Chapter 14: Math of Space — Surface Area and Volume

Part 2: Complete Step-by-Step Solutions with Labeled Diagrams (Exercise Set 14.4 & End-of-Chapter Exercises)

Exercise Set 14.4 (Spheres and Hemispheres)

1. A ball bearing has a radius of 0.7 cm. Find its surface area.
  • Radius of the spherical ball bearing (r) = 0.7 cm.
  • Surface Area of a sphere = 4πr²
  • SA = 4 × (22/7) × 0.7 × 0.7
  • SA = 4 × 22 × 0.1 × 0.7 = 6.16 cm².
Surface Area = 6.16 cm²
r = 0.7 cm
Ball Bearing (r = 0.7 cm)
*2. Two 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.
  • Diameter of smaller sphere (d₂) = 5 cm ⇒ Radius (r₂) = 5 / 2 = 2.5 cm.
  • Let the radius of the larger sphere be r₁.
  • Since both spheres are made of the same metal (same density), the ratio of their volumes equals the ratio of their weights:
  • V₁ / V₂ = 5920 / 740 = 8.
  • [(4/3)πr₁³] / [(4/3)πr₂³] = 8 ⇒ (r₁ / r₂)³ = 8 = 2³
  • Taking cube root: r₁ / r₂ = 2 ⇒ r₁ = 2 × 2.5 cm = 5 cm.
Radius of Larger Sphere = 5 cm
r₁ = 5 cm 5920 g r₂ = 2.5 cm 740 g
Ratio of Volumes = 8 : 1
3. The 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.
  • Let the diameter of the Earth be De and the diameter of the Moon be Dm.
  • Given: Dm = (1/4)De ⇒ Radius of Moon rm = (1/4)re.
  • Ratio of Surface Areas = SA(Moon) / SA(Earth) = (4πrm²) / (4πre²)
  • Ratio = (rm / re)² = (1/4)² = 1 / 16.
Ratio of Surface Areas = 1 : 16
Moon (r) Earth (4r)
Moon and Earth (r_m : r_e = 1 : 4)
4. Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius 21 cm.
  • Radius of the hemisphere (r) = 21 cm.
  • (i) Curved Surface Area (CSA):
    CSA = 2πr² = 2 × (22/7) × 21 × 21 = 44 × 63 = 2772 cm².
  • (ii) Total Surface Area (TSA):
    TSA = 3πr² = 3 × (22/7) × 21 × 21 = 66 × 63 = 4158 cm².
(i) CSA = 2772 cm² | (ii) TSA = 4158 cm²
r = 21 cm
Hemisphere (r = 21 cm)
5. Metal 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.
  • Volume of rectangular box = 16 × 8 × 8 = 1024 cm³.
  • Radius of each metal sphere (r) = 2 cm.
  • Volume of 1 sphere = (4/3)πr³ = (4/3) × π × 2³ = (32/3)π cm³.
  • Volume of 16 spheres = 16 × (32/3) × 3.1416 ≈ 536.17 cm³ (or 536.38 cm³ using π = 22/7).
  • Volume of preservative liquid = Volume of box − Volume of 16 spheres
  • Volume of liquid = 1024 − 536.17 = 487.83 cm³ ≈ 488 cm³.
Volume of Liquid ≈ 488 cm³
Box: 16 × 8 × 8 cm
16 Spheres (r = 2 cm) in Liquid
6. The radius of a sphere is increased by 10%. Show that the volume increases by approximately 33.1%.
  • Let original radius of the sphere = r.
  • Original Volume (V) = (4/3)πr³.
  • New radius after 10% increase: r' = r + 0.10r = 1.1r.
  • New Volume (V') = (4/3)π(1.1r)³ = 1.331 × [(4/3)πr³] = 1.331V.
  • Percentage increase in volume = [(V' − V) / V] × 100% = (1.331 − 1) × 100% = 33.1%. (Hence shown)
Volume Increases by 33.1%
r ⇒ 1.1r
V' = (1.1)³ V = 1.331 V
*7. The radius of a sphere is increased by x%. The volume of the sphere increases by 72.8%. Find the value of x.
  • Let original radius = r, and new radius r' = r(1 + x/100).
  • Given that new volume V' = V(1 + 72.8/100) = 1.728V.
  • Since V' / V = (r' / r)³:
    (1 + x/100)³ = 1.728 = 1728 / 1000 = (12 / 10)³ = (1.2)³
  • Taking the cube root on both sides:
    1 + x/100 = 1.2 ⇒ x/100 = 0.2 ⇒ x = 20.
x = 20 (i.e., 20% increase)
V ⇒ 1.728 V
∛1.728 = 1.2 ⇒ x = 20%
8. The 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.
  • Circumference of base = 2πr = 35.2 m.
  • 2 × (22/7) × r = 35.2 ⇒ r = (35.2 × 7) / 44 = 0.8 × 7 = 5.6 m.
  • Curved Surface Area (CSA) of dome = 2πr² = 2 × (22/7) × 5.6 × 5.6 = 197.12 m².
  • Convert area into cm² (1 m² = 10,000 cm²):
    CSA = 197.12 × 10,000 = 19,71,200 cm².
  • Cost of painting = (19,71,200 / 100) × ₹10 = 19,712 × 10 = ₹1,97,120.
Cost of Painting = ₹1,97,120
2πr = 35.2 m
Hemispherical Dome (r = 5.6 m)

End-of-Chapter Exercises (Questions 1 to 22)

1. Estimate how many scoops of ice cream can be obtained from a cuboidal container of dimensions 10 cm × 15 cm × 20 cm.
  • Intuitive Guess: Around 40 to 50 scoops.
  • Volume of the cuboidal ice-cream container = 10 × 15 × 20 = 3000 cm³.
  • Assumption: Model each scoop of ice cream as a sphere of radius r = 2.5 cm (diameter = 5 cm).
  • Volume of 1 spherical scoop = (4/3)πr³ = (4/3) × 3.14 × (2.5)³ ≈ 65.4 cm³.
  • Estimated number of scoops = 3000 / 65.4 ≈ 45 to 46 scoops (or if radius is taken as 3 cm, V ≈ 113 cm³, giving ≈ 26 to 27 scoops).
Estimate ≈ 45 Scoops (for r = 2.5 cm)
3000 cm³ r ≈ 2.5 cm
Tub (10×15×20 cm) & Spherical Scoop
2. A cube of integer side length a is made of unit cubes. Write an expression giving the number of unit cubes to be added to make a cube of side length a + 1.
  • Number of unit cubes in a cube of side a = Volume = a³.
  • Number of unit cubes in a cube of side (a + 1) = Volume = (a + 1)³.
  • Number of unit cubes to be added = (a + 1)³ − a³
  • = (a³ + 3a² + 3a + 1) − a³ = 3a² + 3a + 1.
Expression = (a + 1)³ − a³ = 3a² + 3a + 1
a³ Side: a → a + 1
Expanding Cube from a to a + 1
*3. Solve the following: (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.
  • (i) Lifetime Water vs. Classroom Volume:
    • Assume a person drinks 3 litres/day for 80 years:
    Total water = 3 × 365 × 80 = 87,600 litres = 87.6 m³ (since 1000 L = 1 m³).
    • A small classroom of dimensions 6 m × 5 m × 3 m has volume = 90 m³ (a larger 9 m × 9 m × 4 m room is 324 m³).
    • Conclusion: Yes, for a small room (~90 m³) it nearly fills the entire room, or fills about 1/3 of a large classroom!
  • (ii) Bricks in Classroom Walls:
    • Assume 4 walls of length 8 m, height 3.5 m, thickness 0.2 m: Gross volume = 4 × 8 × 3.5 × 0.2 = 22.4 m³.
    • Deducting ~20% for doors, windows, and mortar gives net brick volume ≈ 18 m³ = 1,80,00,000 cm³.
    • Volume of 1 standard brick (20 cm × 10 cm × 9 cm) = 1800 cm³.
    • Number of bricks ≈ 1,80,00,000 / 1800 = 10,000 bricks.
(i) ~87.6 m³ (Fills a small room) | (ii) ~10,000 Bricks
Water ≈ 87.6 m³ Walls ≈ 10,000 Bricks
Classroom Water & Brick Estimate
4. You are given two options: (A) one chocolate cube of side 50 mm, (B) hundred chocolate cubes of side 10 mm. Which option gives you more chocolate?
  • Option (A): Volume of 1 cube of side 50 mm = 50³ = 50 × 50 × 50 = 1,25,000 mm³.
  • Option (B): Volume of 100 cubes of side 10 mm = 100 × 10³ = 100 × 1000 = 1,00,000 mm³.
  • Comparing the two volumes: 1,25,000 mm³ > 1,00,000 mm³.
  • Therefore, Option (A) gives 25,000 mm³ (25%) more chocolate.
Option (A) gives more chocolate
50 mm (A) 125,000 100 × (10 mm) (B) 100,000
Option A (125 cm³) > Option B (100 cm³)
5. If all the people in the world are crowded together in one location, how much area would it cover?
  • World population ≈ 8 billion (8 × 10⁹ people).
  • Assumption: In a crowd, about 4 people can stand in 1 m² of floor space (0.25 m² per person).
  • Total area required = (8 × 10⁹) / 4 = 2 × 10⁹ m².
  • Since 1 km² = 10⁶ m², Total Area = 2000 km² (a square of side ≈ 44.7 km, roughly the area of a single metropolitan city like Delhi).
Area ≈ 2000 km² (or 2 × 10⁹ m²)
8 Billion People ≈ 2000 km² 44.7 km × 44.7 km
Square of Side ~44.7 km
6. A school provides milk to its students in cylindrical glasses, each with a diameter of 7 cm. If each glass is filled with milk to a height of 12 cm, how many litres of milk are needed to serve 1600 students? Use 1000 cm³ = 1 litre.
  • Diameter of glass = 7 cm ⇒ Radius (r) = 7 / 2 = 3.5 cm.
  • Height of milk in each glass (h) = 12 cm.
  • Volume of milk in 1 glass = πr²h = (22/7) × 3.5 × 3.5 × 12 = 462 cm³.
  • Total volume of milk for 1600 students = 1600 × 462 = 7,39,200 cm³.
  • Volume in litres = 7,39,200 / 1000 = 739.2 litres.
Milk Needed = 739.2 Litres
h = 12 cm d = 7 cm
1 Glass = 462 cm³ (× 1600 Students)
7. The surface area of a sphere of radius 5 cm is five times the area of the curved surface of a cone of radius 4 cm. Find the height and the volume of the cone.
  • Radius of sphere (R) = 5 cm ⇒ Surface area of sphere = 4πR² = 4π(5)² = 100π cm².
  • Radius of cone (r) = 4 cm ⇒ Curved surface area of cone = πrl = 4πl cm².
  • Given: 100π = 5 × (4πl) ⇒ 100π = 20πl ⇒ Slant height l = 5 cm.
  • Height of the cone (h):
    h = √(l² − r²) = √(5² − 4²) = √(25 − 16) = √9 = 3 cm.
  • Volume of the cone (V):
    V = (1/3)πr²h = (1/3) × (22/7) × 4² × 3 = 352 / 7 ≈ 50.29 cm³ (or 16π ≈ 50.24 cm³).
Height = 3 cm | Volume = 50.29 cm³
R = 5 h=3 r=4 l=5
Sphere (R=5) & Cone (r=4, h=3, l=5)
8. Take Earth to be a perfect sphere with radius 6370 km. Take Jupiter to be a perfect sphere with radius 69,900 km. Take the Sun to be a perfect sphere with radius 6,95,700 km. Compute approximately: (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.
  • Since Volume V = (4/3)πr³, the ratio of volumes of two spheres is the cube of the ratio of their radii: V₁ / V₂ = (r₁ / r₂)³.
  • (i) Volume of Jupiter to Volume of Earth:
    VJupiter / VEarth = (69,900 / 6370)³ ≈ (10.9733)³ ≈ 1321 : 1 (approx. 1300 Earths fit in Jupiter).
  • (ii) Volume of Sun to Volume of Earth:
    VSun / VEarth = (6,95,700 / 6370)³ ≈ (109.215)³ ≈ 13,02,700 : 1 (approx. 1.3 million Earths fit in the Sun).
(i) ≈ 1321 : 1 | (ii) ≈ 1.3 × 10⁶ : 1
Sun Jupiter Earth
Relative Planetary Volumes
9. Show that the volume of a sphere is equal to 2/3 of the volume of the smallest cylinder which encloses it.
  • Let the sphere have radius r.
  • Volume of the sphere: Vsphere = (4/3)πr³.
  • The smallest cylinder that encloses the sphere touches it at the sides, top, and bottom, so:
    • Radius of cylinder = r
    • Height of cylinder (h) = 2r
  • Volume of the enclosing cylinder: Vcylinder = πr²h = πr²(2r) = 2πr³.
  • Now, (2/3) × Vcylinder = (2/3) × (2πr³) = (4/3)πr³ = Vsphere. (Hence proved)
V_sphere = (2/3) V_cylinder (Proved)
r h=2r
Archimedes' Sphere in Cylinder
*10. Suppose the Earth is perfectly spherical. A string is wrapped tightly around the equator of the Earth. Another string, 1 metre longer, is placed around the Earth so that it forms a larger circle, staying the same distance above the ground everywhere. How high above the ground is the second string? 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 of the sphere (Earth) be R metres, and the height of the second string above the ground be h metres.
  • Length of first string = 2πR.
  • Length of second string = 2π(R + h) = 2πR + 1.
  • Simplifying: 2πR + 2πh = 2πR + 1 ⇒ 2πh = 1 ⇒ h = 1 / (2π) ≈ 0.159 m ≈ 15.9 cm.
  • For (i) the Moon, (ii) Jupiter, and (iii) a volleyball: The height is still 15.9 cm in all three cases!
  • Why? Because the radius R cancels out completely in 2π(R + h) − 2πR = 2πh = 1. Thus, the gap h = 1/(2π) is independent of the size of the sphere.
Height h = 1/(2π) m ≈ 15.9 cm (Same for all!)
R h
h = 1/(2π) ≈ 15.9 cm for ANY Sphere
*11. We have a cylinder with a base radius of r cm and height h cm. A square pyramid is fitted inside it. The square base of the pyramid lies on the base of the cylinder, its corners on the boundary of the cylinder. The apex of the pyramid lies on the top of the cylinder. Find the ratio of the volume of this pyramid to the volume of the cylinder.
  • The square base is inscribed in a circle of radius r, so the diagonal of the square is d = 2r.
  • Side of the square (s) = 2r / √2 = r√2 cm.
  • Area of the square base = s² = (r√2)² = 2r² cm².
  • Volume of the square pyramid = (1/3) × Base Area × Height = (1/3) × 2r² × h = (2/3)r²h.
  • Volume of the cylinder = πr²h.
  • Required Ratio = [(2/3)r²h] / [πr²h] = 2 : 3π (or 7 : 33 taking π = 22/7).
Ratio = 2 : 3π (or 7 : 33)
Square Pyramid Inside Cylinder
12. What is 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?
  • (i) Cuboid length increased by 1 unit:
    ΔV = (l + 1)wh − lwh = lwh + wh − lwh = wh cubic units → Option (c).
  • (ii) Cylinder radius increased by 1 unit:
    ΔV = π(r + 1)²h − πr²h = π(r² + 2r + 1)h − πr²h = (2πrh + πh) cubic units → Option (e).
  • *(iii) Sphere radius decreased by 1 unit:
    Volume decreases by: ΔV = (4/3)πr³ − (4/3)π(r − 1)³ = (4/3)π(3r² − 3r + 1) cubic units.
(i) (c) wh | (ii) (e) 2πrh + πh | (iii) (4/3)π(3r² − 3r + 1)
lwh wh ΔV = wh (Slice)
Volume Change When Dimension Changes by 1
*13. Given a cube with volume V, express its total surface area S in terms of V.
  • Let the side length of the cube be a.
  • Volume of the cube: V = a³ ⇒ a = V1/3 (or ∛V).
  • Total Surface Area of the cube: S = 6a².
  • Substituting a = V1/3:
    S = 6(V1/3)² = 6V2/3.
S = 6V^(2/3)
a = V^(1/3)
S = 6a² = 6V^(2/3)
*14. Given a cube with total surface area S, express its volume V in terms of S.
  • Let the side length of the cube be a.
  • Total Surface Area: S = 6a² ⇒ a² = S / 6 ⇒ a = (S / 6)1/2.
  • Volume of the cube: V = a³.
  • Substituting a = (S / 6)1/2:
    V = [(S / 6)1/2]³ = (S / 6)3/2.
V = (S / 6)^(3/2)
a = √(S/6)
V = a³ = (S / 6)^(3/2)
15. A cylindrical glass of height 25 cm and radius 4 cm has water up to a height of 16 cm. A crow wants to drink water from this glass. The water must be at a height of 20 cm for the crow to reach it. There are some marbles lying around. How many marbles of radius 1 cm should the crow drop into the glass to make the water reach the required height?
  • Radius of the cylindrical glass (R) = 4 cm.
  • Required rise in water level (Δh) = 20 cm − 16 cm = 4 cm.
  • Volume of water to be displaced = πR²(Δh) = π × 4² × 4 = 64π cm³.
  • Radius of 1 spherical marble (r) = 1 cm.
  • Volume of 1 marble = (4/3)πr³ = (4/3)π(1)³ = (4/3)π cm³.
  • Number of marbles required = (64π) / [(4/3)π] = 64 × (3/4) = 48 marbles.
Number of Marbles = 48
Δh=4 48 Marbles
Water Rises from 16 cm to 20 cm
16. Find the volume of ink in a new ball point pen. Take the necessary measurements and make approximations, as needed.
  • Model the ink column inside a standard plastic pen refill as a narrow right circular cylinder.
  • Measurements:
    • Inner diameter of refill tube ≈ 3 mm = 0.3 cm ⇒ Radius (r) = 0.15 cm.
    • Height of ink column (h) ≈ 10 cm.
  • Volume of ink = πr²h = 3.14 × (0.15)² × 10 = 3.14 × 0.0225 × 10 ≈ 0.71 cm³ (or 0.71 mL).
Volume of Ink ≈ 0.71 cm³ (0.71 mL)
h ≈ 10 cm r ≈ 0.15 cm
Cylindrical Pen Refill
17. Solve: (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.
  • (i) Number of Chapatis:
    • Volume of dough ball (R = 6 cm) = (4/3)π(6)³ = 288π ≈ 904.78 cm³.
    • Model 1 rolled chapati as a thin cylinder of radius r = 7.5 cm (diameter 15 cm) and thickness h = 0.2 cm:
    Volume of 1 chapati = π(7.5)²(0.2) = 11.25π ≈ 35.34 cm³.
    • Number of chapatis = 288π / 11.25π ≈ 25 to 26 chapatis.
  • *(ii) Edible Volume of Half a Coconut/Muskmelon:
    • Model as a hollow hemisphere with outer radius of flesh R = 6 cm and inner cavity radius r = 4.5 cm (thickness = 1.5 cm).
    • Volume of flesh in one half = (2/3)π(R³ − r³) = (2/3) × 3.14 × (6³ − 4.5³) = 2.093 × (216 − 91.125) ≈ 261.4 cm³.
(i) ~25–26 Chapatis | (ii) ~261 cm³ per half
R = 6 Hollow Hemisphere
Dough Sphere & Hemispherical Fruit
18. Looking at the Ganita Manjari, Grade 9, Part 2 textbook, Sheela wonders: (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?
  • (i) Textbook Pages vs. Classroom Floor:
    • Standard textbook page size ≈ 28 cm × 21 cm = 588 cm² = 0.0588 m².
    • For a Part 2 textbook with ~100 paper sheets (200 pages), total area laid out side-by-side = 100 × 0.0588 = 5.88 m².
    • Since a standard classroom floor is about 6 m × 6 m = 36 m² (or larger), no, the sheets will only cover about 1/6th of the floor.
  • (ii) Maximum Textbooks in Storeroom:
    • Volume of storeroom = 15 × 20 × 30 = 9000 ft³ ≈ 9000 × 28,000 cm³ = 25,20,00,000 cm³.
    • Volume of 1 textbook (28 cm × 21 cm × 1 cm) = 588 cm³.
    • Max number of textbooks = 25,20,00,000 / 588 ≈ 4,28,570 textbooks.
(i) No (~5.88 m²) | (ii) ~4,28,570 Textbooks
5.88 m² Floor ≈ 36 m²
Pages Area vs. Classroom Floor
*19. If the entire human population decided to climb into one giant cube, how long would the side have to be?
  • World population ≈ 8 × 10⁹ people.
  • Assumption: Average volume of a human body is ~0.0625 m³ (62.5 kg), or allowing space between people, ~0.125 m³ per person (8 people per m³).
  • Total volume of the giant cube: V = (8 × 10⁹) × 0.125 m³ = 10⁹ m³.
  • Side length of the cube: a = ∛(10⁹ m³) = 1000 m = 1 km (or ~794 m if packed with zero gaps at 0.0625 m³/person).
Side of Giant Cube ≈ 1 km (1000 m)
V = 1 km³ Side a ≈ 1 km
Giant Cube of Side ~1 km
*20. The Earth's surface has an estimated volume of 1.38 billion km³ of water. Suppose the Earth is a perfect sphere and all this water forms a uniform layer completely covering the Earth's surface, like a thin water bubble. Estimate the thickness of this water layer. The Earth's radius is 6371 km. (i) Write an expression that gives the thickness of this water layer. (ii) Simplify the expression in (i) using a calculator.
  • Volume of water (V) = 1.38 billion km³ = 1.38 × 10⁹ km³.
  • Earth's radius (R) = 6371 km.
  • (i) Expression for thickness (t):
    Since the water layer is very thin compared to R, Volume ≈ Surface Area of Earth × Thickness (t):
    t = V / (4πR²) = (1.38 × 10⁹) / [4π × (6371)²] km.
  • (ii) Simplifying with a calculator:
    4π × (6371)² ≈ 12.5664 × 4,05,89,641 ≈ 5.1006 × 10⁸ km².
    t = (1.38 × 10⁹) / (5.1006 × 10⁸) ≈ 2.705 km ≈ 2.7 km.
Thickness t ≈ 2.7 km
6371 km t ≈ 2.7 km
Uniform Water Layer Around Earth
*21. Give the dimension of a cuboid whose volume is halved when its surface area is doubled.
  • Consider an initial Cuboid 1 (a cube of side 2 cm):
    • Dimensions: 2 cm × 2 cm × 2 cm
    • Original Volume (V₁) = 2 × 2 × 2 = 8 cm³
    • Original Surface Area (S₁) = 6 × 2² = 24 cm²
  • We need a transformed Cuboid 2 with half the volume (V₂ = 4 cm³) and double the surface area (S₂ = 48 cm², so lw + wh + hl = 24):
  • Choose height h = 0.2 cm (1/5 cm), so lw = 4 / 0.2 = 20, and 20 + 0.2(l + w) = 24 ⇒ l + w = 20.
  • Solving l + w = 20 and lw = 20 gives exact dimensions:
    • l = 10 + 4√5 ≈ 18.94 cm, w = 10 − 4√5 ≈ 1.06 cm, h = 0.2 cm.
    • Check: Volume = 20 × 0.2 = 4 cm³ (halved), Surface Area = 2(20 + 0.2 × 20) = 48 cm² (doubled).
2×2×2 cm → 18.94 × 1.06 × 0.2 cm
V=8, SA=24 ⇒ V=4, SA=48
Flattening a Cuboid Halves V & Doubles SA
22. Project: Find the volume of your house making necessary approximations. Present how you solved it.
  • Step 1 (Model the House as a Cuboid): Measure the outer plinth length, width, and floor-to-ceiling height of the house.
  • Step 2 (Measurements):
    • Length (L) = 12 m (approx. 40 ft)
    • Width (W) = 9 m (approx. 30 ft)
    • Height of 1 floor (H) = 3.2 m (approx. 10.5 ft)
  • Step 3 (Calculation):
    • Floor Area = 12 m × 9 m = 108 m².
    • Total Volume of the house = L × W × H = 108 m² × 3.2 m = 345.6 m³.
Estimated House Volume ≈ 345.6 m³
12 m × 9 m × 3.2 m
House Modeled as a Cuboid

Chapter 14 Formula Summary

3D Solid Shape Surface Area Formulas Volume Formula
1. Cuboid (length l, width w, height h) Total Surface Area (TSA) = 2(wl + hw + lh) V = l × w × h
2. Cube (side a) Total Surface Area (TSA) = 6a² V = a³
3. Right Circular Cylinder (radius r, height h) • Curved Surface Area (CSA) = 2πrh
• TSA (open at one end) = 2πrh + πr² = πr(2h + r)
• TSA (closed at both ends) = 2πrh + 2πr² = 2πr(h + r) 		V = πr²h
4. Right Circular Cone (radius r, height h, slant height l = √(h² + r²)) • Curved Surface Area (CSA) = πrl
• Total Surface Area (TSA) = πrl + πr² = πr(l + r) 		V = (1/3)πr²h
5. Pyramid (base area B, height h) Sum of areas of base and all triangular side faces V = (1/3) × Base Area × h
6. Sphere (radius r) Surface Area = 4πr² V = (4/3)πr³
7. Hemisphere (radius r) • Curved Surface Area (CSA) = 2πr²
• Total Surface Area (TSA) = 3πr² 		V = (2/3)πr³