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Surface Areas and Volumes

NCERT Class 10 · Mathematics Based on NCERT Class 10 Mathematics textbook · Free CBSE study kit

Chapter Notes

**SURFACE AREAS AND VOLUMES — CHAPTER 12 CHEAT SHEET**

**KEY CONCEPT: Combination of Solids**

• A combination of solids is formed by joining two or more basic solids (cuboid, cone, cylinder, sphere, hemisphere)

• Common examples: cylinder + two hemispheres (oil tank), cylinder + hemisphere (test tube), cone + hemisphere (toy), cylinder + cone (rocket)

• To find surface area or volume: BREAK into individual solids → calculate each part → combine appropriately

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**FORMULA BANK — BASIC SOLIDS (Review)**

**HEMISPHERE:**

• Curved Surface Area (CSA) = 2πr²

• Total Surface Area (TSA) = 3πr²

• Volume = (2/3)πr³

**CONE:**

• Curved Surface Area (CSA) = πrl (where l = slant height)

• Total Surface Area (TSA) = πrl + πr² = πr(l + r)

• Slant height: l = √(r² + h²)

• Volume = (1/3)πr²h

**CYLINDER:**

• Curved Surface Area (CSA) = 2πrh

• Total Surface Area (TSA) = 2πrh + 2πr² = 2πr(h + r)

• Volume = πr²h

**CUBE:**

• TSA = 6a² (where a = edge)

• Volume = a³

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**SECTION 12.2: SURFACE AREA OF COMBINATION OF SOLIDS**

**CRITICAL PRINCIPLE: When Two Solids Are Joined**

  • Identify which surfaces are VISIBLE and which are HIDDEN (attached/joined)
  • Visible surfaces: count their CSA or TSA
  • Hidden surfaces: EXCLUDE from total calculation
  • Total Surface Area = Sum of CSAs of visible parts ONLY

    • **WORKED EXAMPLE PATTERNS:**

    **Pattern 1: Hemisphere + Cone (Playing Top)**

    • TSA = CSA of hemisphere + CSA of cone

    • Height of cone = Total height − radius of hemisphere

    • Find slant height: l = √(r² + h²)

    • Substitute into formulas

    • **KEY MISTAKE:** TSA ≠ (TSA of cone) + (TSA of hemisphere) — the base areas cancel because they are joined

    **Pattern 2: Cube + Hemisphere (Decorative Block)**

    • TSA = TSA of cube − (base area of hemisphere) + (CSA of hemisphere)

    • Simplifies to: TSA = 6a² − πr² + 2πr² = 6a² + πr²

    • The hemisphere's circular base replaces part of the cube's top face

    **Pattern 3: Cylinder + Cone (Rocket)**

    • When cone base diameter > cylinder base diameter: a RING is exposed

    • Area painted = CSA of cone + (base area of cone − base area of cylinder)

    • Area painted = CSA of cone + π(r₁² − r₂²)

    • For cylinder: CSA + one base area (if the other base is hidden or covered)

    **Pattern 4: Cylinder + Hemisphere (Bird Bath)**

    • TSA = CSA of cylinder + CSA of hemisphere

    • = 2πrh + 2πr² = 2πr(h + r)

    • Both solids have same radius r

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    **STEP-BY-STEP SOLUTION METHOD FOR ANY COMBINATION**

    1. **Identify the solids** → Draw/visualize clearly which parts touch

    2. **List given data** → radius(es), height(s), diameter(s)

    3. **Calculate missing dimensions** → slant height, height of each part

    4. **Determine visible surfaces** → Mark what is exposed, what is joined

    5. **Apply formulas** → CSA of each visible component

    6. **Add CSAs** → Do NOT include base areas of joined surfaces

    7. **Handle special cases** → Ring areas, exposed bases, etc.

    8. **Final answer** → With correct units and approximation

    ---

    **COMMON MISTAKES TO AVOID**

    ❌ **MISTAKE 1:** Adding TSAs of individual solids

    ✓ CORRECT: Add only CSAs of visible parts; subtract base areas that are internal

    ❌ **MISTAKE 2:** Forgetting to convert diameter to radius

    ✓ CORRECT: Always use r = d/2; check units (cm/m)

    ❌ **MISTAKE 3:** Not calculating slant height for cone

    ✓ CORRECT: Must find l = √(r² + h²) before using CSA = πrl

    ❌ **MISTAKE 4:** Miscalculating height of cone when part of total height is hemisphere

    ✓ CORRECT: h_cone = total height − radius_hemisphere (NOT − diameter)

    ❌ **MISTAKE 5:** Missing the ring area when bases have different radii

    ✓ CORRECT: Include π(r₁² − r₂²) as painted area

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    **FORMULA REFERENCE TABLE**

    | Solid | CSA | TSA | Slant Height |

    |-------|-----|-----|---------------|

    | Hemisphere | 2πr² | 3πr² | — |

    | Cone | πrl | πr(l+r) | √(r²+h²) |

    | Cylinder | 2πrh | 2πr(h+r) | — |

    | Cube face | — | 6a² | — |

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    **EXAMPLE 1 ANALYSIS: Playing Top (Cone + Hemisphere)**

    Given: Total height = 5 cm, diameter = 3.5 cm

  • r = 1.75 cm
  • Height of hemisphere = 1.75 cm
  • Height of cone = 5 − 1.75 = 3.25 cm
  • Slant height l = √(1.75² + 3.25²) = 3.7 cm
  • CSA hemisphere = 2π(1.75)² = 2π(3.0625)
  • CSA cone = π(1.75)(3.7)
  • TSA = Sum of both CSAs

    • ---

    **EXAMPLE 2 ANALYSIS: Cube + Hemisphere Block**

    Cube edge = 5 cm, hemisphere diameter = 4.2 cm (r = 2.1 cm)

  • TSA of cube alone = 6(5)² = 150 cm²
  • Hemisphere base area = π(2.1)² (this is HIDDEN)
  • Hemisphere CSA = 2π(2.1)² (this is VISIBLE)
  • Block TSA = 150 − π(2.1)² + 2π(2.1)² = 150 + π(2.1)²

    • ---

    **EXAMPLE 3 ANALYSIS: Cylinder + Cone Rocket**

    Cone: r₁ = 2.5 cm, h = 6 cm → l = 6.5 cm

    Cylinder: r₂ = 1.5 cm, h = 20 cm

  • Since r₁ > r₂, a ring is exposed at junction
  • Orange area = πr₁l + π(r₁² − r₂²) = π[r₁l + r₁² − r₂²]
  • Yellow area = 2πr₂h + πr₂²

    • ---

    **EXAMPLE 4 ANALYSIS: Cylinder + Hemisphere Bird Bath**

    Cylinder height = 1.45 m = 145 cm, r = 30 cm (common radius)

  • TSA = 2πrh + 2πr² = 2πr(h + r)
  • = 2π(30)(145 + 30) = 2π(30)(175) = 10500π

    • ---

    **MUST-REMEMBER POINTS**

    • **π approximation:** Use π = 22/7 unless stated π = 3.14

    • **Units consistency:** Convert all measurements to same unit before calculating

    • **Visible only:** Only count surfaces that are exposed to outside

    • **Base area handling:** When two solids join at their bases, subtract those base areas

    • **Ring concept:** When different-sized bases meet, the annulus (ring) area matters

    • **Slant height:** Only for cone/pyramid; use l = √(r² + h²)

    • **Total surface area ≠ Sum of individual TSAs:** This is the #1 student error

    MCQs — 10 Questions with Answers

    Q1. A container is made by joining a cylinder of radius 7 cm with two hemispheres of the same radius at both ends. When calculating the total surface area, a student claims that the flat circular faces of the hemispheres must be included. Is this claim correct?

    • A. Yes, because hemispheres have two circular faces
    • B. No, because the flat faces are interior to the combined solid and not part of the external surface ✓
    • C. Yes, because we must count all faces of each component
    • D. No, because hemispheres have no flat faces

    Answer: B — When two solids are joined, their interface faces become interior and are excluded from the external surface area; only curved surfaces of the hemispheres and cylinder are exposed.

    Q2. Assertion (A): When a hemisphere is placed on top of a cube, the total surface area of the combined solid equals the total surface area of the cube plus the curved surface area of the hemisphere. Reason (R): The base of the hemisphere and the top face of the cube overlap when the solids are joined. Choose the correct option:

    • A. Both A and R are true and R is the correct explanation of A ✓
    • B. Both A and R are true but R is not the correct explanation of A
    • C. A is true but R is false
    • D. A is false but R is true

    Answer: A — Both statements are true: the overlapping circular base of the hemisphere with the cube's top face means neither contributes to external surface area, making the assertion correct and R logically explains why.

    Q3. A toy is made by surmounting a cone on top of a hemisphere such that their bases coincide exactly. When finding the painted surface area, should the base area of the cone be included?

    • A. Yes, because the base is part of the cone
    • B. No, because the base is internal to the combined solid and completely covered by the hemisphere ✓
    • C. Yes, because we must paint all surfaces of both solids
    • D. No, because the cone's base is smaller than the hemisphere's base

    Answer: B — When the cone and hemisphere are joined with coinciding bases, the circular base becomes an internal interface and is completely hidden, so it cannot be painted or included in surface area.

    Q4. A student argues that the total surface area of a cylinder with two hemispheres at its ends can be calculated by adding the curved surface area of the cylinder to the total surface area of both hemispheres (including their flat circular bases). Is this reasoning valid?

    • A. Yes, this method always works for composite solids
    • B. No, because the flat circular bases of the hemispheres become interior surfaces and must be subtracted ✓
    • C. Yes, because hemispheres are complete solids with circular bases
    • D. No, because hemispheres do not have curved surface areas

    Answer: B — The flat bases of the hemispheres are glued to the cylinder's ends and become interior surfaces; they must be excluded by subtracting them from the total.

    Q5. Assertion (A): In a composite solid made of a cone mounted on a cylinder, if the cone's base diameter is larger than the cylinder's base diameter, then part of the cone's base will be visible and must be included in the painted surface area. Reason (R): The exposed part of the cone's base forms an annular (ring) region that contributes to the external surface. Choose the correct option:

    • A. Both A and R are true and R is the correct explanation of A ✓
    • B. Both A and R are true but R is not the correct explanation of A
    • C. A is true but R is false
    • D. A is false but R is true

    Answer: A — When the cone's base is larger, the annular region (outer circle of cone minus inner circle of cylinder) is indeed exposed and must be painted, making both A and R true with R correctly explaining A.

    Q6. A decorative block consists of a cube with a hemispherical dome on its top face. The hemisphere's diameter is smaller than the cube's edge length. A student calculates the total surface area as: TSA = 6(edge)² − πr² + 2πr², where r is the hemisphere's radius. Is this formula correct?

    • A. Yes, because it accounts for all exposed surfaces of both solids ✓
    • B. No, because the formula subtracts and adds the same term, yielding TSA = 6(edge)² + πr², which is incorrect
    • C. Yes, because the first term represents the cube and the second represents the hemisphere
    • D. No, because it fails to account for the curved surface of the hemisphere

    Answer: A — The formula correctly removes the contact area (πr²) from the cube's total and adds the curved surface area of the hemisphere (2πr²), giving TSA = 6(edge)² + πr².

    Q7. Assertion (A): For a composite solid made by attaching a cylinder and a hemisphere with the same radius, the total surface area equals the sum of the curved surface areas of both components only. Reason (R): The flat circular base of the hemisphere coincides with one of the bases of the cylinder, so both these faces are interior to the combined solid. Choose the correct option:

    • A. Both A and R are true and R is the correct explanation of A ✓
    • B. Both A and R are true but R is not the correct explanation of A
    • C. A is true but R is false
    • D. A is false but R is true

    Answer: A — Assertion A is correct because only curved surfaces are external; Reason R correctly explains this by identifying that both flat faces become interior interfaces.

    Q8. A cone and a hemisphere have the same radius r. When the cone is placed with its base on the flat face of the hemisphere, does the entire circular base of the cone need to be painted if the combined toy is painted on all external surfaces?

    • A. Yes, because the base of the cone is always external
    • B. No, because the entire base is internal and fully covered by the hemisphere ✓
    • C. Partially, only if the cone's base is larger than the hemisphere's circular face
    • D. Only the curved surface of the cone needs painting, not its base

    Answer: B — Since both solids have equal radius and their bases coincide, the cone's base is completely covered and becomes an interior interface, so it is not painted.

    Q9. Assertion (A): When calculating the surface area of a cylinder with a hemispherical depression at one end (like a bird-bath), the curved surface area of the hemisphere is added to the lateral surface area of the cylinder and the area of the exposed base. Reason (R): The hemispherical depression is carved inward, so its curved inner surface becomes part of the external surface area of the bird-bath. Choose the correct option:

    • A. Both A and R are true and R is the correct explanation of A ✓
    • B. Both A and R are true but R is not the correct explanation of A
    • C. A is true but R is false
    • D. A is false but R is true

    Answer: A — Both statements correctly describe a bird-bath: the inner curved surface of the hemispherical depression is exposed and must be included, along with the cylinder's lateral surface and one base.

    Q10. A solid is formed by joining two cones base-to-base (forming a double cone or bicone). To find its total surface area, should the area of the common base be counted?

    • A. Yes, because each cone has a base area
    • B. No, because the base of each cone is internal and completely hidden at the junction ✓
    • C. Yes, because the double cone has a visible widest section
    • D. No, because double cones have no base area by definition

    Answer: B — When two cones are joined base-to-base, their common base becomes an interior surface and is completely concealed, so only the two curved surfaces of the cones are part of the external surface area.

    Flashcards

    What is the curved surface area of a hemisphere with radius r?

    CSA of hemisphere = 2πr²

    When combining two solids, what surfaces do we count in total surface area?

    Only the exposed curved surfaces; hidden bases where solids touch are NOT counted.

    A cone is mounted on a cylinder. What three areas must we calculate?

    CSA of cone + CSA of cylinder + ring area (base of cone minus base of cylinder).

    Formula for curved surface area of a cone with radius r and slant height l?

    CSA of cone = πrl

    A hemisphere sits on top of a cube. Which part of the cube's surface is removed?

    The circular base of the hemisphere where it touches the cube is subtracted from the cube's top face.

    How do you find the slant height of a cone given radius r and height h?

    Slant height l = √(r² + h²)

    What is the total surface area formula for a cylinder with radius r and height h?

    TSA of cylinder = 2πr(r + h) or 2πrh + 2πr²

    In a composite solid, how do you find the height of one component if total height is given?

    Subtract the known heights of other components from the total height.

    A hemispherical depression means the curved surface is inside. Do we count it?

    Yes, because it is an exposed internal surface that needs to be painted or treated.

    Why is TSA of a composite solid NOT equal to sum of individual TSAs?

    Because the base areas where solids are joined together are removed from the total.

    Important Board Questions

    A hemisphere of radius 3.5 cm is placed on top of a cube of side 5 cm. State the formula for the total surface area of the resulting solid and identify which area from the cube is NOT counted. [2 marks]

    TSA = (Total surface of cube) − (Base area of hemisphere) + (Curved surface of hemisphere) = 6(5²) − π(3.5)² + 2π(3.5)². The circular base where hemisphere sits is subtracted because it becomes internal.

    A toy is in the form of a cone surmounted on a hemisphere, both having radius 3.5 cm. The total height of the toy is 15.5 cm. Find (i) the height of the cone, and (ii) the slant height of the cone. [3 marks]

    Height of cone = 15.5 − 3.5 = 12 cm (subtract radius of hemisphere from total height). Use l = √(r² + h²) = √(3.5² + 12²) to find slant height ≈ 12.5 cm. Show both calculations with substitution.

    A wooden rocket consists of a cone mounted on a cylinder. The height of the entire rocket is 26 cm, with the conical part 6 cm high. The cone base diameter is 5 cm (radius 2.5 cm) and the cylinder base diameter is 3 cm (radius 1.5 cm). Find (i) the curved surface area of the cone, (ii) the curved surface area of the cylinder, and (iii) the exposed ring area at the junction. Use π = 3.14. [5 marks]

    First calculate slant height of cone: l = √(2.5² + 6²) ≈ 6.5 cm. CSA of cone = πrl = 3.14 × 2.5 × 6.5. Height of cylinder = 26 − 6 = 20 cm. CSA of cylinder = 2πrh = 2 × 3.14 × 1.5 × 20. Ring area = π(R² − r²) = 3.14(2.5² − 1.5²). Show all three calculations separately and identify which surfaces are painted orange and which yellow based on the problem context.

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