Download File

Details of Class: Lesson 4
Date:
Time: 50mins
Learning Intentions/Topic:
Year: 8
Geometry: Polygons
To recognise different types of polygons and
their properties
To be able to draw all different types of
polygons, labelling sides and angles correctly
To be able to calculate the angles inside different
polygons
Room:
No. of students:
Relationship to VELS:
Classification of polygons with reference to a
definition or key property
Using language relevant to the topic
Resources Required:
Students to each have their own workbook and stationery
Maths Quest 8 textbook
Whiteboard & Markers
Introduction/Housekeeping/Warm-up:
Link to last lesson :
So far we have learnt how to calculate the interior and exterior angles of a triangle, and last lesson
we learnt about different types of quadrilaterals and how to calculate the angles inside a
quadrilateral. Today we are learning how to calculate angles inside any polygon.
2 min
Content
Resources
Description:
So far we have learnt that the sum of the angles in a triangle is
180˚ and the sum of the angles inside a quadrilateral is 360˚.
We will now discover a method of finding the sum of the angles
in any polygon.
Clarification:
What is a polygon?
Origin: poly = many
Therefore, a polygon is a shape with many sides. This can be
anything from 3 sides to 1000 sides, or more.
Activity: Complete table provided ‘Sum of Angles in a Polygon’
Time
5 min
Sum of Angles in a
Polygon table &
instructions
10 min
Examples of how to find angles in polygons (sheet provided)
Write these examples on board, going step-by-step through each
example with the class.
Example sheet
provided
10 min
Activity: Students to complete Ex 7F p. 262 – 263, questions 1 - 9
Maths Quest 8 p.
262 - 263
17 min
Go through answers to activity. Highlight the patterns evident.
Formula: Sum of angles = 180 x (number of sides – 2)
3 min
Reinforcement of Ideas/Questioning/Homework:
Questions?
Homework: Finish any unfinished questions from today’s activities.
Overall reflection of lesson:
3 min
Sum of Angles in a Polygon
Instructions:
1. For every polygon shown in the table, draw as many diagonals as possible from the vertex.
Marked X. (This will divide the polygon into a number of triangles.)
2. Complete the table.
3. Can you see a pattern?
 What would be the sum of the angles in a dodecagon (12 sides)?
 Can you predict the angle sum of an icosagon (20 sides)?
 What about a polygon with 100 sides?
Polygon
Name
Number of Sides
Triangle
3
Number of
Triangles
1
Quadrilateral
4
2
Sum of Angles
180˚
X
X
Pentagon
X
Hexagon
X
Heptagon
X
Octagon
X
Decagon
X
2 x 180˚ = 360˚
Answers:
Sum of Angles in a Polygon
Instructions:
4. For every polygon shown in the table, draw as many diagonals as possible from the vertex.
Marked X. (This will divide the polygon into a number of triangles.)
5. Complete the table.
6. Can you see a pattern?
The number of triangles is 2 less than the number of sides. Sum of angles is 180˚ x number of
triangles. Sum of angles = 180˚ x (number of sides – 2)
 What would be the sum of the angles in a dodecagon (12 sides)?
(12 – 2) x 180˚ = 1800˚
 Can you predict the angle sum of an icosagon (20 sides)?
(20 – 2) x 180˚ = 3240˚
 What about a polygon with 100 sides?
(100 – 2) x 180˚ = 17 640
Polygon
Name
Number of Sides
Triangle
3
Number of
Triangles
1
Quadrilateral
4
2
2 x 180˚ = 360˚
Pentagon
5
3
3 x 180˚ = 540˚
Hexagon
6
4
4 x 180˚ = 720˚
Heptagon
7
5
5 x 180˚ = 900˚
Octagon
8
6
6 x 180˚ = 1080˚
Decagon
10
8
8 x 180˚ = 1440˚
Sum of Angles
180˚
X
X
X
X
X
X
X
Examples:
1. Find the sum of the interior angles of the polygon shown:
Sum of angles = 180˚ x (n – 2)
n = 13
Sum of angles = 180˚ x (13 – 2)
= 180˚ x 11
= 1980˚
2. For the polygon shown, find:
a) the sum of its interior angles
b) the value of the pronumeral
a) Sum of angles = 180˚ x (n – 2)
120˚
130˚
n=5
Sum of angles = 180˚ x (5 – 2)
= 180˚ x 3
= 540˚
95˚
95˚
b) p + 120˚ + 130˚ + 95˚ + 95˚ = 540˚
p + 440˚ = 540˚
p = 540˚ – 440˚
p = 100˚
3. Find the value of the pronumeral in this regular polygon
Sum of angles = 180˚ x (n – 2)
n=6
Sum of angles = 180˚ x (6 - 2)
= 180˚ x 4
= 720˚
a = 720˚ ÷ 6
= 120˚
a
p