Class 8 Mathematics Notes Chapter 11 (Mensuration) – Mathematics Book
Alright class, let's focus on Chapter 11: Mensuration. This is a crucial chapter, not just for your Class 8 exams, but it forms the foundation for many quantitative aptitude questions in government exams. Mensuration deals with the measurement of geometric figures – their perimeter, area, surface area, and volume. Pay close attention to the formulas and understand when and how to apply them.
Chapter 11: Mensuration - Detailed Notes
1. Introduction
Mensuration is the branch of geometry that deals with the measurement of length, area, and volume of various geometric shapes.
2. Review of Plane Figures (2D Shapes)
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Perimeter: The total length of the boundary of a closed plane figure. Unit: unit of length (cm, m, km).
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Area: The measure of the region enclosed by a closed plane figure. Unit: square unit (cm², m², km²).
-
Key Formulas (Revision):
- Rectangle:
- Perimeter = 2 × (length + breadth) = 2(l + b)
- Area = length × breadth = l × b
- Square: (Special rectangle where l = b = side = a)
- Perimeter = 4 × side = 4a
- Area = side × side = a²
- Triangle:
- Perimeter = Sum of all three sides
- Area = ½ × base × height = ½ × b × h
- Parallelogram:
- Perimeter = 2 × (sum of adjacent sides)
- Area = base × corresponding height = b × h
- Circle:
- Circumference (Perimeter) = 2πr (where r is the radius)
- Area = πr²
- Rectangle:
3. Area of Specific Quadrilaterals
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Trapezium:
- A quadrilateral with exactly one pair of parallel sides.
- Let the parallel sides be 'a' and 'b', and the perpendicular distance (height) between them be 'h'.
- Area of Trapezium = ½ × (sum of parallel sides) × height = ½ × (a + b) × h
- Remember: The height 'h' must be perpendicular to the parallel sides.
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General Quadrilateral:
- To find the area, we can divide it into two triangles using one of its diagonals.
- Let the diagonal be 'd'. Let the perpendiculars (heights) from the other two vertices to this diagonal be h₁ and h₂.
- Area of General Quadrilateral = Area of Triangle 1 + Area of Triangle 2
= (½ × d × h₁) + (½ × d × h₂)
= **½ × d × (h₁ + h₂) **
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Rhombus:
- A parallelogram with all sides equal. Its diagonals bisect each other at right angles (90°).
- Let the lengths of the diagonals be d₁ and d₂.
- Since the diagonals divide the rhombus into 4 congruent right-angled triangles, we can find the area.
- Area of Rhombus = ½ × (product of diagonals) = ½ × d₁ × d₂
- (Alternatively, since a rhombus is a parallelogram, Area = base × height if base and height are known).
4. Area of Polygons
- To find the area of irregular polygons (like pentagons, hexagons), we divide the polygon into simpler shapes whose areas we know how to calculate (e.g., triangles, rectangles, trapeziums).
- Method: Draw diagonals or perpendiculars strategically to split the polygon. Calculate the area of each part and add them up.
5. Solid Shapes (3D Shapes)
- These shapes have length, breadth, and height (or depth). They occupy space.
- Key solid shapes in this chapter: Cube, Cuboid, Cylinder.
6. Surface Area of Solid Shapes
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Surface Area (Total Surface Area - TSA): The sum of the areas of all the faces (surfaces) of a solid shape. Unit: square unit (cm², m²).
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Lateral Surface Area (LSA) / Curved Surface Area (CSA): The area of the faces excluding the top and bottom faces. Unit: square unit (cm², m²).
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Cuboid: (Dimensions: length 'l', breadth 'b', height 'h')
- It has 6 rectangular faces.
- Total Surface Area (TSA) = 2 × (lb + bh + hl)
- Lateral Surface Area (LSA) = Area of four walls = 2 × (l + b) × h
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Cube: (Special cuboid where l = b = h = side = a)
- It has 6 square faces.
- Total Surface Area (TSA) = 6 × (area of one face) = 6 × a² = 6a²
- Lateral Surface Area (LSA) = 4 × (area of one face) = 4 × a² = 4a²
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Right Circular Cylinder: (Base radius 'r', height 'h')
- It has 2 circular bases (top and bottom) and one curved surface.
- Curved Surface Area (CSA) = (Circumference of base) × height = (2πr) × h = 2πrh
- Total Surface Area (TSA) = CSA + Area of two circular bases
= 2πrh + 2(πr²)
= 2πr(h + r) or 2πr(r + h)
7. Volume of Solid Shapes
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Volume: The measure of the space occupied by a solid shape. Unit: cubic unit (cm³, m³).
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Generally, Volume = Area of Base × Height (for prisms and cylinders).
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Cuboid:
- Area of base = l × b
- Volume = Area of base × height = (l × b) × h = lbh
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Cube:
- Area of base = a × a = a²
- Volume = Area of base × height = a² × a = a³
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Cylinder:
- Area of base (circle) = πr²
- Volume = Area of base × height = (πr²) × h = πr²h
8. Units and Conversions
- Volume and Capacity: Capacity often refers to the quantity a container can hold.
- 1 Litre (L) = 1000 cm³
- 1 millilitre (mL) = 1 cm³
- 1 m³ = 1000 L = 10,00,000 cm³ (1 million cm³)
- Area:
- 1 cm² = 100 mm²
- 1 m² = 10,000 cm²
- 1 Hectare (ha) = 10,000 m²
Key Takeaways for Exams:
- Memorize the formulas accurately.
- Understand the difference between Perimeter/Circumference, Area, LSA/CSA, TSA, and Volume.
- Identify the shape correctly from the problem description.
- Pay close attention to the units used in the question and required in the answer. Be prepared for unit conversions.
- Practice word problems to understand how these concepts are applied in real-world scenarios (e.g., cost of painting walls, amount of water in a tank, area of a field).
Multiple Choice Questions (MCQs)
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The area of a trapezium is 34 cm² and the length of one of the parallel sides is 10 cm and its height is 4 cm. The length of the other parallel side is:
(a) 7 cm
(b) 8 cm
(c) 6 cm
(d) 10 cm -
The area of a rhombus is 240 cm² and one of the diagonals is 16 cm. The length of the other diagonal is:
(a) 15 cm
(b) 30 cm
(c) 20 cm
(d) 40 cm -
If the side of a cube is doubled, its total surface area will become:
(a) Double
(b) Four times
(c) Six times
(d) Eight times -
The lateral surface area of a cuboid with dimensions 10 cm × 8 cm × 5 cm is:
(a) 400 cm²
(b) 360 cm²
(c) 180 cm²
(d) 500 cm² -
A cylindrical tank has a radius of 7 m and height of 3 m. How much water can it hold? (Use π = 22/7)
(a) 462 m³
(b) 132 m³
(c) 308 m³
(d) 44 m³ -
The diagonal of a quadrilateral is 20 cm. The lengths of perpendiculars dropped on it from the opposite vertices are 8.5 cm and 11.5 cm. The area of the quadrilateral is:
(a) 400 cm²
(b) 200 cm²
(c) 100 cm²
(d) 300 cm² -
The curved surface area of a right circular cylinder of height 14 cm is 88 cm². The radius of its base is: (Use π = 22/7)
(a) 1 cm
(b) 2 cm
(c) 0.5 cm
(d) 4 cm -
The volume of a cube is 729 cm³. Its total surface area is:
(a) 81 cm²
(b) 486 cm²
(c) 324 cm²
(d) 6561 cm² -
A rectangular paper of width 14 cm is rolled along its width to form a cylinder of radius 20 cm. The volume of the cylinder is: (Use π = 22/7)
(a) 17600 cm³
(b) 1760 cm³
(c) 8800 cm³
(d) 1960 cm³ -
How many litres of water can a cubical tank of side length 1.2 m hold?
(a) 1728 L
(b) 172.8 L
(c) 17.28 L
(d) 1.728 L
Answer Key for MCQs:
- (a) 7 cm [Area = ½(a+b)h => 34 = ½(10+b)4 => 34 = 2(10+b) => 17 = 10+b => b=7]
- (b) 30 cm [Area = ½ × d₁ × d₂ => 240 = ½ × 16 × d₂ => 240 = 8 × d₂ => d₂ = 30]
- (b) Four times [Original TSA = 6a². New side = 2a. New TSA = 6(2a)² = 6(4a²) = 4 × (6a²)]
- (c) 180 cm² [LSA = 2(l+b)h = 2(10+8)5 = 2(18)5 = 10 × 18 = 180]
- (a) 462 m³ [Volume = πr²h = (22/7) × 7² × 3 = (22/7) × 49 × 3 = 22 × 7 × 3 = 462]
- (b) 200 cm² [Area = ½ × d × (h₁+h₂) = ½ × 20 × (8.5 + 11.5) = ½ × 20 × 20 = 10 × 20 = 200]
- (a) 1 cm [CSA = 2πrh => 88 = 2 × (22/7) × r × 14 => 88 = 2 × 22 × r × 2 => 88 = 88r => r=1]
- (b) 486 cm² [Volume = a³ = 729 => a = ∛729 = 9 cm. TSA = 6a² = 6 × 9² = 6 × 81 = 486]
- (a) 17600 cm³ [When rolled along width, width becomes height (h=14cm). Radius r=20cm. Volume = πr²h = (22/7) × 20² × 14 = (22/7) × 400 × 14 = 22 × 400 × 2 = 17600]
- (a) 1728 L [Side = 1.2 m. Volume = a³ = (1.2)³ m³ = 1.728 m³. Since 1 m³ = 1000 L, Volume = 1.728 × 1000 L = 1728 L]
Make sure you practice more problems based on these concepts. Good luck!