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8. Calculate the area of each square.

9. What can you deduce from this exercise?

10. Deduction: Write out Pythagoras’ theorem in the space below by making use of the triangle that is provided.

11. Solve x in each of the following triangles:(You may make use of your calculator.)

11.3

1.4

12. Do the calculations to determine whether the following is a right-angled triangle or not:

12.1 Δ size 12{Δ} {} DEF with DE = 8 cm, EF = 10 cm, DF = 6 cm

13. AREA OF TRIANGLES

13.1 Construct rectangle ABCD with AB = 45 mm and AD = 25 mm on a sheet of paper and cut it out. Draw diagonal AC .

13.2 Calculate the area of rectangle ABCD .

13.3 Cut out Δ size 12{Δ} {} ABC . What is the area of Δ size 12{Δ} {} ABC ?Paste it here.

  • Area of Δ size 12{Δ} {} ABC = ................. mm²

13.4 Are you able to develop a formula for determining the area any triangle?

Write it here:

13.5 Calculate the area of Δ size 12{Δ} {} ABC .

13.6 In the figure SQ = 15 cm, QR = 7 cm and PR = 9 cm.

Important : Provide all necessary information on your sketch. Check to see what you may need to complete the instructions fully.

(a) Calculate the area of Δ size 12{Δ} {} PSQ (accurate to 2 decimals).

(b) Now calculate the area of Δ size 12{Δ} {} PSR . Suggestion : You will first have to calculate the area of another triangle.

13.7 Calculate the area of ABCD .

14. Calculate the length of the unknown sides of each of the following:

14.1

14.2

14.3

15. Playing in a park is a necessary aspect of the development of a child.

  • You have been asked to supply slides. The problem that is involved requires calculating the length of the poles that are needed. Make use of the knowledge that you have accumulated to supply a plan to erect the slides.

The following is required:

15.1 a sketch

15.2 a scale, e.g. 1 cm = 1 m

15.3 Calculations must be completed fully.

Assessment

LO 3
Space and Shape (Geometry)The learner will be able to describe and represent characteristics and relationships between two-dimensional shapes and three-dimensional objects in a variety of orientations and positions.
We know this when the learner:
3.2 in context that include those that may be used to build awareness of social, cultural and environmental issues, describes and classifies geometric figures and solids in terms of properties, including:
3.2.1 sides, angles and diagonals and their inter­relationships, with focus on triangles and quadrilaterals (e.g. types of triangles and quadrilaterals).
LO4
MeasurementThe learner will be able to use appropriate measuring units, instruments and formulae in a variety of contexts.
We know this when the learner:
4.2 solves problems involving:
4.2.1 length;
4.2.2 perimeter and area of polygonals and circles;
4.3 solves problems using a range of strategies including:
4.3.1 estimating;
4.3.2 calculating to at least two decimal positions;
4.3.3 using and converting between appropriate SI units;
4.4 describes the meaning of and uses π size 12{π} {} in calculations involving circles and discusses its historical development in measurement;
4.5 calculates, by selecting and using appropriate formulae:
4.5.1 perimeter of polygons and circles;
4.5.2 area of triangles, rectangles circles and polygons by decomposition into triangles and rectangles;
  • investigates (alone and / or as a member of a group or team) the relationship between the sides of a right-angled triangle to develop the Theorem of Pythagoras;
4.9 uses the Theorem of Pythagoras to calculate a missing length in a right-angled triangle leaving irrational answers in surd form (√);
4.10 describes and illustrates ways of measuring in different cultures throughout history (e.g. determining right angles using knotted string leading to the Theorem of Pythagoras).

Memorandum

ACTIVITY 1

1.1 a) all 3 Acute-angled

b) one 90 o angled

c) one obtuse-angled

1.2 a) 2 even sides

b) 3 even sides

c) sides differ in length

2. The sum of the interior angles of any triangle is 180º

ACTIVITY 2

10. r 2 = p 2 + q 2

  • x 2 = 12 2 + 5 2

= 144 + 25

= 169

size 12{∴} {} x = 13

  • 20 2 = 8 2 + x 2

x 2 = 400 – 64

= 336

size 12{∴} {} x size 12{ approx } {} 18,3 cm

11.3 size 12{ nabla } {} ABC : x 2 = 70 2 – 29 2

= 4 900 – 841

= 4 059

size 12{∴} {} x size 12{ approx } {} 63,7 mm

11.4 y 2 = 4 2 + 3 2

= 16 + 9

= 25

size 12{∴} {} x size 12{ approx } {} 9,4cm

12. DE 2 + DF 2 = 100 = EF 2

size 12{∴} {} DEF right angled

(Pythagoras)

  • ½ x b x h
  • BC 2 = 13 2 – 5 2

= 169 – 25

= 144

size 12{∴} {} BC = 12 cm

Area ABC = ½ x b x h

= ½ x 12 x 5

= 30cm 2

13.6 (a) PS 2 = 9 2 – 8 2

= 81 – 64

= 17

size 12{∴} {} PS = 4,12 cm

Area PSQ = ½ x b x h

= ½ x 15 x 4,12

= 30,9cm 2

13.6 ( b ) Area PSR = ½ x 8 x 4,12

= 16,4 cm 2

Area PRQ = area PSQ PSR

= 30,9 – 16,4

= 14,5 cm 2

13.7 AC 2 = 12 2 + 8 2

= 208

size 12{∴} {} AC size 12{ approx } {} 14,4

AD 2 = 16 2 – 14,4 2

= 256 – 207,36

= 48,64

size 12{∴} {} AD = 6,97

Area ABCD = area ABC + area ACD

= (½ x 12 x 8) + (6,97 x 14,4 x ½)

= 48 + 50,18

= 98,18 square units

  • a 2 = 8 2 – 7 2

= 15

size 12{∴} {} a size 12{ approx } {} 3,9

b 2 = (3,9) 2 + 4 2

= 15,21 + 16

= 31,21

size 12{∴} {} b size 12{ approx } {} 5,6

  • x = 18 (radius)

y 2 = 36 2 – 13 2

= 1 296 – 169

= 1 127

size 12{∴} {} y = 33,6

  • UV 2 = 12 2 – 7 2

= 95

size 12{∴} {} UV = 9,8

VS 2 = 14 2 + ( size 12{ approx } {} 9,8) 2

= 196 + 95

= 291

size 12{∴} {} VS = 17,1

y 2 = ( size 12{ approx } {} 17,1) 2 + 5 2

= 291 + 25

= 316

size 12{∴} {} y = 17,8

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Source:  OpenStax, Mathematics grade 8. OpenStax CNX. Sep 11, 2009 Download for free at http://cnx.org/content/col11034/1.1
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