2.1 Geometry of lines and triangles

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 Δ A BC :

As b ² = a ² + c ² , then $\hat{B}$ = 90°

As b ² > a ² + c ² , then $\hat{B}$ >90°

As b ² < a ² + c ² , then $\hat{B}$ <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:

A  O (SSS or RHS from calculated hypotenuse, Pythagoras)

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

C  F  N ( SSS or RHS from calculated hypotenuse)

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

E  H ( AA S ) 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)

Δ ABC  Δ DEF ( S )

2. BC = 12 = FE (Pythagoras)

AC = 15 = DE (given)

Right–angled triangles

Δ ABC  Δ D FE ( RHS )

3. BC = BC (common or shared side)

 B = 55° =  C (given)

 A =  D (given; shown by little arc)

Δ ABC  Δ D CB ( 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°

Δ PQR  Δ XYZ (equiangular)

213 = 3(71)

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

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

Δ ABC  Δ DEF (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.