Chapter 14: Math of Space — Surface Area and Volume

Part 1: Complete Solutions with Labeled Diagrams (Exercise Sets 14.1, 14.2 & 14.3)

Exercise Set 14.1 (Cuboids and Cubes)

1. The volume of a cube is 64 cm³. What is its total surface area?
  • Let the side length of the cube be a cm.
  • Given Volume of the cube: V = a³ = 64 cm³.
  • Taking the cube root on both sides: a = ∛64 = 4 cm.
  • Total Surface Area (TSA) of a cube = 6a²
  • TSA = 6 × (4 cm)² = 6 × 16 cm² = 96 cm².
Total Surface Area = 96 cm²
a = 4 cm V = 64 cm³
Cube of side a = 4 cm
2. How many small cubes with side 20 cm can be packed tight in a cubical box with side 2 m?
  • Side of the large cubical box (A) = 2 m = 2 × 100 cm = 200 cm.
  • Side of each small cube (a) = 20 cm.
  • Volume of large cubical box = A³ = 200 × 200 × 200 cm³.
  • Volume of 1 small cube = a³ = 20 × 20 × 20 cm³.
  • Number of small cubes = (Volume of box) / (Volume of 1 small cube)
  • Number of cubes = (200 / 20) × (200 / 20) × (200 / 20) = 10 × 10 × 10 = 1000.
Number of small cubes = 1000
Box = 200 cm 20 cm
Packing 20 cm cubes in 2 m box
3. The 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, then find the number of boxes.
  • Dimensions of the godown: L = 40 m, W = 25 m, H = 10 m.
  • Volume of the godown = 40 × 25 × 10 = 10,000 m³.
  • Dimensions of one cuboidal box: l = 2 m, w = 1.25 m, h = 1 m.
  • Volume of one box = 2 × 1.25 × 1 = 2.5 m³.
  • Number of boxes = (Volume of godown) / (Volume of one box) = 10,000 / 2.5 = 4000 boxes.
  • (Note: If the third dimension is taken as 11 m due to a typographical print artifact, Volume of box = 2 × 1.25 × 11 = 27.5 m³, giving ⌊10000 / 27.5⌋ = 363 boxes.)
Number of boxes = 4000
40 m × 25 m × 10 m Box: 2×1.25×1 m
Godown and Cuboidal Box
4. Two cubes each of volume 125 cm³ are joined end to end. Find the surface area of the resulting cuboid.
  • Volume of each cube = a³ = 125 cm³ ⇒ a = ∛125 = 5 cm.
  • When two cubes of side 5 cm are joined end to end, the dimensions of the resulting cuboid are:
    • Length (l) = 5 + 5 = 10 cm
    • Width (w) = 5 cm
    • Height (h) = 5 cm
  • Total Surface Area of resulting cuboid = 2(lw + wh + hl)
  • TSA = 2[(10 × 5) + (5 × 5) + (5 × 10)] = 2[50 + 25 + 50] = 2 × 125 = 250 cm².
Surface Area = 250 cm²
5 cm 5 cm 5 cm
Resulting Cuboid: 10 cm × 5 cm × 5 cm
5. A cube of side 4 cm is cut into cubes of side 1 cm. What is the ratio of the surface areas of the original cube and all the cut-out cubes?
  • Side of original cube (A) = 4 cm.
  • Surface area of original cube (S₁) = 6A² = 6 × 4² = 96 cm².
  • Number of 1 cm cubes obtained = (Volume of 4 cm cube) / (Volume of 1 cm cube) = 4³ / 1³ = 64 cubes.
  • Surface area of 1 small cube = 6 × 1² = 6 cm².
  • Total surface area of all 64 cut-out cubes (S₂) = 64 × 6 = 384 cm².
  • Required Ratio = S₁ : S₂ = 96 : 384 = 1 : 4.
Ratio = 1 : 4
⇒ Side = 4 cm 64 × (1 cm)
4 cm Cube Cut into 64 Unit Cubes
6. The surface areas of the three faces of a cuboid that meet at one of the corners of the cuboid are 6 cm², 15 cm², and 10 cm² respectively. What is the volume of the cuboid?
  • Let the length, width, and height of the cuboid be l, w, and h respectively.
  • The areas of the three adjacent faces meeting at a corner are:
    • l × w = 6 cm²
    • w × h = 15 cm²
    • h × l = 10 cm²
  • Multiplying all three equations together:
    (lw)(wh)(hl) = 6 × 15 × 10
    l²w²h² = (lwh)² = 900
  • Since Volume V = lwh:
    V = √900 = 30 cm³.
Volume = 30 cm³
15 cm² 6 cm² 10 cm²
Three Faces Meeting at a Corner
*7. A cube of side 5 cm is painted on all its faces. If it is sliced into 1 cm³ cubes, how many of these 1 cm³ cubes have: (i) exactly three faces painted? (ii) exactly two faces painted? (iii) exactly one face painted? (iv) no face painted?
  • Here, n = 5 (since a 5 cm cube is sliced into 1 cm cubes, giving 5³ = 125 unit cubes).
  • (i) Exactly three faces painted: Located at the 8 corners of the cube.
    Count = 8 cubes.
  • (ii) Exactly two faces painted: Located along the 12 edges (excluding corners).
    Count = 12 × (n − 2) = 12 × (5 − 2) = 12 × 3 = 36 cubes.
  • (iii) Exactly one face painted: Located in the interior of the 6 faces.
    Count = 6 × (n − 2)² = 6 × (5 − 2)² = 6 × 9 = 54 cubes.
  • (iv) No face painted: Located inside the central core of the cube.
    Count = (n − 2)³ = (5 − 2)³ = 3³ = 27 cubes.
  • Check: 8 + 36 + 54 + 27 = 125 cubes.
(i) 8 | (ii) 36 | (iii) 54 | (iv) 27
5 cm × 5 cm Face
Red: 3-face | Pink: 2-face | Yellow: 1-face
*8. Find a cuboid with edges whose lengths are integers (in cm), given that it has a total surface area of exactly 100 cm². (i) Is there more than one such cuboid? (ii) Can you find them all? (iii) Show that you have found them all.
  • Let the integer edges be 1 ≤ l ≤ w ≤ h.
  • Given: 2(lw + wh + hl) = 100 ⇒ lw + wh + hl = 50.
  • Since l ≤ w ≤ h, we have 3l² ≤ lw + wh + hl = 50 ⇒ l² ≤ 16.67 ⇒ l ∈ {1, 2, 3, 4}.
  • Case 1 (l = 1): w + wh + h = 50 ⇒ (w + 1)(h + 1) = 51.
    Since 51 = 3 × 17 and w ≥ 1, we get w + 1 = 3, h + 1 = 17 ⇒ (1 cm, 2 cm, 16 cm).
  • Case 2 (l = 2): 2w + wh + 2h = 50 ⇒ (w + 2)(h + 2) = 54.
    Since w ≥ 2 ⇒ w + 2 ≥ 4. Factors of 54 ≥ 4 are 6 × 9 ⇒ w + 2 = 6, h + 2 = 9 ⇒ (2 cm, 4 cm, 7 cm).
  • Case 3 (l = 3): 3w + wh + 3h = 50 ⇒ (w + 3)(h + 3) = 59. Since 59 is prime, no solution exists.
  • Case 4 (l = 4): 3w² ≤ 50 ⇒ w = 4. Testing l = 4, w = 4 gives 16 + 8h = 50 ⇒ 8h = 34 (no integer solution).
  • Answers: (i) Yes, there are two such cuboids. (ii) 1 cm × 2 cm × 16 cm and 2 cm × 4 cm × 7 cm. (iii) Exhaustive case check for l ∈ {1, 2, 3, 4} proves no other integer solutions exist.
Cuboids: (1, 2, 16) cm and (2, 4, 7) cm
7 cm 4 cm 2 cm TSA = 100 cm²
Integer Cuboid (2 cm × 4 cm × 7 cm)

Exercise Set 14.2 (Right Circular Cylinder)

1. Two cylinders, A and B, are given. The radius of cylinder B is twice that of cylinder A, and the height of cylinder B is half that of cylinder A. Find the ratio of the curved surface area of A to the curved surface area of B. Also find the ratio of the volume of A to the volume of B.
  • Let Cylinder A have radius rA = r and height hA = h.
  • Then Cylinder B has radius rB = 2r and height hB = h / 2.
  • Ratio of Curved Surface Areas:
    CSA(A) / CSA(B) = (2πrh) / [2π(2r)(h/2)] = (2πrh) / (2πrh) = 1 : 1.
  • Ratio of Volumes:
    V(A) / V(B) = (πr²h) / [π(2r)²(h/2)] = (πr²h) / (2πr²h) = 1 : 2.
CSA Ratio = 1 : 1 | Volume Ratio = 1 : 2
r, h A 2r, h/2 B
Cylinders A and B
2. The 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.
  • Let the radii be r₁ = 2x and r₂ = 3x.
  • Let the heights be h₁ = 3y and h₂ = 2y.
  • (a) Ratio of Volumes:
    V₁ / V₂ = [π(2x)²(3y)] / [π(3x)²(2y)] = (4 × 3) / (9 × 2) = 12 / 18 = 2 : 3.
  • (b) Ratio of Curved Surface Areas:
    CSA₁ / CSA₂ = [2π(2x)(3y)] / [2π(3x)(2y)] = 12 / 12 = 1 : 1.
(a) Volume Ratio = 2 : 3 | (b) CSA Ratio = 1 : 1
2x, 3y 3x, 2y
Cylinders with r₁:r₂ = 2:3, h₁:h₂ = 3:2
3. The edge of a cube measures r cm. The largest possible right circular cylinder is cut out of the cube. What do you think is the volume of the cylinder (in cm³)?
  • Edge of the cube = r cm.
  • For the largest possible right circular cylinder cut out of the cube:
    • Diameter of the cylinder's base = Edge of the cube = r cm
    • Radius of the cylinder (R) = r / 2 cm
    • Height of the cylinder (H) = r cm
  • Volume of the cylinder = πR²H = π × (r / 2)² × r = (πr³) / 4 cm³ (or approx. (11/14)r³ cm³ taking π = 22/7).
Volume = πr³ / 4 cm³ (or 11r³ / 14 cm³)
H = r R = r/2
Largest Cylinder Inside Cube of Edge r
*4. The radius of the base of a cylinder is increased by 10%. At the same time, the height of the cylinder is decreased by x%. Given that the volume of the cylinder remains unchanged, find the value of x.
  • Let original radius = r and original height = h. Original Volume V = πr²h.
  • New radius r' = r(1 + 10/100) = 1.1r.
  • New height h' = h(1 − x/100).
  • Since the volume remains unchanged:
    π(1.1r)² × h(1 − x/100) = πr²h
    1.21 × (1 − x/100) = 1
    1 − x/100 = 1 / 1.21 = 100 / 121
  • x/100 = 1 − 100/121 = 21/121 ⇒ x = 2100 / 121 = 17(43/121)% ≈ 17.36%.
x = 2100/121 ≈ 17.36
V₁ V₂ = V₁
r' = 1.1r, h' = h(1 − x/100)
5. A solid metallic cube of side 12 cm is melted and recast into solid cylindrical rods, each having radius 2 cm and height 12 cm. Find: (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)
  • (i) Volume of the cube:
    Vcube = a³ = (12 cm)³ = 1728 cm³.
  • (ii) Volume of one cylindrical rod:
    Vrod = πr²h = (22/7) × 2² × 12 = 1056/7 cm³ ≈ 150.86 cm³.
  • (iii) Number of complete cylindrical rods:
    N = Vcube / Vrod = 1728 / (1056/7) = (1728 × 7) / 1056 = 126 / 11 ≈ 11.45.
    Therefore, 11 complete cylindrical rods can be formed.
(i) 1728 cm³ | (ii) 150.86 cm³ | (iii) 11 rods
12 cm ⇒ r=2, h=12 cm
Cube Melted into 11 Complete Rods

Exercise Set 14.3 (Right Circular Cones)

1. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
  • Slant height (l) = 21 m.
  • Diameter of base (d) = 24 m ⇒ Radius (r) = 24 / 2 = 12 m.
  • Total Surface Area (TSA) of a cone = πr(l + r)
  • TSA = (22/7) × 12 × (21 + 12) = (22/7) × 12 × 33 = 8712 / 7 = 1244.57 m².
Total Surface Area = 1244.57 m²
r = 12 m l = 21 m
Cone with r = 12 m, l = 21 m
2. Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.
  • Slant height (l) = 10 cm, Base radius (r) = 7 cm.
  • Curved Surface Area (CSA) = πrl
  • CSA = (22/7) × 7 × 10 = 22 × 10 = 220 cm².
Curved Surface Area = 220 cm²
r = 7 cm l = 10 cm
Cone with r = 7 cm, l = 10 cm
3. The height of a cone is 16 cm, and its base radius is 12 cm. Find the curved surface area and the total surface area of the cone.
  • Height (h) = 16 cm, Radius (r) = 12 cm.
  • Slant height (l) = √(h² + r²) = √(16² + 12²) = √(256 + 144) = √400 = 20 cm.
  • Curved Surface Area (CSA) = πrl = 3.14 × 12 × 20 = 753.6 cm² (or 754.29 cm² using π = 22/7).
  • Total Surface Area (TSA) = πr(l + r) = 3.14 × 12 × (20 + 12) = 3.14 × 12 × 32 = 1205.76 cm² (or 1206.86 cm² using π = 22/7).
CSA = 753.6 cm² | TSA = 1205.76 cm²
h=16 r=12 l=20
h = 16 cm, r = 12 cm, l = 20 cm
4. A cone has a height of 15 cm. If its volume is 1570 cm³, find the radius of the base.
  • Height (h) = 15 cm, Volume (V) = 1570 cm³. (Taking π = 3.14)
  • Formula: V = (1/3)πr²h
  • 1570 = (1/3) × 3.14 × r² × 15
  • 1570 = 15.7 × r² ⇒ r² = 1570 / 15.7 = 100
  • r = √100 = 10 cm.
Radius of the base = 10 cm
h=15 r = 10 cm
V = 1570 cm³, h = 15 cm ⇒ r = 10 cm
5. The curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find (i) the radius of the base, (ii) the total surface area of the cone.
  • Given CSA = 308 cm², slant height (l) = 14 cm.
  • (i) Radius of the base:
    πrl = 308 ⇒ (22/7) × r × 14 = 308
    44r = 308 ⇒ r = 308 / 44 = 7 cm.
  • (ii) Total Surface Area of the cone:
    TSA = CSA + πr² = 308 + (22/7) × 7 × 7 = 308 + 154 = 462 cm².
(i) Radius = 7 cm | (ii) TSA = 462 cm²
r = 7 cm l = 14 cm
CSA = 308 cm², l = 14 cm
6. A 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.
  • Radius (r) = 7 cm, Height (h) = 24 cm.
  • Slant height (l) = √(r² + h²) = √(7² + 24²) = √(49 + 576) = √625 = 25 cm.
  • Sheet required for 1 cap = Curved Surface Area = πrl
  • CSA of 1 cap = (22/7) × 7 × 25 = 550 cm².
  • Area of sheet required for 10 caps = 10 × 550 = 5500 cm².
Area for 10 caps = 5500 cm²
h=24 r=7 l=25
Joker's Cap (r = 7 cm, h = 24 cm, l = 25 cm)
*7. What length of tarpaulin 3 m wide is required to make a conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that is required for stitching margins and wastage in cutting is 20 cm.
  • Height of conical tent (h) = 8 m, Base radius (r) = 6 m.
  • Slant height (l) = √(8² + 6²) = √(64 + 36) = √100 = 10 m.
  • Curved Surface Area of the tent = πrl = 3.14 × 6 × 10 = 188.4 m².
  • Width of tarpaulin = 3 m. Length required (L₀) = Area / Width = 188.4 / 3 = 62.8 m.
  • Extra length for stitching margins & wastage = 20 cm = 0.2 m.
  • Total length of tarpaulin required = 62.8 + 0.2 = 63 m.
Total Length of Tarpaulin = 63 m
h=8m r=6m l=10m
Conical Tent (h = 8 m, r = 6 m, l = 10 m)
8. A right triangle with sides 6 cm, 8 cm and 10 cm is rotated through 360° about the side of 8 cm. Find the volume and the curved surface area of the solid so formed.
  • When the right triangle (6 cm, 8 cm, 10 cm) is rotated about the 8 cm side, it forms a right circular cone with:
    • Height (h) = 8 cm
    • Base radius (r) = 6 cm
    • Slant height (l) = 10 cm
  • Volume of the cone:
    V = (1/3)πr²h = (1/3) × 3.14 × 6² × 8 = 96π ≈ 301.44 cm³ (or 301.71 cm³ using π = 22/7).
  • Curved Surface Area (CSA):
    CSA = πrl = 3.14 × 6 × 10 = 60π ≈ 188.4 cm² (or 188.57 cm² using π = 22/7).
Volume = 301.44 cm³ | CSA = 188.4 cm²
8 cm 6 cm 10 cm
Right Triangle Rotated About 8 cm Side
*9. Suppose you have a cup in the shape of a right circular cone. Fill it with water to half the depth of the cone. What fraction of the volume of the cup is occupied by the water?
  • Let the full conical cup have base radius R and height H.
  • Volume of the full cup: Vcup = (1/3)πR²H.
  • When filled to half the depth, the height of the water cone is h = H / 2.
  • By similar triangles, the radius of the water surface is proportional to the height:
    r / R = h / H = 1 / 2 ⇒ r = R / 2.
  • Volume of water: Vwater = (1/3)πr²h = (1/3)π(R/2)²(H/2) = (1/8) × [(1/3)πR²H] = (1/8)Vcup.
Fraction of Volume Occupied = 1/8
R R/2 H/2
Water at Half Depth Occupies 1/8 Volume