Class 10 Notes

Class 10 Mathematics Notes Chapter 13 (Surface areas and volumes) – Mathematics Book

Philoid

Philoid

— 7 min read

Mathematics
Alright class, let's focus on a very important chapter for your competitive exams – Chapter 13: Surface Areas and Volumes from your Class 10 NCERT Mathematics textbook. This chapter deals with calculating the surface area (the 'skin' of an object) and volume (the space it occupies) for various three-dimensional shapes, both individually and in combination. Mastery of the formulas and concepts here is crucial.

Chapter 13: Surface Areas and Volumes - Detailed Notes

1. Introduction

  • Surface Area: The total area of the surface of a three-dimensional object. It's measured in square units (like cm², m²).
  • Volume: The amount of space occupied by a three-dimensional object. It's measured in cubic units (like cm³, m³).
  • Key Solids: We will primarily deal with the Cuboid, Cube, Right Circular Cylinder, Right Circular Cone, Sphere, Hemisphere, and combinations of these solids, including the Frustum of a Cone.

2. Basic Solids: Formulas

(a) Cuboid

  • A solid bounded by six rectangular faces.
  • Let length = l, breadth = b, height = h.
  • Lateral Surface Area (LSA): Area of the four walls = 2h(l + b)
  • Total Surface Area (TSA): Area of all six faces = 2(lb + bh + hl)
  • Volume (V): l × b × h
  • Length of Diagonal: √(l² + b² + h²)

(b) Cube

  • A special cuboid where all edges are equal.
  • Let edge = a (or l).
  • Lateral Surface Area (LSA): 4a²
  • Total Surface Area (TSA): 6a²
  • Volume (V): a³
  • Length of Diagonal: a√3

(c) Right Circular Cylinder

  • A solid with two parallel circular bases connected by a curved surface, where the axis is perpendicular to the bases.
  • Let radius of base = r, height = h.
  • Curved Surface Area (CSA): Area of the curved part = 2πrh
  • Total Surface Area (TSA): CSA + Area of two circular bases = 2πrh + 2πr² = 2πr(h + r)
  • Volume (V): Area of base × height = πr²h

(d) Right Circular Cone

  • A solid with a circular base, a vertex, and a curved surface connecting them. The height is the perpendicular distance from the vertex to the center of the base.
  • Let radius of base = r, height = h, slant height = l.
  • Slant Height (l): l = √(r² + h²) or l² = r² + h² (This relationship is very important!)
  • Curved Surface Area (CSA): πrl
  • Total Surface Area (TSA): CSA + Area of circular base = πrl + πr² = πr(l + r)
  • Volume (V): (1/3) × Area of base × height = (1/3)πr²h

(e) Sphere

  • A perfectly round geometrical object in three-dimensional space. All points on the surface are equidistant from the center.
  • Let radius = r.
  • Surface Area (SA): 4πr² (Note: There's no distinction between CSA and TSA for a sphere)
  • Volume (V): (4/3)πr³

(f) Hemisphere

  • Exactly half of a sphere.
  • Let radius = r.
  • Curved Surface Area (CSA): Half the SA of the sphere = 2πr²
  • Total Surface Area (TSA): CSA + Area of the circular top = 2πr² + πr² = 3πr²
  • Volume (V): Half the volume of the sphere = (2/3)πr³

(g) Frustum of a Cone

  • A part of a cone obtained by cutting it with a plane parallel to its base. It has two circular bases of different radii.
  • Let radii of the two circular ends be r₁ and r₂ (r₁ > r₂), height = h, slant height = l.
  • Slant Height (l): l = √(h² + (r₁ - r₂)²)
  • Curved Surface Area (CSA): π(r₁ + r₂)l
  • Total Surface Area (TSA): CSA + Area of top base + Area of bottom base = π(r₁ + r₂)l + πr₁² + πr₂²
  • Volume (V): (1/3)πh(r₁² + r₂² + r₁r₂)

3. Combination of Solids

  • Many real-world objects are combinations of these basic shapes (e.g., a tent as a cylinder surmounted by a cone, an ice-cream cone as a cone topped with a hemisphere, a capsule as a cylinder with hemispherical ends).
  • Surface Area of Combined Solids:
    • Visualize the combined solid.
    • Identify the exposed surfaces.
    • Add the areas of these exposed surfaces ONLY.
    • Crucial Point: When combining solids, the surfaces where they join are not included in the total surface area of the combined solid. Do NOT simply add the TSAs of the individual solids.
    • Example: Capsule (Cylinder + 2 Hemispheres) -> TSA = CSA of Cylinder + CSA of Hemisphere 1 + CSA of Hemisphere 2 = 2πrh + 2πr² + 2πr² = 2πrh + 4πr².
  • Volume of Combined Solids:
    • This is generally simpler.
    • Just add the volumes of the individual solids that make up the combined object.
    • Example: Ice-cream (Cone + Hemisphere) -> Volume = Volume of Cone + Volume of Hemisphere = (1/3)πr²h + (2/3)πr³.

4. Conversion of Solid from One Shape to Another

  • This involves melting a solid of one shape and recasting it into another shape (or multiple smaller shapes).
  • Key Principle: The volume of the material remains constant during the conversion process (assuming no wastage).
  • Method: Calculate the volume of the original solid and equate it to the volume of the new solid(s).
  • Example: A metallic sphere is melted and recast into a cylinder. -> Volume of Sphere = Volume of Cylinder -> (4/3)πR³ = πr²h (where R is sphere radius, r and h are cylinder radius and height). You can then solve for the unknown dimension.
  • Example: A large cone is melted into several small spherical balls. -> Volume of Cone = n × Volume of one Spherical ball (where n is the number of balls).

5. Important Tips for Exams

  • Memorize Formulas: There's no substitute for knowing the formulas accurately. Write them down multiple times.
  • Units: Pay close attention to units (cm, m). Ensure consistency. If radii are in cm and height in m, convert them to the same unit before calculation. Remember Area units (cm², m²) and Volume units (cm³, m³).
  • π Value: Use π = 22/7 unless specified otherwise or if the radii/dimensions are multiples of 7. Use π = 3.14 if specified or if calculations seem easier with it (e.g., radius = 10).
  • Visualization: Draw a rough diagram for combination/conversion problems. It helps clarify which surfaces/volumes are involved.
  • Calculations: Be careful with arithmetic, especially when dealing with fractions and π. Double-check your calculations.
  • Read Carefully: Understand whether the question asks for LSA/CSA, TSA, or Volume. Identify the shapes involved correctly.

Multiple Choice Questions (MCQs)

  1. A cylindrical pencil sharpened at one edge is a combination of which two shapes?
    (A) A cone and a cylinder
    (B) A frustum of a cone and a cylinder
    (C) A hemisphere and a cylinder
    (D) Two cylinders

  2. The total surface area of a cube is 96 cm². The volume of the cube is:
    (A) 8 cm³
    (B) 512 cm³
    (C) 64 cm³
    (D) 27 cm³

  3. A cone of height 24 cm has a curved surface area of 550 cm². Taking π = 22/7, find the radius of its base.
    (A) 7 cm
    (B) 10 cm
    (C) 14 cm
    (D) 12 cm

  4. If a metallic sphere of radius 6 cm is melted and drawn into a wire (cylinder) of radius 0.2 cm, what is the length of the wire?
    (A) 36 m
    (B) 72 m
    (C) 144 m
    (D) 288 m

  5. The radii of the top and bottom ends of a bucket (frustum of a cone) are 14 cm and 7 cm respectively. If its height is 24 cm, what is its slant height?
    (A) 24 cm
    (B) 25 cm
    (C) 26 cm
    (D) 30 cm

  6. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 7 cm and the height of the cone is equal to its radius. Find the volume of the solid. (Use π = 22/7)
    (A) 1078 cm³
    (B) 1232 cm³
    (C) 718.67 cm³
    (D) 1540 cm³

  7. If the radius of a sphere is doubled, its volume becomes:
    (A) Double
    (B) Four times
    (C) Six times
    (D) Eight times

  8. The curved surface area of a right circular cylinder of height 14 cm is 88 cm². The diameter of the base is:
    (A) 0.5 cm
    (B) 1.0 cm
    (C) 1.5 cm
    (D) 2.0 cm

  9. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have?
    (A) 14 cm
    (B) 3.5 cm
    (C) 7 cm
    (D) 10 cm

  10. The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is:
    (A) 9.7 cm³
    (B) 77.6 cm³
    (C) 19.4 cm³
    (D) 38.8 cm³

Answer Key for MCQs:

  1. (A) A cone and a cylinder
  2. (C) 64 cm³ (TSA = 6a² = 96 => a² = 16 => a = 4 cm. Volume = a³ = 4³ = 64 cm³)
  3. (A) 7 cm (CSA = πrl = 550. l = √(h² + r²) = √(24² + r²) = √(576 + r²). (22/7) * r * √(576 + r²) = 550. r√(576+r²) = (5507)/22 = 257 = 175. Squaring: r²(576+r²) = 175². By inspection or solving, if r=7, l=√(576+49)=√625=25. CSA = (22/7)725 = 550. So r=7 cm.)
  4. (B) 72 m (Volume of Sphere = (4/3)π(6)³ = 288π cm³. Volume of wire = π(0.2)²h = 0.04πh cm³. Equating: 288π = 0.04πh => h = 288 / 0.04 = 28800 / 4 = 7200 cm = 72 m)
  5. (B) 25 cm (l = √(h² + (r₁ - r₂)²) = √(24² + (14 - 7)²) = √(576 + 7²) = √(576 + 49) = √625 = 25 cm)
  6. (A) 1078 cm³ (r = 7 cm, h_cone = 7 cm. Volume = Vol(Cone) + Vol(Hemisphere) = (1/3)πr²h + (2/3)πr³ = (1/3)πr³ + (2/3)πr³ = πr³ = (22/7) * 7³ = 22 * 49 = 1078 cm³)
  7. (D) Eight times (V = (4/3)πr³. New V' = (4/3)π(2r)³ = (4/3)π(8r³) = 8 * [(4/3)πr³] = 8V)
  8. (D) 2.0 cm (CSA = 2πrh = 88. 2 * (22/7) * r * 14 = 88. 2 * 22 * r * 2 = 88. 88r = 88 => r = 1 cm. Diameter = 2r = 2 cm)
  9. (C) 7 cm (The hemisphere sits on top face. Its base circle must fit within the square face. Max diameter = side of the cube = 7 cm)
  10. (C) 19.4 cm³ (Largest cone: Base diameter = edge of cube = 4.2 cm => radius r = 2.1 cm. Height of cone = edge of cube = h = 4.2 cm. Volume = (1/3)πr²h = (1/3) * (22/7) * (2.1)² * 4.2 = (1/3) * (22/7) * 4.41 * 4.2 = (1/3) * 22 * 0.63 * 4.2 = 22 * 0.21 * 4.2 ≈ 19.404 cm³)

Make sure you practice a wide variety of problems from NCERT and reference books to get comfortable with these concepts and calculations. Good luck!

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