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Classification of triangles

Module by: Siyavula Uploaders. E-mail the author

MATHEMATICS

Grade 8

INTEGERS, EQUATIONS AND GEOMETRY

Module 10

CLASSIFICATION OF TRIANGLES

CLASS ASSIGNMENT 1

2. Classify the following triangles according to their angles (without the use of a protractor)

Figure 1
Figure 1 (Picture 143.png)

4. Classify the following triangles according to their sides.

Figure 2
Figure 2 (Picture 144.png)

CLASS ASSIGNMENT 3

Try and complete the theorems and explain the theorem on the basis of your own example (with the help of a sketch)

1.1 Theorem 1:

The sum of the angles on a straight line

Example:

1.2 Theorem 2:

Example:

1.3 Theorem 3:

The sum of the interior angles of any triangle is

Example: to prove the theorem, carry out the following instructions:

b) Mark the angles of the triangle with the letters A, B and C.

c) Tear off the angles of the triangle.

d) Paste the angles of the triangle next to one another on the line below so that the vertices face the point on the line.

Figure 3
Figure 3 (Picture 89.png)

Complete the following equation: AˆAˆ size 12{ { hat {A}}} {} + BˆBˆ size 12{ { hat {B}}} {} + CˆCˆ size 12{ { hat {C}}} {} = ………°

(Note how each angle is written.)

1.4 Theorem 4:

1.4.1 Before we look at theorem 4, it is important for you to understand the following terms. Explain the following terms with the aid of sketches:

  • exterior angle of a triangle

  • interior angle of a triangle

1.4.2 Complete:

The exterior angle of a triangle is

Example: (Use degrees in your sketch)

  • The above four theorems will serve as reasons when you calculate the sizes of unknown angles.
  • When calculating the size of any angle, you must always give a reason for your explanation.

2. Calculate the sizes of the unknown angles and provide reasons.(Your teacher will help you with the more difficult examples.)

Figure 4
Figure 4 (Picture 145.png)
Figure 5
Figure 5 (Picture 146.png)

HOMEWORK ASSIGNMENTS 2 AND 3

  1. Complete each of the following and give reasons for the following theorems:
Figure 6
Figure 6 (Picture 147.png)

Figure 7
Figure 7 (Picture 150.png)

Figure 8
Figure 8 (Picture 153.png)
  1. Calculate the sizes of each of the unknown angles and provide reasons for each.
Figure 9
Figure 9 (Picture 155.png)

Memorandum

CLASSWORK ASSIGNMENT 1

  • acute-angled
  • right-angled / acute-angled
  • obtuse-angled
  • right-angled
  • equilateral
  • isosceles
  • scalene
  • scalene

CLASSWORK ASSIGNMENT 1

  • = 1800
  • same size
  • 1800
  • Exterior ∠∠ size 12{∠} {}

Interior

  • Equal to the sum of the 2 subtended interior angles

x = 1300 – 500

= 800

  • a = 1800 – 1260 (straight line)

= 540

2.2 1800 – (900 + 390) (straight line)

= 510

2.3 b = 1800 – (630 + 340) (3∠s∠s size 12{∠s} {} = 1800)

= 830

a = 1800 – 83 (straight line)

= 970 / ext ∠∠ size 12{∠} {} = sum of opp

2 int. ∠s∠s size 12{∠s} {}

2.4 3a + 75 = 1800 (straight line)

3a = 1050

a = 350

  • b = 1800 – 1050 (straight line)

= 750

a = 1800 – (650 + 750) (3∠s∠s size 12{∠s} {} = 1800)

= 400

  • 2a – 100 = 300 - a (vert. opp ∠s∠s size 12{∠s} {})

3a = 40

a = 403403 size 12{ { {"40"} over {3} } } {}

a = 13,30

HOMEWORK ASSIGNMENT 1 AND 2

  • 1ˆ1ˆ size 12{ {1} cSup { size 8{ widehat } } } {} + 2ˆ2ˆ size 12{ {2} cSup { size 8{ widehat } } } {} = 1800 (str. line)
  • 1ˆ1ˆ size 12{ {1} cSup { size 8{ widehat } } } {} + 2ˆ2ˆ size 12{ {2} cSup { size 8{ widehat } } } {} + 3ˆ3ˆ size 12{ {3} cSup { size 8{ widehat } } } {} = 1800 (3∠s∠s size 12{∠s} {} = 1800)
  • 4ˆ4ˆ size 12{ {4} cSup { size 8{ widehat } } } {} = 1ˆ1ˆ size 12{ {1} cSup { size 8{ widehat } } } {} + 2ˆ2ˆ size 12{ {2} cSup { size 8{ widehat } } } {} (ext ∠∠ size 12{∠} {} of = sum of 2 opp int. ∠s∠s size 12{∠s} {})
  • 3ˆ3ˆ size 12{ {3} cSup { size 8{ widehat } } } {} = 2ˆ2ˆ size 12{ {2} cSup { size 8{ widehat } } } {} (isc )
  • 1ˆ1ˆ size 12{ {1} cSup { size 8{ widehat } } } {} = 3ˆ3ˆ size 12{ {3} cSup { size 8{ widehat } } } {} = 1ˆ1ˆ size 12{ {1} cSup { size 8{ widehat } } } {} = 4ˆ4ˆ size 12{ {4} cSup { size 8{ widehat } } } {} ( 1ˆ1ˆ size 12{ {1} cSup { size 8{ widehat } } } {} = 3ˆ3ˆ size 12{ {3} cSup { size 8{ widehat } } } {} (isc ); 1ˆ1ˆ size 12{ {1} cSup { size 8{ widehat } } } {} = 4ˆ4ˆ size 12{ {4} cSup { size 8{ widehat } } } {} (vert opp ∠s∠s size 12{∠s} {})

2.1 p = 890 + 200 (vert opp ∠s∠s size 12{∠s} {})

= 1090

  • 2x + 4x + 3x = 1800 (straight line)

9x = 1800

x = 200

  • b = 1800 – (1150 + 300) (straight line)

= 350

a = 1800 – (1150 + 350) (straight line)

= 300

  • a + a + 1400 = 1800 (straight line)

2a = 400

a = 200

2.5 x + 100 + 3 x – 500 = 2 x + 360 (ext ∠∠ size 12{∠} {}of )

x + 3 x – 2 x = 360 + 500 – 100

2 x = 760

x = 380

2.6 p = r = (1800 – 1100) (straight line)

= 700 (p = r, isc )

a = 1800 – 1400) (3∠s∠s size 12{∠s} {} = 1800)

= 400

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