mathematician Archimedes (c. 225 BCE).
Solved Example 3 (Page 133) — Cone from Rotated Triangle:

A right triangle with sides 5cm, 12cm, 13cm is rotated through 360^° about the 12cm leg:

  • Height h = 12cm, radius r = 5cm, slant height l = 13cm.
  • Volume V = ⅓π r² h = ⅓π (5²)(12) = 25 × 4π = 100πcm³ ≈ 314.16cm³.
PracticeExercise 14.3 Q1–Q9

4Pyramidal Shapes

A pyramid is a solid whose base is any rectilinear polygon (triangle, square, rectangle, pentagon) and whose lateral faces are triangles meeting at a common apex in another plane.

  • Universal Volume Formula: Exactly like a cone (which is a pyramid with a circular base):
    V = ⅓ × Base Area × Height
  • Surface Area: Sum the area of the polygonal base and all triangular lateral faces.
  • Real-World Pyramids: The Great Pyramids of Giza (square base), the Rubik's Pyraminx (triangular base / tetrahedron), and glacial mountain peaks such as Shivling Peak in Uttarakhand and the Matterhorn in the Swiss Alps.

5Spheres & Archimedes' Cylinder Theorem

A sphere is the set of all points in 3D space equidistant from a fixed centre O by radius r. Rotating a circle about any diameter sweeps out a sphere.

Archimedes' Breakthrough Theorem on Spheres (c. 225 BCE):

Archimedes proved that the surface area and volume of a sphere of radius r are directly proportional to the circumscribing cylinder of radius r and height h = 2r:

Surface Area of SphereEquals the curved surface area of the circumscribing cylinder:
SA = 2π r(2r) = 4π r²
Volume of SphereEquals ⅔ of the volume of the circumscribing cylinder:
V = ⅔(π r² × 2r) = ⁴⁄₃π r³

Visualising Volume via Infinite Tiny Cones:

  1. Divide the entire surface of the sphere into a vast number of tiny patches of base areas Δ A_1, Δ A_2, ….
  2. Connect each patch to the sphere's centre. This partitions the sphere into a collection of thin cones, each with apex at the centre and height equal to radius r.
  3. Summing their volumes:
    V = Σ ⅓(Δ A_i)r = (r / 3) Σ Δ A_i = (r / 3) (4π r²) = ⁴⁄₃πr³
Hands-on Activity — String Winding Verification: Tightly wind string around a rubber ball of radius r. Unwinding the string and coiling it flat on paper fills exactly 4 circles of radius r. Therefore, Surface Area = 4 × (π r²) = 4π r²!
PracticeExercise 14.4 Q1–Q8End Q7–Q10

6Hemispheres

When a sphere of radius r is sliced in half through its centre, two identical hemispheres are created. A hemisphere possesses one curved surface and one flat circular base.

  • Curved Surface Area (CSA): Half of the sphere's total surface:
    CSA = ½(4π r²) = 2π r²
  • Total Surface Area (TSA): Curved surface plus flat circular base:
    TSA = 2π r² + π r² = 3π r²
  • Volume (V): Half of the sphere's volume:
    V = ½(⁴⁄₃π r³) = ⅔πr³
Solved Example 4 (Page 140) — Hemispherical Bowl:

A hemispherical bowl has a radius r = 3.5cm. Find the volume of water it holds (π = 22/7):

V = ⅔π r³ = ⅔ × ²²⁄₇ × (3.5)³ = ⅔ × ²²⁄₇ × 42.875 = 89.83cm³
PracticeExercise 14.4 Q4, Q8

7Scientific Guesstimates & Real-Life Modeling

Guesstimates (Fermi problems) are structured estimation problems where exact numbers are unavailable. Rather than guessing randomly, one models the situation using physical geometry, dimensional analysis, reasonable assumptions, and bounds.

Solved Example 5 (Page 141–142) — Lallan's Golgappa Stall:

A cylindrical vessel of radius r = 10cm has flavored water up to height h = 50cm. Total pani = π r² h = 5000π ≈ 15,708cm³.

ParameterStudent 1 AssumptionStudent 2 Assumption
Average Golgappa Radiusr = 1.5cm (V ≈ 14.14cm³)r = 1.75cm (V ≈ 22.43cm³)
Pani Fill Percentage60% filled (≈ 8.5cm³)80% filled (≈ 18cm³)
Estimated Servings(15708 / 8.5) ≈ 1,850 golgappas(15708 / 18) ≈ 870 golgappas
Key Insight: Guesstimates depend upon justified assumptions and show the realistic range of outcomes.
Solved Example 6 (Page 142–143) — Tennis Balls in a Classroom:

Classroom dimensions: 30ft × 30ft × 15ft = 13,500ft³ ≈ 378,000,000cm³. Tennis ball radius r = 3cm (V ≈ 113cm³).

  • Naïve Division Upper Bound: (378,000,000 / 113) ≈ 3,345,132 balls. (Impossible because spheres cannot pack without gaps!).
  • Cubic Lattice Packing (Diameter = 6cm):
    Length: 900 / 6 = 150 balls; Width: 900 / 6 = 150 balls; Height: 450 / 6 = 75 balls.
    Total = 150 × 150 × 75 = 1,687,500 balls (packing efficiency ≈ 52.4%).
  • Why Cubes are the Base Unit of Volume: Cubes tile 3D space with 100% packing efficiency (zero gaps), whereas spheres always leave void spaces.
PracticeEnd Q1, Q3, Q5, Q15–Q20

8Chapter Formula Summary Table

3D SolidDimensionsCurved / Lateral Surface AreaTotal Surface Area (TSA)Volume (V)
Cuboidl, w, h2h(l + w)2(lw + wh + hl)lwh
Cubea4a²6a²a³
Cylinderr, h2π rh2π r(h + r)π r² h
Coner, h, l = √(h²+r²)π rlπ r(l + r)⅓π r² h
PyramidBase, hSum of triangular facesBase Area + Lateral Area⅓ × Base Area × h
Spherer4π r²4π r²⁴⁄₃π r³
Hemispherer2π r²3π r²⅔π r³

Exercise Set 14.1 — Solutions

Cuboids, cubes, packing, painted cubes and integer dimensions.

1The volume of a cube is 64 cm³. What is its total surface area?
Answer / method
Step 1: For a cube, a³ = 64, so a = 4 cm. Step 2: Total surface area = 6a² = 6 × 4² = 96 cm². Answer: 96 cm².
2How many small cubes with side 20 cm can be packed tight in a cubical box with side 2 m?
Answer / method
Step 1: Convert the box side: 2 m = 200 cm. Step 2: Number of small cubes along one edge = 200 ÷ 20 = 10. Step 3: Total cubes = 10 × 10 × 10 = 1000. Answer: 1000 small cubes.
3The dimensions of a godown are 40 m × 25 m × 10 m. If it is filled with cuboidal boxes, each of dimensions 2 m × 1.25 m × 1 m, find the number of boxes.
Answer / method
Step 1: Along the three directions the numbers of boxes are 40 ÷ 2 = 20, 25 ÷ 1.25 = 20 and 10 ÷ 1 = 10. Step 2: Total boxes = 20 × 20 × 10 = 4000. Answer: 4000 boxes.
4Two cubes each of volume 125 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Answer / method
Step 1: Each cube has volume 125 cm³, so its side is 5 cm. Step 2: Joining two cubes end to end gives a cuboid 10 cm × 5 cm × 5 cm. Step 3: Surface area = 2(10×5 + 5×5 + 5×10) = 2(50 + 25 + 50) = 250 cm². Answer: 250 cm².
5A cube of side 4 cm is cut into cubes of side 1 cm. What is the surface area of the original cube and all the smaller cubes? What changes although the volume is unchanged?
Answer / method
Step 1: Original cube surface area = 6 × 4² = 96 cm². Step 2: Number of 1 cm cubes = 4³ = 64. Step 3: Surface area of all small cubes = 64 × 6 × 1² = 384 cm². Step 4: Volume remains 4³ = 64 cm³, but surface area increases because more faces are exposed. Answer: Original 96 cm²; all small cubes 384 cm².
6The surface areas of the three faces of a cuboid that meet at one vertex are 6 cm², 15 cm² and 10 cm² respectively. What is the volume of the cuboid?
Answer / method
Step 1: Let the edges be l, w and h. The three face areas are lw = 6, wh = 15 and hl = 10. Step 2: Multiply: (lw)(wh)(hl) = l²w²h² = (lwh)². Step 3: Therefore V² = 6 × 15 × 10 = 900, so V = 30 cm³. Answer: 30 cm³.
7A cube of side 5 cm is painted on all its faces and then sliced into 1 cm cubes. How many have (i) exactly three faces painted, (ii) exactly two, (iii) exactly one, and (iv) no face painted?
Answer / method
Step 1: There are 5 unit cubes along each edge. Step 2: Exactly 3 painted faces: the 8 corner cubes. Step 3: Exactly 2 painted faces: 12 edges × (5 − 2) = 36. Step 4: Exactly 1 painted face: 6 faces × (5 − 2)² = 54. Step 5: No painted face: (5 − 2)³ = 27. Answer: 8, 36, 54 and 27 respectively.
8Find a cuboid with integer edge lengths in centimetres and total surface area exactly 100 cm². (i) Is there more than one? (ii) Can you find them all? (iii) Show that you have found them all.
Answer / method
Step 1: Let l ≤ w ≤ h be positive integers. TSA = 100 gives lw + wh + hl = 50. Step 2: Since 3l² ≤ 50, l can only be 1, 2, 3 or 4. Step 3: For l = 1, (w + 1)(h + 1) = 51, giving (w,h) = (2,16). Step 4: For l = 2, (w + 2)(h + 2) = 54, giving (w,h) = (4,7). Step 5: l = 3 gives a factorisation of 59, and l = 4 gives no pair with w ≥ l. Answer: The two cuboids are 1×2×16 cm and 2×4×7 cm, apart from rearranging dimensions.

Exercise Set 14.2 — Solutions

Ratios involving cylinders, scaling, and conservation of volume.

1Two cylinders A and B are given. The radius of A is twice the radius of B, and the height of B is half the height of A. Find the ratio of the curved surface area of A to B and the ratio of the volume of A to B.
Answer / method
Step 1: Let rB = r and hB = h. Then rA = 2r and hA = 2h. Step 2: CSA ratio = (2πrA hA):(2πrB hB) = (2r×2h):(r×h) = 4:1. Step 3: Volume ratio = (πrA²hA):(πrB²hB) = (4r²×2h):(r²h) = 8:1. Answer: CSA ratio 4:1; volume ratio 8:1.
2The radii of two cylinders are in the ratio 2:3 and their heights are in the ratio 3:2. Find (a) the ratio of their volumes and (b) the ratio of their curved surface areas.
Answer / method
Step 1: Volume is proportional to r²h. Step 2: Volume ratio = (2²×3):(3²×2) = 12:18 = 2:3. Step 3: Curved surface area is proportional to rh. Step 4: CSA ratio = (2×3):(3×2) = 1:1. Answer: Volumes 2:3; curved surface areas 1:1.
3The edge of a cube measures r cm. The largest possible right circular cylinder is cut out of the cube. What is the volume of the cylinder?
Answer / method
Step 1: The largest cylinder has height r and base diameter r, so its radius is r/2. Step 2: V = π(r/2)²×r = πr³/4. Answer: πr³/4 cm³.
4The radius of the base of a cylinder is increased by 10%. At the same time, the height is decreased by x%. Given that the volume remains unchanged, find x.
Answer / method
Step 1: Original volume = πr²h. Step 2: New radius = 1.1r and new height = (1 − x/100)h. Step 3: Equal volumes give 1.21(1 − x/100) = 1. Step 4: 1 − x/100 = 100/121, so x/100 = 21/121. Answer: x = 2100/121% ≈ 17.36%.
5A solid metallic cube of side 12 cm is melted and recast into solid cylindrical rods, each of radius 2 cm and height 12 cm. Find the volume of the cube, the volume of one rod and the number of complete rods that can be formed.
Answer / method
Step 1: Volume of cube = 12³ = 1728 cm³. Step 2: Volume of one rod = π×2²×12 = 48π cm³ ≈ 150.72 cm³ using π = 3.14. Step 3: Number of complete rods = 1728 ÷ 48π ≈ 11.45. Answer: 1728 cm³; 48π cm³ per rod; 11 complete rods can be made, with material left over.

Exercise Set 14.3 — Solutions

Cones, slant height, curved surface area, total surface area and volume.

1Find the total surface area of a cone if its slant height is 21 m and the diameter of its base is 24 m.
Answer / method
Step 1: Diameter 24 m gives r = 12 m and l = 21 m. Step 2: TSA = πr(l + r) = π×12×(21 + 12) = 396π m². Step 3: Using π = 22/7, TSA = 1244.57 m² approximately. Answer: 396π m² ≈ 1244.57 m².
2Find the curved surface area of a right circular cone whose slant height is 21 cm and base radius is 7 cm.
Answer / method
Step 1: CSA of a cone = πrl. Step 2: CSA = π×7×21 = 147π cm². Step 3: Using π = 22/7, CSA = 462 cm². Answer: 462 cm².
3The height of a cone is 16 cm and its base radius is 12 cm. Find the curved surface area and total surface area of the cone.
Answer / method
Step 1: Slant height l = √(h² + r²) = √(16² + 12²) = √400 = 20 cm. Step 2: CSA = πrl = π×12×20 = 240π cm². Step 3: TSA = πr(l + r) = π×12×(20 + 12) = 384π cm². Answer: CSA = 240π cm²; TSA = 384π cm².
4A cone has height 15 cm and volume 1570 cm³. Find the radius of its base.
Answer / method
Step 1: V = ⅓πr²h = ⅓×3.14×r²×15 = 15.7r². Step 2: 1570 = 15.7r², so r² = 100. Step 3: r = 10 cm. Answer: 10 cm.
5The curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find (i) the radius of its base and (ii) its total surface area.
Answer / method
Step 1: CSA = πrl gives 308 = π×r×14. Step 2: Using π = 22/7, 308 = 44r, so r = 7 cm. Step 3: TSA = πr(l + r) = π×7×(14 + 7) = 147π = 462 cm². Answer: Radius 7 cm; TSA 462 cm².
6A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Answer / method
Step 1: Slant height l = √(24² + 7²) = √625 = 25 cm. Step 2: Area of one cap = πrl = π×7×25 = 175π = 550 cm² using π = 22/7. Step 3: For 10 caps, area = 10×550 = 5500 cm². Answer: 5500 cm² of sheet.
7What length of tarpaulin 3 m wide is required to make a conical tent of height 8 m and base radius 6 m? Assume that an extra length of material 20 cm is required for stitching margins and wastage.
Answer / method
Step 1: Slant height l = √(8² + 6²) = 10 m. Step 2: Curved area of tent = πrl = π×6×10 = 60π m². Step 3: With tarpaulin width 3 m, length needed = 60π ÷ 3 = 20π m. Step 4: Add 20 cm = 0.2 m. Total ≈ 20π + 0.2 = 63.06 m. Answer: Approximately 63.1 m.
8A right triangle with sides 6 cm, 8 cm and 10 cm is rotated through 360° about the side of length 8 cm. Find the volume and curved surface area of the solid formed.
Answer / method
Step 1: Rotation about the 8 cm side gives a cone with h = 8 cm, r = 6 cm and l = 10 cm. Step 2: V = ⅓πr²h = ⅓π×36×8 = 96π cm³. Step 3: CSA = πrl = π×6×10 = 60π cm². Answer: Volume 96π cm³; curved surface area 60π cm².
9A cup has the shape of a right circular cone. It is filled with water to half the depth of the cone. What fraction of the volume of the cup is occupied by water?
Answer / method
Step 1: At half the depth, the radius of a similar smaller cone is half the full radius. Step 2: Volumes of similar solids are in the cube of the linear ratio. Step 3: Water volume : cup volume = (1/2)³ = 1/8. Answer: 1/8 of the cup.

Exercise Set 14.4 — Solutions

Spheres, hemispheres, scaling and approximate volume.

1A ball bearing has a radius of 0.7 cm. Find its surface area.
Answer / method

Step-by-step Calculation:

  1. A ball bearing is spherical with radius r = 0.7cm.
  2. Surface area formula: SA = 4π r².
  3. Substituting r = 0.7cm and π = 22/7:
    SA = 4 × ²²⁄₇ × 0.7 × 0.7 = 4 × 22 × 0.1 × 0.7 = 88 × 0.07 = 6.16cm²
Answer: The surface area of the ball bearing is 6.16 cm².
*2Two solid spheres made of the same metal have weights 5920 g and 740 g. Determine the radius of the larger sphere, if the diameter of the smaller one is 5 cm.
Answer / method

Step-by-step Solution:

  1. Since both spheres are made of the same metal, their mass is directly proportional to their volume: (M_2 / M_1) = (V_2 / V_1).
  2. Given M_2 = 5920g and M_1 = 740g:
    (V_2 / V_1) = (5920 / 740) = 8
  3. Since volume is proportional to the cube of the radius, (V_2 / V_1) = ((r_2 / r_1))³ = 8 ⇒ (r_2 / r_1) = ∛(8) = 2.
  4. Diameter of smaller sphere is 5cm ⇒ r_1 = 2.5cm.
  5. Radius of larger sphere: r_2 = 2 × r_1 = 2 × 2.5 = 5cm.
Answer: The radius of the larger sphere is 5 cm.
3The diameter of the moon is approximately one fourth the diameter of the earth. Given that the moon and earth are both roughly spherical, find the ratio of their surface areas.
Answer / method

Step-by-step Solution:

  1. Let diameter of Earth be D_e ⇒ radius R_e = D_e / 2.
  2. Diameter of Moon D_m = ¼D_e ⇒ radius R_m = ¼R_e.
  3. Ratio of surface areas: (S_m / S_e) = (4π R_m² / 4π R_e²) = ((R_m / R_e))² = (¼)² = 1/16.
Answer: The ratio of their surface areas is 1 : 16.
4Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius 21 cm.
Answer / method

Step-by-step Calculation:

  1. Radius r = 21cm.
  2. (i) Curved Surface Area (CSA):
    CSA = 2π r² = 2 × ²²⁄₇ × 21 × 21 = 2 × 22 × 3 × 21 = 2772cm²
  3. (ii) Total Surface Area (TSA):
    TSA = 3π r² = 3 × ²²⁄₇ × 21 × 21 = 3 × 22 × 3 × 21 = 4158cm²
Answers: (i) CSA = 2772 cm²; (ii) TSA = 4158 cm².
5Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm × 8 cm × 8 cm. When 16 spheres are packed, the box is filled with preservative liquid. Find the volume of this liquid. Round your answer to the nearest integer.
Answer / method

Step-by-step Solution:

  1. Volume of rectangular box: V_{box} = 16 × 8 × 8 = 1024cm³.
  2. Volume of one sphere of radius r = 2cm:
    V_{sphere} = ⁴⁄₃π (2³) = (32 / 3)πcm³
  3. Volume of 16 spheres: V_{16} = 16 × (32 / 3)π = (512 / 3)π ≈ 170.67 × 3.1416 ≈ 536.17cm³.
  4. Volume of preservative liquid = V_{box} - V_{16} = 1024 - 536.17 = 487.83cm³ ≈ 488cm³.
Answer: The volume of preservative liquid is approximately 488 cm³.
6The radius of a sphere is increased by 10%. Show that the volume increases by approximately 33.1%.
Answer / method

Mathematical Proof:

  1. Let original radius be r. Original volume V = ⁴⁄₃π r³.
  2. New radius r' = r + 0.10r = 1.1r.
  3. New volume:
    V' = ⁴⁄₃π (r')³ = ⁴⁄₃π (1.1r)³ = (1.1)³ × ⁴⁄₃π r³ = 1.331 V
  4. Percentage increase in volume:
    (V' - V / V) × 100% = (1.331 - 1) × 100% = 33.1%
Conclusion: Increasing radius by 10% causes the volume to increase by exactly 33.1%.
*7The radius of a sphere is increased by x%. The volume of the sphere increases by 72.8%. Find the value of x.
Answer / method

Step-by-step Solution:

  1. Let initial radius be r and volume V = ⁴⁄₃π r³.
  2. New volume V' = V + 0.728V = 1.728V.
  3. Since volume is proportional to the cube of the radius, (1 + (x / 100))³ = 1.728.
  4. Taking the cube root on both sides:
    1 + (x / 100) = ∛(1.728) = 1.2 (since 12³ = 1728)
  5. Solving for x: (x / 100) = 1.2 - 1 = 0.2 ⇒ x = 20.
Answer: The value of x is 20%.
8The hemispherical dome of a building needs to be painted (see Fig. 14.16). If the circumference of the base of the dome is 35.2 m, find the cost of painting it, given the cost of painting is ₹10 per 100 cm².
Answer / method

Step-by-step Solution:

  1. Base circumference of the dome = 2π r = 35.2m.
    2 × ²²⁄₇ × r = 35.2 ⇒ r = (35.2 × 7 / 44) = 5.6m = 560cm
  2. Curved surface area to be painted:
    CSA = 2π r² = 2 × ²²⁄₇ × 5.6 × 5.6 = 197.12m² = 1,971,200cm²
  3. Cost of painting is ₹10 per 100cm² (which equals ₹1000 per m²):
    Total Cost = (1,971,200 / 100) × 10 = 19,712 × 10 = ₹197,120
Answer: The total cost of painting the dome is ₹197,120.

End-of-Chapter Exercises — Solutions

Estimation, modelling, ratios, composite solids and change in volume.

1Estimate how many scoops of ice cream can be obtained from a cuboidal container of dimensions 10 cm × 15 cm × 20 cm.
Answer / method
Step 1: Make an estimate for one scoop, for example a spherical scoop of radius 2.5 cm. Step 2: Container volume = 10×15×20 = 3000 cm³. Step 3: One scoop volume ≈ ⁴⁄₃π(2.5)³ ≈ 65.45 cm³. Step 4: Estimated number ≈ 3000 ÷ 65.45 ≈ 46. Answer: About 46 scoops under this stated assumption; a different scoop size changes the estimate.
2A cube of integer side length a is made of unit cubes. Write an expression for the number of unit cubes needed to make a cube of side length a + 1.
Answer / method
Step 1: A cube of side a contains a×a×a unit cubes. Step 2: A cube of side a + 1 contains (a + 1)×(a + 1)×(a + 1) unit cubes. Answer: (a + 1)³ unit cubes.
3(i) Could a person drink enough water in a lifetime to fill an entire classroom? (ii) Estimate the number of bricks used to build the walls of your classroom.
Answer / method
Step 1: Estimate classroom volume and average water consumed by one person in a lifetime. Step 2: For example, a room 8×6×3 m has volume 144 m³ = 144,000 litres. Step 3: If one person drinks about 2 litres per day for 70 years, water ≈ 2×365×70 = 51,100 litres. Step 4: Therefore one person cannot fill this example room; the conclusion depends on the chosen dimensions and lifetime estimate. Step 5: For bricks, estimate wall volume and divide by one brick volume, allowing for mortar. Answer: It is an estimation problem; state assumptions and compare volumes.
4You are given two options: (A) one chocolate cube of side 50 mm, or (B) one hundred chocolate cubes of side 10 mm. Which option gives more chocolate?
Answer / method
Step 1: Volume of one 50 mm cube = 50³ = 125,000 mm³. Step 2: Volume of one 10 mm cube = 10³ = 1000 mm³. Step 3: Volume of 100 small cubes = 100×1000 = 100,000 mm³. Step 4: 125,000 > 100,000. Answer: Option A gives more chocolate.
5If all the people in the world are crowded together in one location, how much area would they cover?
Answer / method
Step 1: Estimate the number of people and the area occupied by one person, for example 0.25 m². Step 2: Area ≈ population × area per person. Step 3: Using an assumed population of 8×10⁹ gives approximately 2×10⁹ m², or 2000 km². Answer: About 2000 km² under the stated assumption; the result changes with the population and spacing assumed.
6A school provides milk in cylindrical glasses, each with diameter 7 cm and filled to a height of 12 cm. How many litres of milk are needed to serve 1600 students? Use 1000 cm³ = 1 litre.
Answer / method
Step 1: Radius = 7/2 = 3.5 cm and height = 12 cm. Step 2: Volume of one glass = πr²h = 22/7×3.5²×12 = 462 cm³. Step 3: For 1600 glasses, volume = 462×1600 = 739,200 cm³. Step 4: Convert to litres: 739,200 ÷ 1000 = 739.2 litres. Answer: 739.2 litres.
7The surface area of a sphere of radius 5 cm is five times the curved surface area of a cone of radius 4 cm. Find the height of the cone.
Answer / method
Step 1: Sphere surface area = 4π×5² = 100π. Step 2: Cone curved surface area = π×4×l. Five times this is 20πl. Step 3: 100π = 20πl, so l = 5 cm. Step 4: h² = l² − r² = 5² − 4² = 9, so h = 3 cm. Answer: Height 3 cm.
8Take the Earth, Jupiter and the Sun to be spheres with radii approximately 6379 km, 69,900 km and 695,700 km respectively. Compute the approximate ratios of the volume of Jupiter to Earth and the volume of the Sun to Earth.
Answer / method
Step 1: Volumes of spheres are proportional to the cubes of their radii. Step 2: Jupiter:Earth = (69,900:6379)³ ≈ 1315.75:1. Step 3: Sun:Earth = (695,700:6379)³ ≈ 1,297,204:1. Answer: Jupiter is approximately 1316 Earth-volumes; the Sun is approximately 1.297 million Earth-volumes.
9Show that the volume of a sphere is equal to 2/3 of the volume of the smallest cylinder that encloses it.
Answer / method
Step 1: Volume of a sphere is ⁴⁄₃πR³. Step 2: The smallest enclosing cylinder has radius R and height 2R. Step 3: Cylinder volume = πR²×2R = 2πR³. Step 4: Sphere:cylinder = (⁴⁄₃πR³):(2πR³) = 2:3. Answer: Sphere volume is 2/3 of the cylinder volume.
10A string is wrapped tightly around the equator of a spherical Earth. Another string is 1 m longer and forms a larger circle, staying the same distance above the ground everywhere. How high above the ground is the second string?
Answer / method
Step 1: Let the original radius be R and the new radius be R + h. Step 2: Original circumference = 2πR; new circumference = 2π(R + h). Step 3: Difference is 1 m, so 2πh = 1. Step 4: h = 1/(2π) m ≈ 0.159 m = 15.9 cm. Answer: Approximately 15.9 cm above the ground, independent of Earth’s radius.
11Repeat the string question for (i) the Moon, (ii) Jupiter and (iii) a volleyball. Are you surprised by the answers? Why?
Answer / method
Step 1: The derivation uses only circumference difference 2πh = 1. Step 2: The radius cancels, so the answer is the same for the Moon, Jupiter and a volleyball. Answer: Approximately 15.9 cm in all three cases. The result may seem surprising because it does not depend on the original radius.
12A cylinder has base radius r and height h. A square pyramid is fitted inside it: the square base lies on the cylinder’s base, its corners lie on the circular boundary, and its apex lies on the top of the cylinder. Find the ratio of the volume of the pyramid to the volume of the cylinder.
Answer / method
Step 1: The square base is inscribed in a circle of radius r, so its diagonal is 2r. Step 2: If the square side is s, s√2 = 2r, so s² = 2r². Step 3: Pyramid volume = ⅓×2r²×h = ⅔r²h. Step 4: Cylinder volume = πr²h. Answer: Ratio = (⅔r²h):(πr²h) = 2:3π.
13Find the change in volume when (i) the length of a cuboid with dimensions l, w, h is increased by 1 unit; (ii) the radius of a cylinder with dimensions r, h is increased by 1 unit; (iii) the radius of a sphere is decreased by 1 unit.
Answer / method
Step 1: Cuboid volume changes from lwh to (l + 1)wh. Increase = wh cubic units. Step 2: Cylinder volume changes from πr²h to π(r + 1)²h. Increase = π[(r + 1)² − r²]h = 2πrh + πh. Step 3: Sphere volume decreases from ⁴⁄₃πr³ to ⁴⁄₃π(r − 1)³. Decrease = ⁴⁄₃π[r³ − (r − 1)³] = 4πr² − 4πr + ⁴⁄₃π cubic units. Answer: wh; 2πrh + πh; and the stated sphere decrease.
14Given a cube with total surface area S, express its volume V in terms of S.
Answer / method
Step 1: For a cube, S = 6a². Step 2: a² = S/6, so a = √(S/6). Step 3: V = a³ = (S/6)^(3/2). Answer: V = (S/6)^(3/2) cubic units.
15A cylindrical glass has height 25 cm and radius 4 cm. Water is at a height of 16 cm. How many marbles of radius 1 cm should be dropped into the glass to make the water reach a height of 20 cm?
Answer / method
Step 1: Water rise = 20 − 16 = 4 cm. Step 2: Required displaced volume = cylinder base area × rise = π×4²×4 = 64π cm³. Step 3: One marble volume = ⁴⁄₃π×1³ = ⁴⁄₃π cm³. Step 4: Number = 64π ÷ (⁴⁄₃π) = 48. Answer: 48 marbles.
16Find the volume of ink in a new ball-point pen. Take the necessary measurements and make approximations as needed.
Answer / method
Step 1: Measure or estimate the inner radius r and length h of the ink tube. Step 2: Model the ink as a cylinder, so V = πr²h. Step 3: Convert cm³ to mL using 1 cm³ = 1 mL. Answer: The numerical value depends on the measurements; report the approximation with units and assumptions.
17(i) A ball of chapati dough has radius 6 cm. Estimate how many chapatis can be made from it. (ii) Cut a coconut or muskmelon in half and estimate the volume of edible flesh or fruit by taking necessary measurements.
Answer / method
Step 1: Measure the dough ball radius R and choose an estimated chapati radius r and thickness t. Step 2: Dough volume = ⁴⁄₃πR³. One chapati is approximately πr²t. Step 3: Estimated number = dough volume ÷ one chapati volume. Step 4: For the coconut or muskmelon, estimate the volume of the edible part by a suitable cylinder, sphere or composite shape and subtract the inedible part. Answer: These are measurement-based estimates; write the chosen measurements clearly.
18Looking at the Ganita Manjari Grade 9 Part 2 textbook: (i) If all pages were laid side by side on the floor, would they cover an 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?
Answer / method
Step 1: For the pages, estimate total page area and compare it with the classroom floor area. Include the thickness and page count if estimating a stack. Step 2: For textbooks, estimate one book’s volume and divide the storeroom volume 15×20×30 = 9000 ft³ by it. Step 3: Allow for gaps, shelves and movement space. Answer: No unique numerical answer is possible without page, book and room-use measurements.
19If the entire human population decided to climb into one giant cube, how long would one side of the cube have to be?
Answer / method
Step 1: Estimate population N and average volume occupied by one person, v. Step 2: Total volume ≈ Nv. Step 3: If the cube side is s, then s³ = Nv, so s = ∛(Nv). Answer: The numerical side depends on the population and personal-space assumption.
20The Earth’s surface has an estimated volume of 1.38 billion km³ of water. Suppose the Earth is a perfect sphere and all the water forms a uniform layer covering it. Estimate the thickness of this water layer. Take Earth’s radius as 6371 km.
Answer / method
Step 1: Let Earth’s radius be R = 6371 km and water-layer thickness be t. Step 2: Exact shell model: ⁴⁄₃π[(R + t)³ − R³] = 1.38×10⁹ km³. Step 3: Since t is small compared with R, use shell volume ≈ 4πR²t. Step 4: t ≈ 1.38×10⁹ ÷ (4π×6371²) ≈ 2.71 km. Answer: The water layer is approximately 2.7 km thick.
*21Give the dimension of a cuboid whose volume is halved when its surface area is doubled.
Answer / method

Step-by-step Analysis & Solution:

  1. Let original dimensions be l, w, h. Volume V = lwh and surface area S = 2(lw + wh + hl).
  2. We seek new dimensions l', w', h' such that V' = ½V and S' = 2S.
  3. Key Geometric Principle: Flattening a solid (making it very thin and wide like a pancake, or long and thin like a wire) dramatically increases surface area while simultaneously reducing volume!
  4. Concrete Example:
    • Original Cuboid: A cube of side 4cm ⇒ V = 4³ = 64cm³ and S = 6(4²) = 96cm².
    • Target for New Cuboid: V' = 32cm³ and S' = 192cm².
    • Choosing a thin slab with h' = 0.2cm: l'w' = 32 / 0.2 = 160cm².
    • Then S' = 2(l'w' + w'h' + h'l') = 2(160 + 0.2(l' + w')) = 320 + 0.4(l' + w') > 192.
    • By choosing h' = ⅓cm, l' = 48cm, and w' = 2cm, we achieve V' = 48 × 2 × ⅓ = 32cm³ (V/2) and S' ≈ 209cm² ≈ 2S. Exact dimensions are obtained by solving l'w'h' = V/2 and 2(l'w' + w'h' + h'l') = 2S.
Conclusion: Any cuboid deformed into an extremely thin sheet or wire achieves a doubled surface area while cutting its enclosed volume in half.
22Project: Find the volume of your house making necessary approximations. Present how you solved it.
Answer / method

Project Guide & Estimation Model:

  1. Step 1 — Geometric Decomposition: Model the house as a combination of simple 3D solids — rectangular cuboids for each room (living room, bedrooms, kitchen, bathrooms, hallways) and a triangular prism for a pitched roof/attic.
  2. Step 2 — Measure Dimensions:
    • Measure length, width, and ceiling height of each rectangular section (e.g. living room 6m × 4m × 3m = 72m³).
    • Bedroom 1: 4m × 3.5m × 3m = 42m³.
    • Kitchen & dining: 5m × 3m × 3m = 45m³.
    • Other spaces: ≈ 60m³.
  3. Step 3 — Sum Enclosed Volumes: Total living volume = 72 + 42 + 45 + 60 = 219m³.
  4. Step 4 — Adjust for Walls and Fixtures: Subtract roughly 10% to 15% for internal partition walls, cupboards, and structural pillars.
Answer: A typical 2-bedroom home has an interior air volume of approximately 200 to 250 m³ (around 200,000 to 250,000 litres).
Quick check

Mini Test: 10 Questions

Try these without opening the exercise answers. Show units at every stage.

Q1. Find the volume of a cuboid 8 cm x 5 cm x 3 cm.

Q2. What is the total surface area of a cube of side 5 cm?

Q3. The volume of a cylinder is:

Q4. A cone has r = 3 cm and h = 4 cm. Its slant height is:

Q5. The curved surface area of a cone is:

Q6. The volume of a hemisphere is:

Q7. A 10% increase in the radius of a sphere changes its volume by:

Q8. When a solid is melted and recast, its:

Q9. A diameter of 14 cm gives a radius of:

Q10. The unit of surface area is:

Score: 0 / 10