Lesson Notes By Weeks and Term v5 - Grade 9
Measurement and trigonometry (Grade 9) – Week 8 focus
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Measurement and trigonometry (Grade 9) – Week 8 focus
Download the Lessonotes Mobile South Africa app for faster lesson access on Android and iPhone.
Subject: Mathematics
Class: Grade 9
Term: 3rd Term
Week: 8
Theme: General lesson support
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This week, we delve deeper into measurement and trigonometry, focusing on applying our knowledge of area, surface area, and volume calculations to real-world problems and introducing basic trigonometric ratios. Measurement is a fundamental skill, vital for tasks ranging from calculating the amount of paint needed for your room to understanding the dimensions of land parcels. Trigonometry, derived from Greek words meaning "triangle measurement," builds on this foundation, allowing us to relate angles and side lengths in right-angled triangles, which has applications in surveying, navigation, and engineering.
2.1 Area of Composite 2D Shapes Composite shapes are formed by combining two or more basic geometric shapes. To find the area of a composite shape, we break it down into its simpler components, calculate the area of each component separately, and then add (or subtract) the individual areas to find the total area.
Example 1: A garden is shaped like a rectangle with a semi-circle attached to one of its shorter sides. The rectangle is 8m long and 5m wide. What is the area of the garden?
Step 1: Identify the component shapes. The garden consists of a rectangle and a semicircle.
Step 2: Find the dimensions of each shape. The rectangle has a length of 8m and a width of 5m. The semicircle has a diameter equal to the width of the rectangle, which is 5m.
Therefore, its radius is 5m / 2 = 2.5m.
Step 3: Calculate the area of each shape. Area of rectangle = length × width = 8m × 5m = 40 m² Area of semicircle = (1/2) × π × radius² = (1/2) × π × (2.5m)² ≈ 9.82 m² (using π ≈ 3.14)
Step 4: Add the areas together. Total area = Area of rectangle + Area of semicircle = 40 m² + 9.82 m² ≈ 49.82 m² Therefore, the area of the garden is approximately 49.82 m².
Example 2: Calculate the area of the shaded region in the figure. A square with side length 10 cm has a circle inscribed in it (i.e., the circle touches all four sides of the square).
Step 1: Identify the shapes. We have a square and a circle. The shaded area is the square's area minus the circle's area.
Step 2: Find the dimensions. The square has side length 10 cm. The diameter of the circle is equal to the side length of the square, so the circle's diameter is 10 cm, and its radius is 5 cm.
Step 3: Calculate the areas. Area of square = side × side = 10 cm × 10 cm = 100 cm² Area of circle = π × radius² = π × (5 cm)² ≈ 78.54 cm² (using π ≈ 3.14)
Step 4: Subtract the areas. Shaded area = Area of square - Area of circle = 100 cm² - 78.54 cm² ≈ 21.46 cm² The area of the shaded region is approximately 21.46 cm². 2.2 Surface Area of 3D Objects The surface area of a 3D object is the total area of all its faces.
Cube: A cube has 6 identical square faces. Surface Area = 6 × side² Rectangular Prism: A rectangular prism has 6 rectangular faces. If the length, width, and height are l, w, and h respectively, Surface Area = 2( lw + lh + wh)
Triangular Prism: A triangular prism has two triangular faces and three rectangular faces. The surface area depends on the type of triangle. For a right triangular prism with base b, height h, and length l, the surface area is bh + 2(lside) + lb where side is the hypotenuse of the triangular face.
Cylinder: A cylinder has two circular faces and one curved surface. If the radius is r and the height is h, Surface Area = 2πr² + 2πrh Example 3: A rectangular prism has a length of 5 cm, a width of 3 cm, and a height of 2 cm. Calculate its surface area.
Step 1: Identify the shape and its dimensions. We have a rectangular prism with l = 5 cm, w = 3 cm, and h = 2 cm.
Step 2: Apply the formula. Surface Area = 2(lw + lh + wh) = 2(5 cm × 3 cm + 5 cm × 2 cm + 3 cm × 2 cm) = 2(15 cm² + 10 cm² + 6 cm²) = 2(31 cm²) = 62 cm² The surface area of the rectangular prism is 62 cm². 2.3 Volume of 3D Objects The volume of a 3D object is the amount of space it occupies.
Cube: Volume = side³ Rectangular Prism: Volume = length × width × height Triangular Prism: Volume = (1/2) × base × height × length (where base and height refer to the triangular face)
Cylinder: Volume = π × radius² × height Example 4: A cylinder has a radius of 4 cm and a height of 7 cm. Calculate its volume.
Step 1: Identify the shape and its dimensions. We have a cylinder with r = 4 cm and h = 7 cm.
Step 2: Apply the formula. Volume = π × r² × h = π × (4 cm)² × 7 cm ≈ 351.86 cm³ (using π ≈ 3.14) The volume of the cylinder is approximately 351.86 cm³. 2.4 Trigonometric Ratios In a right-angled triangle, the sides are named relative to a specific acute angle (θ): Hypotenuse: The side opposite the right angle (the longest side).
Opposite: The side opposite to the angle θ.
Adjacent: The side adjacent to the angle θ (and not the hypotenuse).
The three basic trigonometric ratios are: Sine (sin θ): Opposite / Hypotenuse Cosine (cos θ): Adjacent / Hypotenuse Tangent (tan θ): Opposite / Adjacent Mnemonic: SOH CAH TOA (Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent)
Example 5: Consider a right-angled triangle ABC, where angle B is the right angle. AB = 3 cm, BC = 4 cm, and AC = 5 cm. Find sin(A), cos(A), and tan(A).
Step 1: Identify the sides relative to angle
A. Hypotenuse: AC = 5 cm Opposite: BC = 4 cm Adjacent: AB = 3 cm Step 2: Apply the trigonometric ratios. sin(A) = Opposite / Hypotenuse = 4 cm / 5 cm = 0.8 cos(A) = Adjacent / Hypotenuse = 3 cm / 5 cm = 0.6 tan(A) = Opposite / Adjacent = 4 cm / 3 cm ≈ 1.33 2.5 Applications of Trigonometric Ratios Trigonometric ratios can be used to find unknown side lengths or angles in right-angled triangles.