Curriculum
Mathematics Grade 10
Grade 10 Mathematics
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Chapter 1. Real Numbers
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Chapter 2. Polynomials
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Chapter 3: PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
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Chapter 4: QUADRATIC EQUATIONS
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Chapter 5: ARITHMETIC PROGRESSIONS
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Chapter 6: TRIANGLES
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Chapter 7: COORDINATE GEOMETRY
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Chapter 8: INTRODUCTION TO TRIGONOMETRY
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Chapter 9: SOME APPLICATIONS OF TRIGONOMETRY
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Chapter 10: CIRCLES
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Chapter 11: AREAS RELATED TO CIRCLES
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Chapter 12: SURFACE AREAS AND VOLUMES
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Chapter 13: STATISTICS
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Chapter 14: PROBABILITY
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Interactive Worksheets
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Chapter Wise Worksheets
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Solved Problems and solutions
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Chapter 12: SURFACE AREAS AND VOLUMES
Surface Areas and Volumes
Combinations of Solids · Surface Area · Volume · Real-World Applications
| ▢ Cuboid | △ Cone | ◍ Cylinder | ◎ Sphere | ◐ Hemisphere |
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01
Formulae Reference
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02
Combination Rules
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12+
Worked Examples
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A–D
Practice Sets
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8pt
Exam Quick-Check
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Introduction |
In earlier classes you learned to calculate the surface areas and volumes of basic 3-D solids: the cuboid, cone, cylinder, and sphere. In real life, however, most objects are not a single basic shape — they are built from two or more of these shapes joined together.
A water tanker truck is a cylinder with two hemispherical ends. A test tube is a cylinder with a hemispherical closed end. A toy rocket may be a cone sitting on a cylinder. An ice-cream cone has a cone topped with a hemisphere. Understanding how to handle such combinations is the entire focus of this chapter.
The big idea: Break the combined solid into its constituent basic shapes. Then add up the visible surface areas (being careful about faces that disappear when shapes are joined), and add all volumes (volumes always add fully).
| ƒ |
Formulae Reference — All Basic Solids |
| Solid | CSA / LSA | TSA | Volume |
|---|---|---|---|
| Cuboid (l × b × h) | 2(lb + bh + lh) − 2lb LSA = 2h(l + b) |
2(lb + bh + lh) | l × b × h |
| Cube (side a) | 4a² | 6a² | a³ |
| Cylinder (r, h) | 2πrh | 2πr(h + r) | πr²h |
| Cone (r, h, l) | πrl where l = √(r²+h²) | πr(l + r) | (1/3)πr²h |
| Sphere (r) | 4πr² | 4πr² | (4/3)πr³ |
| Hemisphere (r) | 2πr² | 3πr² | (2/3)πr³ |
💡 Key Note on Slant Height
For a cone with base radius r and perpendicular height h: slant height l = √(r² + h²). This is the distance from the apex of the cone down the slant to the edge of the base — always calculated using Pythagoras’ theorem before finding CSA of the cone.
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How to Handle Combinations — The Two Rules |
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Rule 1: Surface Area
Add only the visible (exposed) curved/flat surfaces. When two solids are joined, the faces at the joint disappear from the outer surface. You must subtract those hidden faces. General approach: TSA of combination = Sum of CSAs of each part − Area of faces hidden at each joint + Any exposed flat faces that were not counted |
Rule 2: Volume
Always add volumes — no subtraction needed. Unlike surface areas, volumes are not affected by how the solids are joined. The total volume is simply the sum of the individual volumes. Exception — carved-out shapes: When a solid is hollowed out (e.g., a cone drilled from a cylinder), subtract the volume of the removed portion from the outer solid’s volume. |
Common Combination Types
| Combined Solid | Real-World Example | TSA Rule | Volume Rule |
|---|---|---|---|
| Cylinder + 2 Hemispheres | Water tanker, Capsule | 2×CSA(hemi) + CSA(cyl) | V(cyl) + 2×V(hemi) |
| Cone + Hemisphere | Toy, Ice-cream cone | CSA(cone) + CSA(hemi) | V(cone) + V(hemi) |
| Cone + Cylinder | Toy rocket, Tent | CSA(cone) + CSA(cyl) + base(cyl) | V(cone) + V(cyl) |
| Cube/Cuboid + Hemisphere | Decorative block | TSA(cube) − πr² + 2πr² | V(cube) + V(hemi) |
| Cylinder − Cone (hollow) | Scooped-out object | CSA(cyl) + base(cyl) + CSA(cone) | V(cyl) − V(cone) |
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Worked Examples |
▶ Example 1 — Spinning Top: Cone Surmounted by Hemisphere, Height 5 cm, Diameter 3.5 cm (π = 22/7)
Shape: Hemisphere on top + Cone below. Diameter = 3.5 cm → r = 1.75 cm. Total height = 5 cm.
Visible surface: CSA of hemisphere + CSA of cone (the flat circular base where they join is hidden).
Height of cone = total height − radius of hemisphere = 5 − 1.75 = 3.25 cm
Slant height l = √(r² + h²) = √(1.75² + 3.25²) = √(3.0625 + 10.5625) = √13.625 ≈ 3.7 cm
CSA of hemisphere = 2πr² = 2 × (22/7) × 1.75 × 1.75 = 2 × (22/7) × 3.0625 ≈ 19.25 cm²
CSA of cone = πrl = (22/7) × 1.75 × 3.7 ≈ 20.35 cm²
Total surface area = 19.25 + 20.35 ≈ 39.6 cm²
▶ Example 2 — Decorative Block: Cube (edge 5 cm) + Hemisphere (diameter 4.2 cm) on Top (π = 22/7)
Shape: Cube base with a hemisphere sitting on top face. r = 2.1 cm, a = 5 cm.
TSA of cube = 6 × 5² = 150 cm²
The circle where the hemisphere sits on the cube is hidden, so subtract πr² and add CSA of hemisphere (2πr²).
TSA of block = 150 − πr² + 2πr² = 150 + πr²
= 150 + (22/7) × 2.1 × 2.1 = 150 + (22/7) × 4.41 = 150 + 13.86 = 163.86 cm²
▶ Example 3 — Toy Rocket: Cone (d=5 cm, h=6 cm) on Cylinder (d=3 cm, total h=26 cm) (π = 3.14)
Shape: Cone on top of cylinder. Cone r = 2.5 cm, h = 6 cm. Cylinder r′ = 1.5 cm, h′ = 20 cm.
Slant height of cone l = √(2.5² + 6²) = √(6.25 + 36) = √42.25 = 6.5 cm
Note: The cone base (r=2.5) is wider than cylinder base (r′=1.5). So a ring-shaped annulus on the cone base is visible.
Orange area = CSA(cone) + πr² − π(r′)² = πrl + π(r² − r′²)
= π[(2.5 × 6.5) + (2.5² − 1.5²)] = π[16.25 + 6.25 − 2.25] = π × 20.25 = 3.14 × 20.25 ≈ 63.59 cm²
Yellow area = CSA(cylinder) + base of cylinder = 2πr′h′ + π(r′)²
= πr′(2h′ + r′) = 3.14 × 1.5 × (40 + 1.5) = 4.71 × 41.5 ≈ 195.47 cm²
▶ Example 4 — Bird-Bath: Cylinder (r=30 cm, h=1.45 m) with Hemispherical Depression (π = 22/7)
Shape: Open cylinder with a hemispherical bowl scooped into the top. r = 30 cm, h = 145 cm.
The visible surface = CSA of the cylinder (outer side) + CSA of the hemispherical depression (inner bowl).
TSA = CSA(cylinder) + CSA(hemisphere) = 2πrh + 2πr² = 2πr(h + r)
= 2 × (22/7) × 30 × (145 + 30) = 2 × (22/7) × 30 × 175
= (2 × 22 × 30 × 175)/7 = 231000/7 = 33000 cm² = 3.3 m²
▶ Example 5 — Industrial Shed: Cuboid (7m×15m×8m) + Half-Cylinder on Top; Volume of Air (π = 22/7)
Shape: Cuboid (l=15m, b=7m, h=8m) with a half-cylinder along the 15 m length on top. Diameter of half-cylinder = 7 m, so r = 3.5 m.
Volume of cuboid = 15 × 7 × 8 = 840 m³
Volume of half-cylinder = (1/2) × πr²h = (1/2) × (22/7) × 3.5² × 15 = (1/2) × (22/7) × 12.25 × 15
= (1/2) × 22 × 1.75 × 15 = (1/2) × 577.5 = 288.75 m³
Total volume of shed = 840 + 288.75 = 1128.75 m³
Space used by machinery = 300 m³; by 20 workers = 20 × 0.08 = 1.6 m³
Air remaining = 1128.75 − 300 − 1.6 = 827.15 m³
▶ Example 6 — Juice Glass: Cylinder (d=5 cm, h=10 cm) with Hemispherical Base — Actual vs Apparent Capacity (π = 3.14)
Shape: Cylinder with an upward-facing hemispherical bump at the base. r = 2.5 cm, h = 10 cm.
Apparent capacity (as if a plain cylinder) = πr²h = 3.14 × 2.5² × 10 = 3.14 × 62.5 = 196.25 cm³
Volume of hemispherical bump = (2/3)πr³ = (2/3) × 3.14 × 2.5³ = (2/3) × 3.14 × 15.625 ≈ 32.71 cm³
Actual capacity = 196.25 − 32.71 = 163.54 cm³
▶ Example 7 — Solid Toy: Hemisphere + Cone (r=2 cm, cone h=2 cm); Cylinder Circumscribes — Volume Difference (π = 3.14)
Shape: Hemisphere (r=2 cm) with cone (r=2 cm, h=2 cm) sitting on top. Both share the same circular base.
Volume of toy = V(hemisphere) + V(cone)
= (2/3)πr³ + (1/3)πr²h = (2/3) × 3.14 × 8 + (1/3) × 3.14 × 4 × 2
= 16.75 + 8.37 = 25.12 cm³
Circumscribed cylinder: radius = 2 cm, height = 2 (hemi) + 2 (cone) = 4 cm.
Volume of cylinder = πr²h = 3.14 × 4 × 4 = 50.24 cm³
Difference = 50.24 − 25.12 = 25.12 cm³
▶ Example 8 — Two Cubes (volume 64 cm³ each) Joined End to End — Surface Area of Resulting Cuboid
Two cubes, each of volume 64 cm³, are joined face-to-face.
Side of each cube = ∛64 = 4 cm.
Resulting cuboid: l = 8 cm, b = 4 cm, h = 4 cm.
Surface area = 2(lb + bh + lh) = 2(8×4 + 4×4 + 8×4) = 2(32 + 16 + 32) = 2 × 80 = 160 cm²
Note: Two square faces (each 4×4 = 16 cm²) were hidden when the cubes were joined. TSA of two separate cubes = 2 × 6 × 16 = 192 cm². After joining: 192 − 2 × 16 = 160 cm². ✓
▶ Example 9 — Vessel: Hollow Hemisphere (d=14 cm) + Hollow Cylinder; Total height 13 cm — Inner Surface Area (π = 22/7)
r = 7 cm. Height of cylindrical part = 13 − 7 = 6 cm.
Inner surface area = CSA(hemisphere) + CSA(cylinder)
= 2πr² + 2πrh = 2πr(r + h) = 2 × (22/7) × 7 × (7 + 6)
= 2 × 22 × 13 = 572 cm²
▶ Example 10 — Cylinder with Conical Cavity (both h=2.4 cm, d=1.4 cm) — Total Surface Area of Remaining Solid
r = 0.7 cm, h = 2.4 cm. A cone of same height and radius is hollowed from one end.
Slant height of cone l = √(0.7² + 2.4²) = √(0.49 + 5.76) = √6.25 = 2.5 cm
TSA of remaining solid = CSA of cylinder + base of cylinder + CSA of cone
= 2πrh + πr² + πrl = πr(2h + r + l)
= (22/7) × 0.7 × (4.8 + 0.7 + 2.5) = 2.2 × 8 = 17.6 cm²
▶ Example 11 — Sweet (Gulab Jamun Shape): Cylinder + 2 Hemispheres, Length 5 cm, Diameter 2.8 cm — Volume of Syrup in 45 Pieces (π = 22/7)
r = 1.4 cm. Total length = 5 cm → cylinder height = 5 − 2r = 5 − 2.8 = 2.2 cm.
Volume of one piece = πr²h + (4/3)πr³ (cylinder + full sphere equivalent)
= (22/7) × 1.4² × 2.2 + (4/3) × (22/7) × 1.4³
= (22/7) × 1.96 × 2.2 + (4/3) × (22/7) × 2.744
= 13.55 + 11.50 = 25.05 cm³ (approx.)
Volume of 45 pieces = 45 × 25.05 = 1127.28 cm³
Syrup (30%) = 0.30 × 1127.28 ≈ 338.18 cm³
▶ Example 12 — Iron Pole: Cylinder (h=220 cm, d=24 cm) + Cylinder on Top (h=60 cm, r=8 cm) — Mass (π = 3.14, 1 cm³ = 8 g)
Two cylinders stacked. Lower: r = 12 cm, h = 220 cm. Upper: r = 8 cm, h = 60 cm.
Volume of lower cylinder = 3.14 × 12² × 220 = 3.14 × 144 × 220 = 99475.2 cm³
Volume of upper cylinder = 3.14 × 8² × 60 = 3.14 × 64 × 60 = 12057.6 cm³
Total volume = 99475.2 + 12057.6 = 111532.8 cm³
Mass = 111532.8 × 8 g = 892262.4 g ≈ 892.26 kg
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Practice Sets A – D |
A1. A toy is a cone (r=3.5 cm) mounted on a hemisphere of the same radius. Total height = 15.5 cm. Find TSA. (π = 22/7) [Cone h = 15.5 − 3.5 = 12 cm; l = √(3.5² + 12²) = √156.25 = 12.5 cm; TSA = πrl + 2πr² = (22/7)×3.5×12.5 + 2×(22/7)×12.25 = 137.5 + 77 = 214.5 cm²]
A2. A capsule is a cylinder with two hemispheres at each end. Total length = 14 mm, diameter = 5 mm. Find surface area. (π = 22/7) [r = 2.5 mm; cylinder h = 14 − 5 = 9 mm; TSA = 2πrh + 2×2πr² = 2πr(h + 2r) = 2×(22/7)×2.5×(9+5) = 220 mm²]
A3. A cube of side 7 cm is surmounted by a hemisphere. Find the greatest possible diameter and then find the TSA of the solid. (π = 22/7) [Greatest d = 7 cm, r = 3.5 cm; TSA = 6×49 − π×3.5² + 2π×3.5² = 294 + π×12.25 = 294 + 38.5 = 332.5 cm²]
A4. A tent is a cylinder (h=2.1 m, d=4 m) with a conical top (slant height 2.8 m). Find canvas area (no base). (π = 22/7) [Canvas = CSA(cyl) + CSA(cone) = 2πrh + πrl = πr(2h + l) = (22/7)×2×(4.2+2.8) = 44 m²]
B1. A cone (r=h=1 cm) stands on a hemisphere of same radius. Find total volume in terms of π. [V = (2/3)πr³ + (1/3)πr²h = (2/3)π + (1/3)π = π cm³]
B2. A model is a cylinder (d=3 cm, total length 12 cm) with cones (h=2 cm each) at both ends. Find total volume. (π = 22/7) [Cylinder h = 12 − 4 = 8 cm; r = 1.5 cm; V = πr²h + 2×(1/3)πr²h_cone = π×2.25×8 + 2×(1/3)×π×2.25×2 = 18π + 3π = 21π = 66 cm²]
B3. A wooden article has a cylinder (h=10 cm, r=3.5 cm) with hemispheres scooped from each end. Find total surface area. (π = 22/7) [TSA = CSA(cyl) + 2×CSA(hemi) = 2πrh + 2×2πr² = 2πr(h+2r) = 2×(22/7)×3.5×(10+7) = 374 cm²]
B4. A pen-holder cuboid (15×10×3.5 cm) has four conical depressions (r=0.5 cm, depth=1.4 cm). Find volume of wood. (π = 22/7) [V(cuboid) = 525 cm³; each cone = (1/3)×(22/7)×0.25×1.4 = 11/30 cm³; 4 cones = 44/30 ≈ 1.47 cm³; Wood = 525 − 1.47 ≈ 523.53 cm³]
C1. From a solid cylinder (h=2.4 cm, d=1.4 cm) a cone of same dimensions is hollowed out. Find TSA of remaining solid. (π = 22/7) [l = √(0.7² + 2.4²) = 2.5 cm; TSA = 2πrh + πr² + πrl = πr(2h + r + l) = (22/7)×0.7×(4.8+0.7+2.5) = 2.2×8 = 17.6 cm²]
C2. An inverted cone (h=8 cm, r=5 cm) is full of water. Lead spheres (r=0.5 cm) are dropped in and 1/4 of water flows out. How many spheres were dropped? (π = 3.14) [V(cone) = (1/3)×3.14×25×8 = 209.33 cm³; 1/4 overflows = 52.33 cm³; V(sphere) = (4/3)×3.14×0.125 = 0.524 cm³; Number = 52.33/0.524 ≈ 100]
C3. A hemisphere (r=60 cm) + cone (r=60 cm, h=120 cm) is placed in a cylinder (r=60 cm, h=180 cm) full of water. Find volume of water remaining. (π = 3.14) [V(cyl) = 3.14×3600×180 = 2034720 cm³; V(solid) = (2/3)π×216000 + (1/3)π×3600×120 = 452160 + 452160 = 904320 cm³; Water = 2034720 − 904320 = 1130400 cm³ ≈ 1.131 m³]
C4. A spherical vessel (d=8.5 cm) has a cylindrical neck (length=8 cm, d=2 cm). Find its volume. (π = 3.14) [V(sphere) = (4/3)×3.14×(4.25)³ ≈ 321.39 cm³; V(cyl) = 3.14×1²×8 = 25.12 cm³; Total ≈ 346.51 cm³ — the child’s estimate of 345 cm³ is approximately correct]
D1. A hemispherical depression (diameter = edge of cube = l) is cut from one face of a wooden cube. Find surface area of remaining solid. [TSA = 6l² − πl²/4 + 2π(l/2)² = 6l² − πl²/4 + πl²/2 = 6l² + πl²/4 = l²(6 + π/4)]
D2. A solid cylinder (r=6 cm, h=10 cm) has a cone (same base) drilled from the top. If the cone height = 10 cm, find the volume and TSA of remaining solid. (π = 3.14) [V = πr²h − (1/3)πr²h = (2/3)πr²h = (2/3)×3.14×36×10 = 753.6 cm³; l = √(36+100) ≈ 11.66 cm; TSA = CSA(cyl) + base(cyl) + CSA(cone) = 2πrh + πr² + πrl = 3.14×6×(20+6+11.66) ≈ 708.8 cm²]
D3. A metallic sphere (r=4.2 cm) is melted and recast into small spheres of radius 0.6 cm. How many small spheres are formed? [V(large) = (4/3)π×4.2³; V(small) = (4/3)π×0.6³; Number = 4.2³/0.6³ = 74.088/0.216 = 343]
D4. A cylindrical bucket (h=32 cm, r=18 cm) of sand is emptied on the ground forming a conical heap of radius 24 cm. Find the height and slant height of the heap. [V(cyl) = V(cone): π×324×32 = (1/3)π×576×h → h = (324×32×3)/576 = 54 cm; l = √(576 + 2916) = √3492 ≈ 59.09 cm]
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Chapter Summary |
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▢ Surface Area — Subtract Joints
Combine only the visible surfaces. Faces hidden at joints are subtracted. Add back any newly exposed flat areas. Never add TSAs of components directly. |
△ Volume — Always Additive
Total volume = sum of component volumes. No subtraction needed for joints. Only subtract when a cavity or depression is carved out of the solid. |
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◍ Slant Height l = √(r² + h²)
Always compute slant height before finding CSA of a cone. Given r and h, use Pythagoras. For problems involving total height of a combined solid, first find h of cone separately. |
◎ Hemisphere Formulae
CSA = 2πr² (curved bowl surface). TSA = 3πr² (CSA + flat circular base). Volume = (2/3)πr³. Half a sphere — all values are exactly half those of a full sphere (except flat base). |
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🔄 Conversion of Solids
When a solid is melted and recast: total volume is conserved. Set V(original) = V(new) to find number of new pieces or new dimensions. This is a separate concept from combinations. |
☷ Real-World Applications
Tanker trucks, test tubes, tent canvas area, toy rockets, juice glasses, industrial sheds, bird-baths, iron poles, decorative blocks — all solved using the combination/subtraction approach. |
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8-Point Exam Quick-CheckCommon mistakes and essential rules to score full marks |
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① Never add TSAs when combining shapes
The most common error. When two solids are joined, the joined faces are no longer exposed. Always identify which surfaces are visible and add only those. Blindly adding TSA of cone + TSA of hemisphere gives a wrong answer. |
② Compute slant height before using CSA of cone
CSA of cone = πrl, not πrh. The slant height l = √(r² + h²). In combined solid problems, you must find h of the cone first (by subtracting r from total height), then compute l. This step is skipped by many students. |
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③ Volumes always add — no subtraction at joints
Unlike surface area, volume is not affected by how solids are joined. The total volume is always the sum of individual volumes. Only subtract volume when a hole or cavity is carved out of a solid. |
④ Hemisphere CSA vs TSA — know the difference
CSA of hemisphere = 2πr² (the curved bowl part only). TSA = 3πr² (bowl + flat circle). In combinations, use CSA — because the flat circular base of the hemisphere is hidden at the joint. Using TSA leads to double-counting. |
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⑤ Account for the annular ring when bases differ
In the rocket example, the cone base (r=2.5 cm) is wider than the cylinder base (r=1.5 cm). The exposed ring of the cone base must be added to the orange area. Area of ring = π(R² − r²). Check whether bases are equal or unequal. |
⑥ For hollowed-out solids — subtract volume, add inner CSA
When a shape is scooped out: (a) volume decreases — subtract V(cavity) from V(outer solid), and (b) surface area increases — add CSA of the inner cavity wall (e.g., CSA of the cone drilled into a cylinder). |
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⑦ Use π = 22/7 for multiples of 7; π = 3.14 otherwise
When the problem says “take π = 22/7” and r = 3.5 or 7 or 14, the simplification is designed to give whole numbers. If the problem says “use π = 3.14”, use that. Mixing the two in one problem is a calculation error. |
⑧ Conversion problems: equate volumes
When a solid is melted and reformed into a new shape, set the two volumes equal. This gives you either the number of small shapes, or a missing dimension of the new shape. No surface area formula is needed for pure conversion problems. |
Grade 10 Maths Chapter 12 — Surface Areas and Volumes extends students’ knowledge of basic solid geometry to combinations of two or more 3-D shapes. The central skill is identifying which surfaces are visible on a combined solid and computing their total area — avoiding the common error of adding total surface areas of individual parts directly. Volumes, by contrast, simply add up for any combination. Key combinations studied include a cylinder with hemispherical ends (tanker/capsule), a cone on a cylinder (rocket/tent), a hemisphere on a cube (decorative block), and shapes with conical or hemispherical cavities drilled out. Slant height computation using Pythagoras’ theorem is essential before applying the cone’s curved surface area formula πrl. The chapter also covers conversion of solids, where a shape is melted and recast, and volume conservation is used to find new dimensions or counts. This fully original revision page covers all formulae for basic solids, detailed combination rules, 12 solved examples, four graded practice sets with complete answers, a chapter summary, and an 8-point exam quick-check — comprehensive preparation for any assessment on surface areas and volumes of combined solids in Class 10.
Keywords: surface areas and volumes class 10, combination of solids, CSA of cone cylinder hemisphere, volume of combination grade 10, TSA of combined solids, slant height formula