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    This module is included inLens: Siyavula: Mathematics (Gr. 7-9)
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Geometry of lines and triangles

Module by: Siyavula Uploaders. E-mail the author

MATHEMATICS

Grade 9

ALGEBRA AND GEOMETRY

Module 8

GEOMETRY OF LINES AND TRIANGLES

Activity 1

To learn conventions of naming sides and angles in triangles

Figure 1
Figure 1 (Picture 3.png)
  • Refer to the triangle alongside to understand the terms.
  • The three sides of the triangle are called AB, AD and BD.
  • We often name a side in a triangle by using the small letter of the name of the opposite corner. The corners (vertices) of the triangle are named by using capital letters.
  • When A or AˆAˆ size 12{ { hat {A}}} {}is used, it refers to the angle made by side BA and side AD.
  • For the angle one can also say BAD or BAˆDBAˆD size 12{B { hat {A}}D} {}.
  • But if I say A=38°, it is clear what I mean.
  • We refer to triangle ABD or ΔABD, writing the letters in alphabetical order.

Exercise:

To show that you understand the naming

conventions, draw the following triangle in the space

to the right:

Draw ΔQRT with q=4cm, T=65° and QT=5,5cm.

You should notice that you don’t need to be told

the length of t, nor the sizes of QˆQˆ size 12{ { hat {Q}}} {} or RˆRˆ size 12{ { hat {R}}} {}. First

draw a rough sketch and fill the details in on that

sketch to help you plan your drawing.

Assignment:

You should already know that triangles are classified according to their shape. Make an A4–sized poster for your own use, clearly showing the characteristics of the following types of triangle: equilateral, right–angled, isosceles and scalene. Name the vertices and sides according to the conventions above. You must work as accurately and neatly as you can.

Measure the sides and the angles of your triangles and fill these measurements in on your poster.

Activity 2

To develop the principle of congruence in triangles

[LO 4.4, 3.3, 3.5]

  • In the previous exercise you drew ΔQTR from specifications given to you. Ask the other learners who did this exercise to show you their drawings, and check whether their triangles agree perfectly with the sizes given in the question. Measure the side and the two angles not specified, to see whether they also agree with yours.
  • You should find that all triangles drawn by anybody according to the instructions, are always identical. In fact, it is impossible to draw that triangle so that it is different! Write down, with a partner, why you think this is the case.
  • Here are more descriptions of triangles. See whether the same happens with them – in other words, that it is impossible to draw different triangles that fit the same description. Again, write down your view of each situation. The last one is quite difficult to draw – try it!
  • Draw ΔAGE with a=4cm, E=90° and AG=5cm.
  • Draw ΔNOH with HN=4cm, H=56° and O=72°.
  • Draw ΔBAT with B=48°, T=65° and A=67°.
  • Draw ΔMOD with m=5,5cm, O=65° and DM=4cm.
  • Draw ΔAMP with a=4,2cm, m=5cm and p=5,6cm.
    • In each of the above triangles, only three of the six sizes (three sides and three angles) were specified in the question. And sometimes that was enough to ensure that everyone drew identical triangles. But in ΔBAT and ΔMOD the three items were not enough to ensure identical triangles from everyone.
    • So, when is it enough? Maybe you have already discovered the secrets:

ΔQRT: Two sides and the angle between them were specified.

ΔAGE: A right–angle, the hypotenuse and another side were specified.

ΔNOH: One side and two angles were specified.

ΔAMP: Three sides were specified.

ΔBAT: Three angles were specified, but nothing said how big the triangle could be!

ΔMOD: Two sides and the angle not between the two sides were specified, so that it could happen that some learners drew a short OM side and others drew a longer OM side; because nothing was said how long OM had to be!

  • When two triangles are identical in every way – size and shape – then we call them congruent. This means that if you cut one out, it can be placed exactly on top of the other. As you will see later, the word can be used for other identical shapes as well, but for now we will concern ourselves only with triangles.
  • From the drawing exercise you saw that there are four different ways to ensure that triangles are congruent. Here they are, with helpful sketches:

Figure 2
Figure 2 (Picture 9.png)

Figure 3
Figure 3 (Picture 10.png)

Case 1: Two triangles with two sides and the included angle equal, will be congruent.

Case 2: Two right–angled triangles with the hypotenuse and another side equal, will be congruent.

Case 3: Two triangles with three sides equal will be congruent.

Case 4: Two triangles with two angles and corresponding sides equal, will be congruent. This means that the equal sides must be opposite corresponding angles.

Investigation:

The next exercise shows 15 triangles, named A to O. They are all mixed–up and in strange orientations. Work in a group of 4 or 5 to decide whether any of them are congruent. Group the names of those that are congruent, with explanations and reasons. This is not a straight–forward exercise; it is much more like a puzzle. You will have to use all your experience, common-sense and logic. The sizes are not correct, so that you have to use the information given, and notmeasure anything.

Figure 4
Figure 4 (Picture 11.png)

Activity 3

To apply the four cases of congruence in problems

[LO 4.4, 3.3, 3.4]

  • When you have shown that two triangles are congruent (as you had to do in the previous exercise) you have to do a number of things: First decide which case of congruence will apply. Then say why each of the three items is equal. Then write your conclusions down in the proper order. Here is an example of how it can be done. We use the symbol  to show congruence. So we can see that, if we know that three very special things are equal, we know that everything else must be equal too!

Prove that ΔABC and ΔDEF are congruent.

Figure 5
Figure 5 (Picture 15.png)

1. AˆAˆ size 12{ { hat {A}}} {}= 60° because the angles of a triangle add up to 180°

Therefore, Aˆ=FˆAˆ=Fˆ size 12{ { hat {A}}= { hat {F}}} {} because both are 60°

2. Cˆ=EˆCˆ=Eˆ size 12{ { hat {C}}= { hat {E}}} {} because both are 50°

3. BC = DE because they are both 12 units, and they lie opposite equal angles. We have two equal angles and a corresponding equal side in each of the two triangles.

4. So we write: ΔABC  ΔDEF (AAS) which means: ΔABC is congruent to ΔDEF because two angles and a corresponding side are equal. This means that all other things must be equal too.

From the exercise in the previous section, do at least three congruencies in this way.

Exercise:

Figure 6
Figure 6 (graphics1.png)
Prove that the two triangles in each of the following problems are congruent.

1.

Figure 7
Figure 7 (Picture 16.png)

3.

Figure 8
Figure 8 (Picture 17.png)

Activity 4

To understand the principle of similarity in triangles

[LO 1.2, 1.4, 3.5]

  • In a previous exercise you were asked to draw ΔBAT with B=48°, T=65° and A=67°. As you noticed, it was possible to draw many triangles according to these specifications, but they were not congruent. This is because nothing specified the size of the triangle, so some were bigger and some were smaller.
  • Work in groups of four or five. Measure the sides of your triangle. Each uses his own Δ BAT to complete his row in the table below. Where you divide, give your answer rounded to one decimal place.
Table 1
Learner AB AT BT BT  AT AB  AT BT  AB
             
             
             
             
             
  • In the next exercise you must draw two isosceles triangles with angles 80°, 50° and 50°. Make the one triangle about two or three times as big as the other. Work very accurately.
  • Call the small triangle DEF (F = 80°) and the big one OPT (T = 80°). Measure all the sides and complete the table from your measurements.
Table 2
OP PT OT DE EF DF OP  DE PT  EF OT  DF
                 

Assignment:

  • Study the two tables (especially the last three columns of both tables). What do you notice?
  • Write a very clear explanation of why these calculations work out the way they do.
  • These triangles are not congruent, as their sizes are different, even if their angles agree. When two triangles have equal angles but different sizes, we call them similar.
  • The sign is , so that ΔDEF  ΔOPT from the last table.
  • All triangles with three angles equal are automatically similar. If they have the same size, then they are congruent as well.
  • Similar triangles have sides that are in the same proportion. This is what we see from the two tables we completed. The fractions that were calculated from the side lengths give us the ratios between sides.
  • From the first table we see that the ratios between the sides of similar triangles are the same. From the second table we see that the ratios between corresponding sides of two similar triangles are the same. This ratio is called the proportional constant for the two triangles.
  • We know two things from the facts we have learned about similar triangles:
  • Firstly, if we have two triangles with equal angles (equiangular triangles), then we know they are similar and therefore the sides must be in proportion.
  • Secondly, when we have triangles with sides in proportion, we know they must have equal angles because they must be similar.

Example:

What can you say about the two triangles below? Calculate the values of x and y.

Figure 9
Figure 9 (Picture 18.png)

To find out whether the triangles are similar, we must find either equal angles or sides in proportion. In this problem, we can say the angles are equal but we cannot say the sides are in proportion. In other problems, it may be the other way around. .

We set it out this way:

1. A = 65° because the angles of ΔABC add up to 180°

F = 35° because the angles of ΔDEF add up to 180°

2. The triangles have equal angles, therefore they are similar, so

ΔABC  ΔDEF (equal angles)

3. This means that the sides must be in proportion.

4. Find out what the proportional constant is. AC = 16 and DF = 8.

  • These two sides are both opposite the 80° angles, so that they are corresponding angles.
  • The proportional constant is 168=2168=2 size 12{ { {"16"} over {8} } =2} {}. If we multiply the a side of the small triangle by 2, we get the length of the corresponding side in the large triangle. If you divide a side of the large triangle by 2 then you get the length of the corresponding side in the small triangle.

5. Now the value of x can be calculated by dividing 9 by 2. x = 2  9,4 = 4,7.

6. And y = 2 × 5,5 = 11.

Exercise:

Calculate the values of sides PR and XY in the following triangles.

Figure 10
Figure 10 (Picture 20.png)

Example:

Find all the missing angles in these two triangles, if possible.

Figure 11
Figure 11 (Picture 21.png)

1. The sides are in proportion: 42 × 1,5 = 63 , 38 × 1,5 = 57 and 34 × 1,5 = 51

2. This means that the triangles are similar: ΔEFG  ΔKLM (sides in proportion)

3. So, corresponding angles are equal: L = 68° (corresponds to F)

E = 51° (corresponds to K)

G = M = 61° (sum of the angles of a triangle)

Exercise:

Find all the missing angles in these triangles:

Figure 12
Figure 12 (Picture 22.png)

Activity 5

To apply similarity in problems

[LO 4.4, 1.4, 3.5]

  • In the following problems, you must draw sketches of the given triangles, but you must NOT make accurate drawings.

1. Are the following triangles similar?

1.1 ΔBAG with B = 90°, AG = 15cm and AB = 9cm and

ΔPOT with P = 90°, OT = 5cm and PO = 4cm.

1.2 ΔREM with R = 60° and M = 50° and

ΔSUP with U = 70° and S = 50°.

1.3 ΔLOP with P = 90°, LO = 13cm and OP = 12cm and

ΔCAT with C = 90°, AC = 16cm and CT = 12cm

2. Calculate the proportional constant in similar triangles ΔABC and ΔDEF when

AB = 36cm, EF = 12cm, C = 48° and D = 48°.

3. Two flagpoles (one longer than the other) throw shadows on the ground. The shadow of the longer pole (which is 8 m tall) is 3 m and the shorter flagpole has a 2,5 m shadow. Calculate how tall the short flagpole is.

4. Gloria is designing a logo for her school’s computer club. The design shows a computer next to a pile of books which is 50 % higher than the computer. She is photocopying the design to make it smaller. On the photocopy the computer is 18 cm high and on the original the pile of books is 54 cm high. By what factor is she making the design smaller?

Assessment

Table 3
Learning outcomes(LOs)
 
LO 1
Numbers, Operations and RelationshipsThe learner will be able to recognise, describe and represent numbers and their relationships, and to count, estimate, calculate and check with competence and confidence in solving problems.
Assessment standards(ASs)
 
We know this when the learner :
1.1 describes and illustrates the historical development of number systems in a variety of historical and cultural contexts (including local);
1.2 recognises, uses and represent rational numbers (including very small numbers written in scientific notation), moving flexibly between equivalent forms in appropriate contexts;
1.3 solves problems in context, including contexts that may be used to build awareness of other Learning Areas, as well as human rights, social, economic and environmental issues such as:
1.3.1 financial (including profit and loss, budgets, accounts, loans, simple and compound interest, hire purchase, exchange rates, commission, rentals and banking);
1.3.2 measurements in Natural Sciences and Technology contexts;
1.4 solves problems that involve ratio, rate and proportion (direct and indirect);
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 :
3.1 recognises, visualises and names geometric figures and solids in natural and cultural forms and geometric settings, including:3.1.1 regular and irregular polygons and polyhedra;3.1.2 spheres;3.1.3 cylinders;3.2 in contexts that include those that may be used to build awareness of social, cultural and environmental issues, describes the interrelationships of the properties of geometric figures and solids with justification, including:3.2.1 congruence and straight line geometry;3.3 uses geometry of straight lines and triangles to solve problems and to justify relationships in geometric figures;3.4 draws and/or constructs geometric figures and makes models of solids in order to investigate and compare their properties and model situations in the environment;
3.5 uses transformations, congruence and similarity to investigate, describe and justify (alone and/or as a member of a group or team) properties of geometric figures and solids, including tests for similarity and congruence of triangles.
LO 4
MeasurementThe learner will be able to use appropriate measuring units, instruments and formulae in a variety of contexts.
We know this when the learner :
4.1 solves ratio and rate problems involving time, distance and speed;
4.2 solves problems (including problems in contexts that may be used to develop awareness of human rights, social, economic, cultural and environmental issues) involving known geometric figures and solids in a range of measurement contexts by:
4.2.1 measuring precisely and selecting measuring instruments appropriate to the problem;
4.2.2 estimating and calculating with precision;
4.2.3 selecting and using appropriate formulae and measurements;
4.3 describes and illustrates the development of measuring instruments and conventions in different cultures throughout history;
4.4 uses the Theorem of Pythagoras to solve problems involving missing lengths in known geometric figures and solids.

Memorandum

Discussion

Point out to learners the relationships between angle sizes and lengths of opposite sides, i.e. that the largest angle lies opposite the longest side, etc. In this regard the following theorems are interesting:

In ΔABC:

As b2 = a2 + c2 , then BˆBˆ size 12{ { hat {B}}} {}= 90°

As b2 > a2 + c2 , then BˆBˆ size 12{ { hat {B}}} {}> 90°

As b2 < a2 + c2 , then BˆBˆ size 12{ { hat {B}}} {}< 90°

  • These relationships are easily confirmed by construction. If time allows, this could be a good basic familiarisation exercise.
  • Many learners enter grade 9 without knowing that the relationship between the height and the base of a triangle is crucial in the formula for the area of a triangle. This is a good point to emphasize that the height is measured from the vertex opposite the base, perpendicular to the base. When they construct scalene triangles with three heights and then use these heights and the corresponding three bases to work out the area three times, this usually makes the point very clear.

Congruence

In the first exercise in activity 3.2 the learners should not work in very large groups – pairs would be best. The aim is to obtain as many versions of the triangles as possible, so as to confirm when they are congruent and when not. It would be best to give this exercise as a homework assignment if, in the teacher’s judgement, the learners can do it without help.

When discussing the four cases of congruence, point out that two right–angled triangles are congruent if the two non-hypotenuse sides are respectively equal, not by the RHS-rule, but by the SAS-rule.

Learners have to become comfortable with the idea that figures need not be drawn to scale – and that the information given on the figure must be used. They must not measure attributes unless specifically asked to do so.

Answers to matching exercise:

AO (SSS or RHS from calculated hypotenuse, Pythagoras)

B  G (SAS) They are not congruent to I, as the given angle is not included

CFN (SSS or RHS from calculated hypotenuse)

D, L and K are not congruent as only angles are given

EH (AAS) They are not congruent to M, as the given side is not in a corresponding position

  • In doing proofs, rigour is not required from grade 9 learners. On the other hand, one of the strengths of learning geometry is that it teaches the learners to work in a logical and rigorous order. Make a point of encouraging this style from the type of learner who might benefit from it, particularly in further mathematics.

Congruency proofs:

1. C = 180° – 88° – 43° = 49° = F (angles of Δ sum to 180°)

B = 43° = E (given)

AB = 15 = DE opposite 49°–angles (given)

ΔABC  ΔDEF (S)

2. BC = 12 = FE (Pythagoras)

AC = 15 = DE (given)

Right–angled triangles

ΔABC  ΔDFE (RHS)

3. BC = BC (common or shared side)

B = 55° = C (given)

A = D (given; shown by little arc)

ΔABC  ΔDCB(RHS)

Similarity

If a photocopier is available, teachers can design more exercises that will illustrate the principles of similarity by direct measurement.

Exercise:

Q = 55° en Z = 65°

ΔPQR  ΔXYZ (equiangular)

213 = 3(71)

PR = 201  3 = 67 en XY = 74 × 3 = 222

DE = AB × 4, DF = AC × 4 and EF = BC × 4

ΔABC  ΔDEF (sides are in proportion)

Corresponding angles must be equal

A = D = 62°; E = B = 49° en C = F = 69° (angles of Δ sum to 180°)

Exercise from activity 3.5:

1.1 Yes, Pythagoras gives BG = 12 cm and PT = 3cm; sides in proportion

1.2 Yes, E = 70° = U and P = 60° = R (angles of triangle); equiangular

1.3 No, LP = 5 cm and AT = 20 cm (Pythagoras); but sides are not in proportion

2. Proportional constant = 36 12 = 3

3. Short flagpole is 6,67 m tall

4. Pile of books on photocopy is 18  2 × 3 = 27 cm high

54 27 = 2 is the factor by which the design is made smaller.

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