Classification of triangles

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)

4. Classify the following triangles according to their sides.

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.

Complete the following equation: Aˆ size 12{ { hat {A}}} {} + Bˆ size 12{ { hat {B}}} {} + Cˆ 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.)

HOMEWORK ASSIGNMENTS 2 AND 3 Complete each of the following and give reasons for the following theorems:

Calculate the sizes of each of the unknown angles and provide reasons for each.

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 size 12{∠s} {}} = 180⁰) = 83⁰ a = 180⁰ – 83 (straight line) = 97⁰ / ext ^{∠ size 12{∠} {}} = sum of opp 2 int. ^{∠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 size 12{∠s} {}} = 180⁰) = 40⁰ 2a – 10⁰ = 30⁰ - a (vert. opp ^{∠s size 12{∠s} {}}) 3a = 40 a = 403 size 12{ { {"40"} over {3} } } {} a = 13,3⁰ HOMEWORK ASSIGNMENT 1 AND 2 1ˆ size 12{ {1} cSup { size 8{ widehat } } } {} + 2ˆ size 12{ {2} cSup { size 8{ widehat } } } {} = 180⁰ (str. line) 1ˆ size 12{ {1} cSup { size 8{ widehat } } } {} + 2ˆ size 12{ {2} cSup { size 8{ widehat } } } {} + 3ˆ size 12{ {3} cSup { size 8{ widehat } } } {} = 180⁰ (3^{∠s size 12{∠s} {}} = 180⁰) 4ˆ size 12{ {4} cSup { size 8{ widehat } } } {} = 1ˆ size 12{ {1} cSup { size 8{ widehat } } } {} + 2ˆ size 12{ {2} cSup { size 8{ widehat } } } {} (ext ^{∠ size 12{∠} {}} of = sum of 2 opp int. ^{∠s size 12{∠s} {}}) 3ˆ size 12{ {3} cSup { size 8{ widehat } } } {} = 2ˆ size 12{ {2} cSup { size 8{ widehat } } } {} (isc ) 1ˆ size 12{ {1} cSup { size 8{ widehat } } } {} = 3ˆ size 12{ {3} cSup { size 8{ widehat } } } {} = 1ˆ size 12{ {1} cSup { size 8{ widehat } } } {} = 4ˆ size 12{ {4} cSup { size 8{ widehat } } } {} ( 1ˆ size 12{ {1} cSup { size 8{ widehat } } } {} = 3ˆ size 12{ {3} cSup { size 8{ widehat } } } {} (isc ); 1ˆ size 12{ {1} cSup { size 8{ widehat } } } {} = 4ˆ size 12{ {4} cSup { size 8{ widehat } } } {} (vert opp ^{∠s size 12{∠s} {}}) 2.1 p = 89⁰ + 20⁰ (vert opp ^{∠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 size 12{∠s} {}} = 180⁰) = 40⁰