Lesson Notes By Weeks and Term v5 - Grade 12
Measurement: complex applications in real-life contexts – Week 5 focus
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Measurement: complex applications in real-life contexts – Week 5 focus
Download the Lessonotes Mobile South Africa app for faster lesson access on Android and iPhone.
Subject: Mathematical Literacy
Class: Grade 12
Term: 2nd Term
Week: 5
Theme: General lesson support
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This week, we delve into applying measurement principles in complex real-life scenarios. Measurement is fundamental to numerous aspects of our daily lives, from planning a road trip to understanding municipal bills and even optimizing the use of resources in our communities. In South Africa, with its diverse landscapes and socio-economic realities, understanding and applying measurement accurately is crucial for informed decision-making and problem-solving. We will focus on scenarios that require combining different measurement concepts and performing multi-step calculations to arrive at meaningful solutions.
2. 1. Complex Measurement Scenarios Many real-world measurement problems require a combination of skills and knowledge. These scenarios are "complex" because they involve: Multiple steps: You need to perform several calculations sequentially.
Different units: You need to convert between units (e.g., meters to centimeters, liters to milliliters).
Estimation: You may need to estimate measurements when exact values are not available.
Application of multiple formulas: You need to know which formulas to use in different situations and potentially combine them. 2.
2. Area and Volume of Irregular Shapes Irregular shapes don't have standard formulas. To find their area or volume, we often use: Decomposition: Break the shape into smaller, regular shapes (rectangles, triangles, circles) whose areas/volumes we can calculate. Then, add the areas/volumes of the smaller shapes.
Estimation: Superimpose a grid over the irregular shape. Count the squares (or partial squares) within the shape. Multiply the number of squares by the area of each square. This provides an approximate area. For volumes, imagine filling the space with cubes.
For liquids (volume): Use a measuring jug or container.
Example 1: Calculating the Area of an Irregular Garden Bed A garden bed is shaped roughly like a combination of a rectangle and a semicircle. The rectangular part is 4 meters long and 2 meters wide. The semicircle has a diameter of 2 meters (the width of the rectangle). What is the approximate area of the garden bed?
Solution: Area of the rectangle: Length × Width = 4 m × 2 m = 8 m² Radius of the semicircle: Diameter / 2 = 2 m / 2 = 1 m Area of a full circle (if it existed): πr² = π × (1 m)² ≈ 3.14 m² Area of the semicircle: (1/2) × 3.14 m² ≈ 1.57 m² Total area of the garden bed: 8 m² + 1.57 m² ≈ 9.57 m² Therefore, the approximate area of the garden bed is 9.57 m².
Example 2: Calculating the Volume of a Swimming Pool A swimming pool has an irregular shape. It is 10 meters long, 5 meters wide at one end, and 3 meters wide at the other end. It has a uniform depth of 1.5 meters. Estimate the volume of the pool.
Solution: Approximate shape: Treat the pool as a prism with a trapezoidal base (the irregular width).
Area of the trapezoid: (1/2) × (sum of parallel sides) × height = (1/2) × (5 m + 3 m) × 10 m = (1/2) × 8 m × 10 m = 40 m² Volume of the prism: Area of base × height (depth) = 40 m² × 1.5 m = 60 m³ Therefore, the estimated volume of the swimming pool is 60 m³. 2.
3. Scale Drawings and Maps Scale drawings and maps are representations of real-world objects or areas, reduced or enlarged by a specific scale factor. A scale factor is the ratio of a measurement on the drawing/map to the corresponding measurement in reality.
Example 3: Using a Map Scale to Calculate Distance A map has a scale of 1:50,
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0. This means 1 cm on the map represents 50,000 cm in reality. Two towns are 8 cm apart on the map. What is the actual distance between the towns in kilometers?
Solution: Actual distance in centimeters: 8 cm × 50,000 = 400,000 cm Convert centimeters to meters: 400,000 cm / 100 = 4,000 m Convert meters to kilometers: 4,000 m / 1000 = 4 km Therefore, the actual distance between the towns is 4 kilometers. 2.
4. Cost Estimation Cost estimation involves calculating the total cost of a project by considering all the materials, labor, and other expenses. Accurate measurements are essential for accurate cost estimation.
Example 4: Estimating the Cost of Painting a Room A rectangular room is 5 meters long, 4 meters wide, and 2.5 meters high. You need to paint the walls. One liter of paint covers 10 square meters. Paint costs R80 per liter. Estimate the total cost of the paint.
Solution: Area of the walls: Two walls are 5 m × 2.5 m = 12.5 m² each, and two walls are 4 m × 2.5 m = 10 m² each.
Total area of the walls: 2 × 12.5 m² + 2 × 10 m² = 25 m² + 20 m² = 45 m² Liters of paint needed: 45 m² / 10 m²/liter = 4.5 liters Since you can't buy half a liter, round up: 5 liters Total cost of the paint: 5 liters × R80/liter = R400 Therefore, the estimated total cost of the paint is R
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0. Guided Practice (With Solutions)
Question 1: A farmer wants to fence a rectangular field that is 120 meters long and 85 meters wide. Fencing costs R45 per meter. What will be the total cost of the fencing?
Solution: Calculate the perimeter of the field: Perimeter = 2 × (Length + Width) = 2 × (120 m + 85 m) = 2 × 205 m = 410 m Calculate the total cost: Total Cost = Perimeter × Cost per meter = 410 m × R45/m = R18,450 Therefore, the total cost of the fencing will be R18,
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0. Question 2: A cylindrical water tank has a diameter of 2 meters and a height of 3 meters. How many liters of water can the tank hold?