Class 9 Mathematics Notes Chapter 11 (Chapter 11) – Examplar Problem (Englisha) Book

Alright class, let's get straight into Chapter 11: Surface Areas and Volumes. This chapter is fundamental for understanding 3D geometry and frequently appears in various government examinations. Accuracy with formulas and understanding which 'area' (Lateral/Curved or Total) is required are key. We will be referring to the concepts as covered in your NCERT Class 9 Exemplar book.

Chapter 11: Surface Areas and Volumes - Detailed Notes

1. Introduction
This chapter deals with calculating the surface area (the area of the outer surfaces) and volume (the space occupied) by various three-dimensional shapes.

2. Cuboid
A 3D shape with six rectangular faces. Let length = l, breadth = b, height = h.

• Lateral Surface Area (LSA): Area of the four walls.
  • LSA = 2h(l + b)
• Total Surface Area (TSA): Area of all six faces.
  • TSA = 2(lb + bh + hl)
• Volume (V): Space occupied.
  • V = l × b × h
• Diagonal: Length of the longest rod that can fit inside.
  • Diagonal = √(l² + b² + h²)

3. Cube
A special cuboid where all edges are equal. Let edge = a.

• Lateral Surface Area (LSA): Area of four faces.
  • LSA = 4a²
• Total Surface Area (TSA): Area of all six faces.
  • TSA = 6a²
• Volume (V):
  • V = a³
• Diagonal:
  • Diagonal = a√3

4. Right Circular Cylinder
A solid generated by revolving a rectangle about one of its sides. Let radius of the base = r, height = h.

• Curved Surface Area (CSA): Area of the curved rectangular sheet forming the cylinder.
  • CSA = 2πrh
• Total Surface Area (TSA): CSA + Area of two circular bases.
  • TSA = 2πrh + 2πr² = 2πr(h + r)
• Volume (V): Area of base × height.
  • V = πr²h
• Hollow Cylinder: If R is the external radius and r is the internal radius.
  • Volume of material = π(R² - r²)h
  • CSA = External CSA + Internal CSA = 2πRh + 2πrh
  • TSA = CSA + Area of two rings = 2π(R+r)h + 2π(R² - r²)

5. Right Circular Cone
A solid generated by revolving a right-angled triangle about one of the sides containing the right angle. Let radius = r, height = h, slant height = l.

• Slant Height (l): The distance from the apex to any point on the circumference of the base.
  • l² = r² + h² => l = √(r² + h²)
• Curved Surface Area (CSA): Area of the curved surface.
  • CSA = πrl
• Total Surface Area (TSA): CSA + Area of the circular base.
  • TSA = πrl + πr² = πr(l + r)
• Volume (V):
  • V = (1/3)πr²h (Note: Volume of cone = 1/3 * Volume of cylinder with same base radius and height)

6. Sphere
A perfectly round 3D object, like a ball. Let radius = r.

• Surface Area (SA): It has only one surface (curved).
  • SA = 4πr²
• Volume (V):
  • V = (4/3)πr³

7. Hemisphere
Exactly half of a sphere. Let radius = r.

• Curved Surface Area (CSA): Area of the curved part.
  • CSA = 2πr² (Half of the sphere's surface area)
• Total Surface Area (TSA): CSA + Area of the flat circular base.
  • TSA = 2πr² + πr² = 3πr²
• Volume (V): Half the volume of the sphere.
  • V = (2/3)πr³

Key Concepts & Tips for Exams:

• Units: Be extremely careful with units. Area is in square units (m², cm²) and Volume is in cubic units (m³, cm³). Ensure consistency throughout the calculation.
• Conversions: Remember common volume conversions:
  • 1 m³ = 1000 Litres
  • 1 Litre = 1000 cm³
  • 1 m³ = 10⁶ cm³ (1 million cubic centimetres)
• LSA/CSA vs TSA: Read the question carefully.
  • "Area of four walls", "Cost of painting a room (excluding ceiling/floor)" usually implies LSA (for cuboid) or CSA (for cylinder/cone).
  • "Material required", "Total sheet needed", "Cost of painting an object completely" usually implies TSA.
  • For objects like an open box or a tent, calculate the area of the required surfaces only.
• Melting & Recasting: When one solid shape is melted and recast into another shape (or multiple smaller shapes), the volume remains constant. Equate the initial volume to the final volume(s).
• Ratios: If the ratio of dimensions (like radii or heights or edges) of two similar solids is a:b, then:
  • Ratio of their surface areas = a²:b²
  • Ratio of their volumes = a³:b³
• Pythagoras Theorem: Essential for cones to find slant height (l) when radius (r) and height (h) are known, or vice versa.
• π Value: Use π = 22/7 unless specified otherwise (e.g., 3.14).

Exemplar Focus:
Exemplar problems often involve:

• Finding dimensions when volume or surface area is given.
• Calculating the cost of painting or material based on surface area.
• Determining the capacity of containers (volume).
• Problems involving ratios of dimensions, areas, or volumes.
• Situations requiring careful interpretation (e.g., area of label on a cylindrical can, canvas for a tent which might be conical + cylindrical).

Multiple Choice Questions (MCQs)

Here are 10 MCQs based on the concepts from Surface Areas and Volumes:

1. 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

2. The length, breadth, and height of a cuboid are 15 cm, 10 cm, and 20 cm respectively. Its lateral surface area is:
(A) 1000 cm²
(B) 1300 cm²
(C) 750 cm²
(D) 500 cm²

3. The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratio of the surface areas of the balloon in the two cases is:
(A) 1:4
(B) 1:3
(C) 2:3
(D) 2:1

4. A right circular cylinder just encloses a sphere of radius 'r'. The ratio of the surface area of the sphere to the curved surface area of the cylinder is:
(A) 1:1
(B) 1:2
(C) 2:1
(D) 1:√2

5. The radii of two cylinders are in the ratio 2:3 and their heights are in the ratio 5:3. The ratio of their volumes is:
(A) 10:17
(B) 20:27
(C) 17:27
(D) 20:37

6. The curved surface area of a right circular cone of height 15 cm and base diameter 16 cm is:
(A) 60π cm²
(B) 68π cm²
(C) 120π cm²
(D) 136π cm²

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

8. How many litres of water can a hemispherical bowl of radius 21 cm hold? (Use π = 22/7)
(A) 19.404 Litres
(B) 38.808 Litres
(C) 9.702 Litres
(D) 20.51 Litres

9. The dimensions of a room are 8m x 6m x 4m. The cost of whitewashing the four walls of the room at a rate of ₹ 50 per m² is:
(A) ₹ 5600
(B) ₹ 11200
(C) ₹ 4800
(D) ₹ 9600

10. The volume of a cone is 1570 cm³. If the radius of its base is 10 cm, its height is: (Use π = 3.14)
(A) 10 cm
(B) 15 cm
(C) 18 cm
(D) 20 cm

Answer Key for MCQs:

1. (C) TSA = 6a² = 96 => a² = 16 => a = 4 cm. Volume = a³ = 4³ = 64 cm³.
2. (A) LSA = 2h(l+b) = 2 * 20 (15 + 10) = 40 * 25 = 1000 cm².
3. (A) Surface area of hemisphere ∝ r². Ratio = (r₁/r₂)² = (6/12)² = (1/2)² = 1:4.
4. (A) For cylinder enclosing sphere, h = 2r. SA(Sphere) = 4πr². CSA(Cylinder) = 2πr(h) = 2πr(2r) = 4πr². Ratio = 4πr² / 4πr² = 1:1.
5. (B) V₁/V₂ = (πr₁²h₁)/(πr₂²h₂) = (r₁/r₂)² * (h₁/h₂) = (2/3)² * (5/3) = (4/9) * (5/3) = 20/27. Ratio = 20:27.
6. (D) Diameter = 16 cm => r = 8 cm. h = 15 cm. l = √(r² + h²) = √(8² + 15²) = √(64 + 225) = √289 = 17 cm. CSA = πrl = π * 8 * 17 = 136π cm².
7. (D) V ∝ r³. If r becomes 2r, Volume becomes (2r)³ = 8r³. So, eight times.
8. (A) Volume = (2/3)πr³ = (2/3) * (22/7) * (21)³ = (2/3) * (22/7) * 21 * 21 * 21 = 2 * 22 * 21 * 21 = 19404 cm³. 19404 cm³ = 19404 / 1000 Litres = 19.404 Litres.
9. (A) Area of four walls = LSA = 2h(l+b) = 2 * 4 (8 + 6) = 8 * 14 = 112 m². Cost = Area * Rate = 112 * 50 = ₹ 5600.
10. (B) V = (1/3)πr²h => 1570 = (1/3) * 3.14 * (10)² * h => 1570 = (1/3) * 3.14 * 100 * h => 1570 = (314/3) * h => h = (1570 * 3) / 314 = 5 * 3 = 15 cm.

Study these notes and formulas thoroughly. Practice problems from the Exemplar book to gain confidence. Good luck!