Lesson Notes By Weeks and Term v5 - Grade 9

Lesson Video

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Performance objectives

• Calculate area and perimeter of composite shapes.
• Apply the Pythagorean theorem to find unknown side lengths.
• Define and use sine, cosine, and tangent ratios.
• Solve for unknown sides and angles in right-angled triangles.

Lesson summary

This week, we delve into the exciting world of Measurement and Trigonometry. Measurement is essential for everyday life, from building houses and roads to calculating the cost of groceries. Trigonometry, derived from Greek words meaning "triangle measurement," builds upon our knowledge of geometry and ratios to unlock the secrets of triangles, particularly right-angled triangles. These concepts are crucial for understanding the world around us and are foundational for further studies in mathematics, science, and engineering. In South Africa, understanding measurement is vital in industries such as construction, agriculture (land measurement for farming), and manufacturing.

Lesson notes

2.1 Area and Perimeter of Composite Shapes Many real-world shapes are not simple squares or circles. They are composite shapes, made up of several basic geometric figures. To find the area and perimeter, we break down the shape into its components.

Area: The total surface enclosed by a shape.

Perimeter: The total length of the boundary of a shape.

Formulas to remember: Rectangle: Area = length × width; Perimeter = 2(length + width)

Triangle: Area = (1/2) × base × height; Perimeter = sum of all sides Circle: Area = π × radius²; Circumference (Perimeter) = 2 × π × radius where π (pi) ≈ 3.14159 Example 1: A garden is shaped like a rectangle with a semi-circle on one side. The rectangle is 8m long and 6m wide. What is the area and perimeter of the garden?

Area: Area of rectangle = 8m × 6m = 48 m² Radius of semi-circle = 6m / 2 = 3m Area of semi-circle = (1/2) × π × (3m)² ≈ (1/2) × 3.14159 × 9 m² ≈ 14.14 m² Total area = 48 m² + 14.14 m² = 62.14 m² Perimeter: Three sides of rectangle = 8m + 6m + 8m = 22m Circumference of semi-circle = (1/2) × 2 × π × 3m = π × 3m ≈ 9.42 m Total perimeter = 22m + 9.42m = 31.42m Example 2: A swimming pool is made of a rectangle with two identical isosceles triangles on either end. The rectangle is 15m long and 8m wide. The height of each triangle is 3m. What is the area of the swimming pool?

Area: Area of rectangle = 15m 8m = 120m^2 Area of each triangle = (1/2) 8m * 3m = 12m^2 Total area of both triangles = 2 12m^2 = 24m^2 Total area of the pool = 120m^2 + 24m^2 = 144m^2 2.2 The Pythagorean Theorem The Pythagorean theorem describes the relationship between the sides of a right-angled triangle. A right-angled triangle has one angle that measures 90 degrees. The side opposite the right angle is called the hypotenuse (c), and the other two sides are called the legs (a and b).

Theorem: a² + b² = c² This theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Example 3: A ladder is leaning against a wall. The ladder is 5m long, and the base of the ladder is 3m from the wall. How high up the wall does the ladder reach? We have a right-angled triangle. The ladder is the hypotenuse (c = 5m), and the distance from the wall is one leg (a = 3m). We need to find the other leg (b).

Using the Pythagorean theorem: 3² + b² = 5² 9 + b² = 25 b² = 25 - 9 = 16 b = √16 = 4m Therefore, the ladder reaches 4m up the wall. 2.3 Trigonometric Ratios (SOH CAH TOA) Trigonometric ratios relate the angles and sides of right-angled triangles. The three basic trigonometric ratios are sine (sin), cosine (cos), and tangent (tan).

SOH: Sin (θ) = Opposite / Hypotenuse CAH: Cos (θ) = Adjacent / Hypotenuse TOA: Tan (θ) = Opposite / Adjacent Where: θ (theta) represents the angle.

Opposite: The side opposite to the angle θ.

Adjacent: The side adjacent (next to) to the angle θ (not the hypotenuse).

Hypotenuse: The longest side, opposite the right angle.

Example 4: Consider a right-angled triangle with angle θ = 30 degrees. The hypotenuse is 10cm. Find the length of the opposite side. We know the angle and the hypotenuse and need to find the opposite side. The sine ratio relates these. sin(30°) = Opposite / Hypotenuse sin(30°) = Opposite / 10cm Opposite = sin(30°) × 10cm Using a calculator, sin(30°) = 0.5 Opposite = 0.5 × 10cm = 5cm Example 5: Consider a right-angled triangle with angle θ. The adjacent side is 8cm and the opposite side is 6cm. Find the value of the angle θ. We know the opposite and the adjacent and need to find the angle. The tangent ratio relates these. tan(θ) = Opposite / Adjacent tan(θ) = 6cm / 8cm = 0.75 θ = arctan(0.75) (or tan⁻¹(0.75)) Using a calculator, θ ≈ 36.87 degrees. Guided Practice (With Solutions)

Question 1: A rectangular field is 12m long and 5m wide. A diagonal path cuts across the field. How long is the diagonal path?

Solution: The diagonal path forms the hypotenuse of a right-angled triangle. a = 12m, b = 5m c² = a² + b² = 12² + 5² = 144 + 25 = 169 c = √169 = 13m The diagonal path is 13m long.

Question 2: Calculate the value of x in the following triangle, where the angle is 40 degrees, the hypotenuse is 15cm, and x is the opposite side.

Solution: We use the sine ratio. sin(40°) = x / 15cm x = 15cm sin(40°) x ≈ 15cm 0.6428 ≈ 9.64cm Question 3: A flagpole casts a shadow of 8m. The angle of elevation from the tip of the shadow to the top of the flagpole is 60 degrees. How tall is the flagpole?

Solution: We have a right-angled triangle. The height of the flagpole is the opposite side, and the length of the shadow is the adjacent side. We can use the tangent ratio. tan(60°) = Height / 8m Height = 8m tan(60°) Using a calculator, tan(60°) ≈ 1.732 Height ≈ 8m 1.732 ≈ 13.86m The flagpole is approximately 13.86m tall.

Question 4: A garden bed is shaped like a rectangle (5m x 3m) with a quarter circle attached on one of the 3m sides. Calculate the perimeter of the garden bed.