# Curriculum for Excellence Revising Level 3 and Investigating Level 4 - PowerPoint PPT Presentation

Curriculum for Excellence Revising Level 3 and Investigating Level 4

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Curriculum for Excellence Revising Level 3 and Investigating Level 4

## Curriculum for Excellence Revising Level 3 and Investigating Level 4

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1. Curriculum for Excellence Revising Level 3 and Investigating Level 4

2. Scatter graphs Stem & Leaf charts Mean, median & mode Pie charts Probability Scale Factors Polygons Angles

3. Standard Form Circle Ratio Prime Factors Algebra Pythagoras Order of Operations Speed, Distance & Time Area and Perimeter

4. Evaluate expressions Brackets Equations Algebra is fun – Investigate further

5. Powers and roots Integers Fractions/Decimals/Percentages Mathsrevision.com website Supermathsworld website Password 1712aquin

6. Investigating Polygons Remember The internal angles in a triangle add to 180° The angles at a point on a straight line add to 180°

7. Since any quadrilateral can be split into two triangles its internal angles add to 360°

8. A pentagon can be split into three triangles so its internal angles add to 540° A hexagon can be split into four triangles so its internal angles add to 720°

9. Triangle Internal angles 180°  3 = 60° External angles 180° - 60° = 120° 60° 120° Quadrilateral Internal angles 360°  4 = 90° 90° 90° External angles 180° - 90° = 90°

10. Pentagon Internal angles 540°  5 = 108° External angles 180° - 108° = 72° 108° 72° Hexagon Internal angles 720°  6 = 120° 120° 60° External angles 180° - 120° = 60°

11. Internal angles External angles Sides of Polygon Angle total (n-2)180° n (n-2)180° 180°- n Results 120° 60° 180° 3 90° 90° 360° 4 72° 108° 540° 5 60° 120° 720° 6 n (n-2)180°

12. Drawing Polygons Draw a regular hexagon of side 4cm. Sketch to identify angles 60º 120º 4cm 60º Use this information to accurately draw the hexagon.

13. Drawing Polygons Draw a regular pentagon of side 6cm. Sketch to identify angles 6cm 72º 108º 72º Use this information to accurately draw the pentagon.

14. 8cm 8 cm 6cm Drawing Polygons Draw a rhombus with diagonals 8cm and 6cm. Sketch first now draw 4 cm 3cm 3cm 3cm 3cm 4 cm

16. c b a Pythagoras Hypotenuse In any right angled triangle the square on the hypotenuse is equal to the sum of the squares on the two shorter sides.

17. Calculating the hypotenuse. 9cm 7cm x x 8m 6m c2 = a2 + b2 Examples: Calculate x in these two triangles. to 1 dp Pythagoras

18. Pythagoras theorem can be rearranged so that a shorter side can be calculated. also Remember Write the biggest number first Addto find the hypotenuse Subtractto find a shorter side

19. Calculating a shorter side. x 24cm 27cm x c2 = a2 + b2 5m 13m Examples: Calculate x in these two triangles. to 2 dp Pythagoras

20. B A Find the distance between A and B. to 1 dp Pythagoras

22. Speed - Distance - Time When we talk about the speed of an object we usually mean the average speed. A car may speed up and slow down during a journey but if the distance covered in one hour is 50 miles, we would say its average speed was 50mph. When we are doing calculations using speed, distance and time, it is important to keep the units consistent. If distance is measured in kilometres and time is measured in hours, then the speed is in kilometres per hour (km/h).

23. Speed, Distance and Time D D S = T = T S D D S S T T D = S x T

24. D S T Speed, Distance and Time Problem: Stewart walks 15km in 3 hours. Calculate his average speed. Stewart covers 15km in 3hours So his average speed is 15  3 = 5 km/h Speed = 5km/h distance covered Average speed = time taken

25. D S T Problem Claire cycled at a steady speed of 11 kilometres per hour. How far did she cycle in 3 hours? In 1 hour she covers 11 km So in 3 hours she covers 11 X 3 = 33km Distance = 33km D = S X T Distance = average speed X time taken

26. D S T Problem Paul drives 144 kilometres at an average speed of 48km/h. How long will the journey take? He drives 48 km in 1 hour. 144  48 = 3 (there are three 48s in 144) So the journey takes 3 hours. Time = 3 hours. distance covered Time taken = average speed D T = S

27. D S T Problem A car travelled for 2 hours at an average speed of 90km/h. How far did it travel? D = S X T D = 90 km X 2 hours = 180 km The car travelled 180 km

28. D S T Problem A car on a 240km journey can travel at 60km/h. How long will the journey take? D T = S T = 240 km  60 = 4 hours The journey will take 4 hours

30. Circle Area & Circumference

32. Circumference • The formula for the circumference of a circle is: C = d where C is circumference and d is diameter

33. Area • The formula for the area of a circle is: A = r2 where A is area and r is radius

34. Calculate the circumference of this circle. 8cm C = d = 3·14  8 = 25·12cm An approximation for is 3·14

35. Calculate the area of this circle. A = r2 10cm = 3·14  52 = 3·14  25 = 78·5cm2 Diameter is 10cm  Radius is 5cm

36. Using the formulae for circumference and area to calculate diameter and radius

37. Calculate the diameter and radius of a circle with a circumference of 157m. C = d 157 = 3·14  d d = 157 ÷ 3·14 d = 50m r = 50 ÷ 2 = 25m

38. Calculate the radius and diameter of a circular slab with an area of6280cm2. A = r2 6280 = 3·14  r2 r2 = 6280 ÷ 3·14 = 2000 r = 2000 = 44·72135955  d = 89·4cm  44·7cm

39. Composite Shapes • Calculate the Perimeter of this shape C = d = 3·14  9 9m = 28·26 12m 28·26  2 = 14·13m Perimeter = 14·13 + 12 + 12 + 9 = 47·13m

40. Composite Shapes • Calculate the shaded Area. Area of square = 28  28 = 784cm2 28cm A = r2 = 3·14  142 = 3·14  196 28cm = 615·44cm2 Shaded area = 784  615·44 = 168·56cm2

42. Area of Triangle, Kite, Rhombus & Parallelogram

43. Triangle height base 9cm 4cm 7cm Example: Calculate the area of this triangle.

44. Kite / Rhombus

45. Kite / Rhombus 3m 4m 6m 8m 9m Example: Calculate the area of these shapes.

46. height base base Parallelogram This shows that the area of a parallelogram is similar to the rectangle.

47. Parallelogram height base 5cm 8cm Example: Calculate the area of this parallelogram.

48. 11cm 6cm B 10cm A 6cm A composite shape can be split into parts so that the area can be calculated. Examples: Calculate the area of the following shapes. Area A = 10 × 6 = 60 Area B = 6 × 5 = 30 90cm2

49. 12m 11m A Area B = × 6 × 11 = 33 B 18m Area A = 12 × 11 = 132 165m2 A shape can be split into as many parts as necessary.