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# Classifying and constructing triangles

Module by: Siyavula Uploaders. E-mail the author

## CLASSIFYING AND CONSTRUCTING TRIANGLES

### [LO 3.1, 3.3, 3.4, 4.2.1]

• By the end of this learning unit, you will be able to do the following:
• understand how important the use of triangles is in everyday situations;
• explain how to find the unknown sides of a right-angled triangle (Pythagoras);
• calculate the area of a triangle;
• enjoy the action in geometry;
• use mathematical language to convey mathematical ideas, concepts, generalisations and mental processes.

1. When you classify triangles you can do it according to the angles or according to the sides.

1.1 Classification on the basis of the angles of a triangle:Are you able to complete the following?

a) Acute-angled triangles are triangles with

b) Right-angled triangles have

c) Obtuse-angled triangles have

1.2 Classification on the basis of the sides of the triangle:Are you able to complete the following?

a) An isosceles triangle has

b) An equilateral triangle has

c) A scalene triangle's

2. Are you able to complete the following theorems about triangles? Use a sketch to illustrate each of the theorems graphically.

THEOREM 1:

• The sum of the interior angles of any triangle is.........................

Sketch:

THEOREM 2:

• The exterior angle of a triangle is

Sketch:

3. Constructing triangles:

• Equipment: compasses, protractor, pencil and ruler

Remember this:

• Begin by drawing a rough sketch of the possible appearance.
• Begin by drawing the base line.

3.1 Construct ΔΔ size 12{Δ} {}PQR with PQ = 7 cm, PR = 5 cm and PˆPˆ size 12{ { hat {P}}} {} = 70°.

a) Sketch:

b) Measure the following:

1. QR = ........ 2. RˆRˆ size 12{ { hat {R}}} {} = ........ 3. QˆQˆ size 12{ { hat {Q}}} {} = ........ 4. Pˆ+Qˆ+Rˆ=Pˆ+Qˆ+Rˆ= size 12{ { hat {P}}+ { hat {Q}}+ { hat {R}}={}} {} ........

3.2 Construct ΔΔ size 12{Δ} {}KLM , an equilateral triangle. KM = 40 mm, KL=LMand KˆKˆ size 12{ { hat {K}}} {} = 75°.Indicate the sizes of all the angles in your sketch.

Sketch:

## [LO 4.2.1, 4.8, 4.9, 4.10]

• The following could be done in groups.

Practical exercise: Making you own tangram.

1. Cut out a cardboard square (10 cm x 10 cm).

2. Draw both diagonals, because they form part of the bases of some figures.

3. Divide the square in such a way that the complete figure consists of the following:

3.1 two large equilateral triangles with bases of 10 cm in length;

3.2 two smaller equilateral triangles, each with base 5 cm in length;

3.3 one medium equilateral triangle with adjacent sides 5 cm in length;

3.4 one square with diagonals of 5cm;

3.5 one parallelogram with opposite sides of 5 cm.

• Make two of these. Cut along all the lines so that you will have two sets of the above shapes.

4. Now trace the largest triangle of your tangram in your workbook as a right-angled triangle.

5. Arrange the seven pieces to form a square and place this on the hypotenuse of the traced triangle.

6. Now arrange the two largest triangles to form a square and place this on one of the sides adja­cent to the right angle of the traced triangle.

7. Arrange the remaining pieces to form a square and place this on the other adjacent side.

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

• Area of ΔΔ 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 ΔΔ 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 ΔΔ size 12{Δ} {}PSQ (accurate to 2 decimals).

(b) Now calculate the area of ΔΔ 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 ππ 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 90o 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. r2 = p2 + q2

• x2 = 122 + 52

= 144 + 25

= 169

size 12{∴} {}x = 13

• 202 = 82 + x2

x2 = 400 – 64

= 336

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

11.3 size 12{ nabla } {}ABC: x2 = 702 – 292

= 4 900 – 841

= 4 059

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

11.4 y2 = 42 + 32

= 16 + 9

= 25

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

12. DE2 + DF2 = 100 = EF2

size 12{∴} {}DEF right angled

(Pythagoras)

• ½ x b x h
• BC2 = 132 – 52

= 169 – 25

= 144

size 12{∴} {}BC = 12 cm

Area ABC = ½ x b x h

= ½ x 12 x 5

= 30cm2

13.6 (a) PS2 = 92 – 82

= 81 – 64

= 17

size 12{∴} {}PS = 4,12 cm

Area PSQ = ½ x b x h

= ½ x 15 x 4,12

= 30,9cm2

13.6 (b) Area PSR = ½ x 8 x 4,12

= 16,4 cm2

Area PRQ = area PSQPSR

= 30,9 – 16,4

= 14,5 cm2

13.7 AC2 = 122 + 82

= 208

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

= 256 – 207,36

= 48,64

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

• a2 = 82 – 72

= 15

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

b2 = (3,9)2 + 42

= 15,21 + 16

= 31,21

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

y2 = 362 – 132

= 1 296 – 169

= 1 127

size 12{∴} {}y = 33,6

• UV2 = 122 – 72

= 95

size 12{∴} {}UV = 9,8

VS2 = 142 + ( size 12{ approx } {} 9,8)2

= 196 + 95

= 291

size 12{∴} {}VS = 17,1

y2 = ( size 12{ approx } {} 17,1)2 + 52

= 291 + 25

= 316

size 12{∴} {}y = 17,8

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