1.1.2 0,56 =56% (1) 1.1.3 1,044 =104,4% (1) 1.2.4 0,004 = 0,4% (1) 1.2.5 0,47 = 47% (1) QUESTION 2 2.1.1 2.1.2 2.1.3 2 = (2,4,6) 7 = (7,14,21) 11= (11,22,33) (2) (2) (2) QUESTION 3 3.1 (2) 3.2 (2) 3.3 (2) 3.4 (2) 3.5 (3) QUESTION 5 (3) (3) QUESTION 5 5.1 (3)
Figure 1 Figure 2 2. Use your answers to fill in the angle sizes below b) Which pair of angles are adjacent angles in figure 2? 4. Define the concept supplementary angles. LESSON PRESENTATION/ DEVELOPMENT 15min ACTIVITY 1. Two lines that intersect and form right angles are called perpendicular lines. The symbol ⊥ is used to denote perpendicular lines. In Figure 1, line l ⊥ line m. 2. Intersecting lines: Intersecting lines are two lines that share exactly one point. This shared point is called the point of intersection. , lines AB and CD intersect at point M. Figure 2 Intersecting lines 3. Use figure 2 to answer the following questions: lines AB and redraw the diagram and calculate with reasons the value of m.
CLASSWORK 10 Min 1.1. Place your compass on the given point (P). Draw an arc across the line on each side of the given point. Do not adjust the compass width when drawing the second arc. 1. 2. Open your compass so that it is wider than the distance from one of the arcs. 1.3. Use your ruler to join the given point (P) and the point where the arcs intersect (Q). Use the compass and ruler to draw a perpendicular at the given point on each line. Use your compass and your ruler to bisect the right angle to create an angle of 45˚. Constructing a 30º Angle We know that: ½ of 60o = 30o So, to construct an angle of 30º, first construct a 60º angle and then it. Often, we apply the following steps. Draw the arm PQ.
10. REFLECTION ANSWERS: TERM 2 GRADE 8 WEEK 3 LESSON 3 and 4 (MENTAL MATHS) CLASSWORK HOMEWORK An equilateral triangle is a triangle that has all three sides equal and each angle is 600. An isosceles triangle has two equal sides and two equal angles. A right-angled triangle is a triangle that has a angle By or use the symbol: (interior angles of a ) ==29° (angles opp equal sides) = 122° Alternatively, ( But (angles on a straight Line=180) ==38° (angles opp equal sides) 38 = 104°
CONSOLIDATION/CONCLUSION AND OR HOMEWORK 6min Find the length of all the unknown sides and angles of the trapezium below: REFLECTION ANSWERS: TERM 2 GRADE 8 WEEK 4 LESSON 5 MENTAL MATHS CLASSWORK HOMEWORK Properties of a Kite Two pairs of adjacent sides are equal. One pair of opposite angles are equal. Property of a kite The quadrilateral is a kite hence a kite is made up of two isosceles triangles (sum of interior angles of a trapezium) SUBJECT: MATHEMATICS GRADE 8 WEEK 4, LESSON 6 TOPIC: GEOMETRY OF 2D SHAPES CONCEPTS AND SKILLS TO BE ACHIEVED By the end of the lesson, learners should know and be able to: Investigate sides and angles in quadrilaterals focusing on:
The sum of the interior angles of quadrilaterals The sides, diagonals and opposite angles of parallelograms Discuss properties of a parallelogram Solve for unknown angles and sides RESOURCES DBE workbook, Sasol-Inzalo workbook 1 (199-204), textbook, pencil, eraser, calculators, graph paper, ruler (GDE 19 05 2014 GDE 21 05 2014) PRIOR KNOWLEDGE 2D shapes Angles Quadrilaterals COMPONENTS TIME TASKS/ACTIVITIES CAPS INTRODUCTION/(MENTAL MATHS) 3min Give all the characteristics of a trapezium Page 96 REVIEW AND CORRECTION OF HOMEWORK 7min (sum of interior angles of a trapezium) LESSON PRESENTATION/DEVELOPMENT AND CLASSWORK 15min Calculate the value of : Find and : CONSOLIDATION/CONCLUSION AND OR HOMEWORK 5min With the aid of a diagram, write down the properties of a rhombus. REFLECTION
DEVELOPMENT Mins Ask learners to draw the following: angles on a straight line perpendicular lines intersecting lines and parallel lines cut by a transversal. Ask the learners to identify any difference in the lines they have drawn. Ask the learners to draw quadrilaterals. Give their properties. Work through different problems starting with problems that will help learners understand the concept, e.g. Solve problems of line and perpendicular on a straight line and highlight to the learners the properties of angles on a straight line. Next, show learners the difference between parallel lines and intersecting lines. Solve problems on parallel lines and learners observe the problem-solving strategy. Do the same with intersecting lines, and again, learners observe the problem-solving strategy. Solve problems that require application of properties of quadrilateral. Introduce problems with algebra and help the learner solve and make substitutions in an expression to determine the angles. 1. Fill in the sizes of the marked angles on the diagram. Make sure you give reasons for your answer. (4)
MENTAL MATHS CLASSWORK HOMEWORK State whether the following statements are true or false. If false, explain why it is false. (a) All equilateral triangles are similar. b) All scalene triangles are similar. c) All rectangles are squares. d) All squares are rhombi. (interior of a ) Alternative methods can be used to solve the problem. (opposite of a rhombus are =) (interior of a ) but ( ABC is isosceles)
MENTAL MATHS CLASSWORK HOMEWORK 72° = 180° (interior angle sum of triangle) but = (AD = BD) = 1800 54° 180° 54° (adjacent suppl. ) 126° c ∠BDC c (540)27° Breadth = Length = cm SUBJECT: MATHEMATICS GRADE 8 WEEK 6, LESSON 4 TOPIC: Measurement: The Theorem of Pythagoras CONCEPTS AND SKILLS TO BE ACHIEVED By the end of the lesson, learners should be able to:
INTRODUCTION (MENTAL MATHS) 3Min Write down the formula of area of the following shapes Square Rectangle Triangle In the discussion, ask learners to explain the terms “perimeter” and “area”. Do the following exercise with the learners practically to show how the formula to find an area of a kite is established. To calculate the area of a kite, we use one of its properties: that the diagonals of a kite are perpendicular. LESSON PRESENTATION/ DEVELOPMENT 10 Min Area of a kite DEFG = Area of Area of = + = + = (Taking out a common factor) = Explain the above statement practically using flash cards to show how to derive a formula for the area of a kite: KEYWORDS Squares Square roots Inverse square
triangles circles. RESOURCES DBE workbook 2(42-64), Sasol-Inzalo workbook 2 (53-66), textbook, pencil, eraser, calculators, graph paper, ruler( DVD GDE 20 08 2014) PRIOR KNOWLEDGE Properties of 2-D shapes substitution equations COMPONENTS TIME TASKS/ACTIVITIES CAPS REVIEW AND CORRECTION OF HOMEWORK 4Min Perimeter of the square = 12 x 4 = 48cm Area of shaded part = Area of square – area of circle = Solving equations using formulae • The use of formulae provides a context to practise solving equations by inspection or using additive or multiplicative inverses. INTRODUCTION (MENTAL MATHS) 3Min Write down the formula for perimeter of a circle and the formula for area of a circle. LESSON PRESENTATION/ 10 Min In the discussion, ask learners to explain the terms “perimeter” and “area”. KEYWORDS Area
COMPONENTS TIME TASKS/ACTIVITIES CAPS REVIEW AND CORRECTION OF HOMEWORK 4 Min From B draw BP perpendicular DC Therefore, AB = DP = 20 cm So, PC = DC - DP = (25 - 20) cm = 5 cm Now, area of trapezium ABCD = Area of rectangle ABPD + Area of △BPC △BPC is right-angled at ∠BPC Therefore, using Pythagoras theorem, BC² = BP² + PC² 13² = BP² + 5² ⇒ 169 = BP² + 25 ⇒ 169 - 25 = BP² ⇒ 144 = BP² ⇒ BP = 12 Now, area of trapezium ABCD = Area of rectangle ABPD + Area of ∆BPC = AB × BP + 1/2 × PC × BP = 20 × 12 + 1/2 × 5 × 12 = 240 + 30 = 270 cm² Circles • Make sure learners can identify the centre, radius, diameter and circumference of the circle . • Spend time investigating the relationship between radius, circumference and diameter, so that learners develop a sense of where the irrational number Pi (π) is derived from. • Develop an understanding of π, making sure learners understand that: - π represents the value of the circumference divided by the diameter, for any circle - π is an irrational number
Section B Question 1 1.1 2406 6 14436 - 12 24 -24 3 (long division method) - 0 36 36 0 (3) 1.2.1 0, 12(3 + ) ÷ (1, 35 + 0, 65) = () ÷ 2 = () × = = (4) 1.2.2 ) ÷ = (2 + 4) ÷ 3 = 6 ÷ 3 = 2 (3) 1.3 = -140 C + 50 C = -90 C (2)
3.1 70% of R125 = R125 = R37, 50 (2) 3.2.1 £1 = R13,14 £2000 = = R26 280 (3) 3.2.2 R10,46 = $1 R6 000 = = $ 573,61 (3) 3.3 SI = P×i×n = R10 000 × 0,12 × 3 = R3 600 (4) 3.4 10% of R7 500 = × R7 500 = R750 R350 × 24 = R8 400 Total amount paid = R750 + R8 400 = R9 150 (4) [16]
C = x + y + 2(x + y) C = 3x + 3y (3) [16] Question 6 6.1 i + 550 + 620 = 1800 [sum of ∠s of Δ] i + 1170 = 1800 i = 1800 - 1170 i = 630 h = 620 j = i = 630 [Corresponding ∠s] (6) 6.2 x = 600 [Corresponding ∠s] x + y = 1800 [adjacent ∠s] y = 1800 - 600 = 1200 (4) 6.3 p = q = 220 [remaining ∠s] ∴ ΔABC ΔDEF [AAA] (2) 6.4 x = 15cm and z = 10cm [corresponding sides of ≡ Δs] (3)