Lesson Notes By Weeks and Term v5 - Grade 11
Measurement – Week 2 focus
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Measurement – Week 2 focus
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Subject: Mathematics
Class: Grade 11
Term: 3rd Term
Week: 2
Theme: General lesson support
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This week, we will delve deeper into the world of measurement, focusing on the application of perimeter, area, surface area, and volume to more complex 2D and 3D shapes. Understanding measurement is crucial not only for success in mathematics but also for practical applications in everyday life, from calculating the amount of paint needed for a room to designing efficient packaging for products. In a South African context, these skills are vital for entrepreneurs, builders, architects, farmers, and many other professions. We will be working towards solving problems that require a deeper understanding of geometric properties and the relationships between different measurements.
2.1 Perimeter and Area of Composite Shapes: A composite shape is formed by combining two or more basic shapes (e.g., triangles, rectangles, circles, and sectors). To find the perimeter of a composite shape, we add up the lengths of all the outer boundaries. To find the area, we typically break the composite shape down into simpler shapes, calculate the area of each simpler shape, and then add or subtract the areas as needed.
Example 1: Consider a shape formed by a rectangle (length 10cm, breadth 5cm) with a semi-circle attached to one of the length sides. Calculate the perimeter and area.
Perimeter: The perimeter consists of three sides of the rectangle (5cm + 10cm + 5cm = 20cm) and the curved part of the semi-circle. The diameter of the semi-circle is 10cm, so the radius is 5cm. The circumference of a full circle is 2πr, so the length of the semi-circular arc is (1/2) 2 π 5 = 5π cm.
Therefore, the total perimeter is 20 + 5π ≈ 20 + 15.71 ≈ 35.71 cm.
Area: The area is the sum of the area of the rectangle and the area of the semi-circle. Area of rectangle = length breadth = 10cm * 5cm = 50 cm². Area of semi-circle = (1/2) π r² = (1/2) π * (5cm)² = (25/2)π cm².
Therefore, the total area is 50 + (25/2)π ≈ 50 + 39.27 ≈ 89.27 cm². 2.2 Surface Area and Volume of 3D Objects: Surface Area: The total area of all the surfaces of a 3D object. It's like wrapping paper covering the entire object.
Volume: The amount of space a 3D object occupies. It's like the amount of water the object could hold.
Formulas: Prism: Surface Area = 2 (Area of base) + (Perimeter of base) height; Volume = (Area of base) height.
Cylinder: Surface Area = 2πr² + 2πrh; Volume = πr²h.
Pyramid: Surface Area = (Area of base) + (1/2) (Perimeter of base) slant height; Volume = (1/3) (Area of base) * height.
Cone: Surface Area = πr² + πrl (where l is the slant height); Volume = (1/3)πr²h.
Sphere: Surface Area = 4πr²; Volume = (4/3)πr³.
Example 2: A water tank consists of a cylinder (radius 2m, height 3m) topped by a hemisphere (radius 2m). Find the total surface area and volume.
Surface Area: Cylinder (excluding top): 2πrh + πr² = 2π(2)(3) + π(2)² = 12π + 4π = 16π m² Hemisphere: (1/2) 4πr² = 2π(2)² = 8π m² Total Surface Area = 16π + 8π = 24π ≈ 75.40 m² Volume: Cylinder: πr²h = π(2)²(3) = 12π m³ Hemisphere: (1/2) (4/3)πr³ = (2/3)π(2)³ = (16/3)π m³ Total Volume = 12π + (16/3)π = (36π + 16π)/3 = (52/3)π ≈ 54.45 m³ 2.3 Optimisation Problems: Optimisation involves finding the maximum or minimum value of a function (e.g., surface area or volume) subject to certain constraints (e.g., fixed perimeter or volume). This often involves using calculus techniques (derivatives), which will be covered in more detail later. For this section, we'll focus on geometric reasoning and problem-solving.
Example 3: A farmer wants to build a rectangular kraal for his sheep. He has 40m of fencing. What dimensions should the kraal be to maximise the area? Let the length be l and the breadth be b. The perimeter is 2l + 2b = 40, so l + b = 20, and b = 20 - l. The area is A = l b = l(20 - l) = 20l - l*². To maximise the area, we look for a symmetrical solution, where the rectangle is as close to a square as possible. Since l + b = 20, if l = b then 2l = 20, and l =
1
0. So, b = 10 as well. The maximum area occurs when the kraal is a square with sides of 10m. Area = 10m 10m = 100 m². 2.4 Unit Conversions: It's important to be able to convert between different units of measurement (e.g., cm to m, m² to hectares, cm³ to litres).
Remember: 1 m = 100 cm 1 m² = 10000 cm² 1 hectare = 10000 m² 1 litre = 1000 cm³ 1 m³ = 1000 litres Example 4: Convert 5 hectares to m² and then to km². 5 hectares = 5 10000 m² = 50000 m² 1 km = 1000 m, so 1 km² = (1000 m)² = 1000000 m² Therefore, 50000 m² = 50000 / 1000000 km² = 0.05 km² Guided Practice (With Solutions)
Question 1: A circular garden (radius 4m) is surrounded by a path of width 1m. Calculate the area of the path.
Solution: Area of the garden = πr² = π(4)² = 16π m² Radius of the garden plus the path = 4m + 1m = 5m Area of the garden plus the path = π(5)² = 25π m² Area of the path = (Area of garden plus path) - (Area of garden) = 25π - 16π = 9π ≈ 28.27 m² Question 2: A rectangular box (length 12cm, width 8cm, height 5cm) is to be filled with cylindrical tins of diameter 4cm and height 5cm. What is the maximum number of tins that can fit into the box?
Solution: Radius of the tins = 4cm / 2 = 2cm Number of tins that can fit along the length = 12cm / 4cm = 3 Number of tins that can fit along the width = 8cm / 4cm = 2 Number of tins that can fit along the height = 5cm / 5cm = 1 Total number of tins = 3 2 * 1 = 6 tins.
Question 3: A cone has a slant height of 13cm and a base radius of 5cm. Calculate its volume.