1. 2D Applications of Pythagoras- Part 1
Slideshow 41, Mathematics
Mr. Richard Sasaki

2. Objectives Review polygon properties and interior angles
Review circle properties
Be able to apply the Pythagorean Theorem to other polygons, circles and graphs
Note: Please bring a compass, ruler and tools for decoration next lesson.

3. Pythagorean Theorem
We know how to find missing edges of triangles now. Let’s apply this to other polygons. Oh wait, what is a polygon?
A polygon is a 2D shape with straight edges only.
Example
Find the unknown value on the parallelogram below.
𝑎 2 + 𝑏 2 = 𝑐 2
𝑐= 58
𝑥 𝑐𝑚
7 𝑐𝑚
= 𝑐 2
𝑥= 58
7 𝑐𝑚
9+49= 𝑐 2
58= 𝑐 2
3 𝑐𝑚

4. Answers (Question 1) 𝑎 2 + 𝑏 2 = 𝑐 2 3 3 2 𝑐𝑚 3 𝑐𝑚 3 2 𝑥 𝑐𝑚𝑐𝑚
3 𝑐𝑚
3 2
𝑥 𝑐𝑚
(____) 2 + 𝑥 2 = 3 2
60 o
30 o
9 4 + 𝑥 2 =9
𝑦 𝑐𝑚
𝑥 2 = 36 4 − 9 4 𝑥= = =
𝑏 2 = (_____) 2
𝑎 2 + 𝑏 2 = 𝑐 2 ⇒
3 3
⇒ 𝑏 2 =27
⇒ 𝑏 2 = − 27 4
⇒𝑏=
= 9 2 ⇒ = 𝑦=
6 𝑐𝑚

5. Answers (Question 2) 1
We know interior angles in a hexagon add up to 1
720 𝑜
1 2 𝑥
120 o
720 𝑜 6 =
60 o
120 o
Let’s calculate 𝑥 2 first.
30 o
3 2
As we have a 30−60−90 triangle, 𝑥 2 =
So 𝑥= 3 .

6. Answers (Question 3) We know interior angles in an octagon add up to .1
1080 𝑜 1 𝑎
1080 𝑜 8 =
135 o 𝑥 1
We have a 45−45−90 triangle.
45 o
2 2
𝑎= = 𝑎
135 o
45 𝑜
45 𝑜 2 1
1 2 𝑥= 1
1 2
=1+ 2
45 𝑜
45 𝑜 1

7. Circles Let’s review some parts of the circle. Chord Centre Diameter
Radius
Tangent
Question types with circles are simple, but it’s important that we understand the vocabulary.

8. Circles
Example
Consider a circle with radius 6 𝑐𝑚. The circle has a chord 8 𝑐𝑚 long. Find the distance between the centre and the chord.
𝑎 2 + 𝑏 2 = 𝑐 2
8 𝑐𝑚
𝑥 = 6 2
𝑥 𝑐𝑚
𝑥 2 +16=36
6 𝑐𝑚
𝑥 2 =20
𝑥=2 5
Note: It doesn’t matter where the chord is, as it is a fixed length, it is always the same distance from the centre. So it may as well touch the radius.

9. 𝑥=
𝑥=2 6
Radii touch tangents (or segments of tangents) at 90 𝑜 .
𝑐𝑚 or 𝑚𝑚
The length required is the shortest distance from the chord, so must be at 90 𝑜 .
2 5 𝑐𝑚

10. Graphs
Distances between points can be calculated, based on their coordinates.
Note: The shortest distance between two points is always the of a triangle.
hypotenuse
Example
Find the distance between two points, 𝐴 2, 7 and 𝐵(5, 5).
We should visualise the triangle like…
or…
How did we get 3 and 2? 3 3=
5−2 2=
7−5 2
Let’s find the hypotenuse.
𝑎 2 + 𝑏 2 = 𝑐 2 ⇒
= 𝑐 2
⇒13= 𝑐 2
⇒𝑐= 13

11. Answers - Easy Answers - Medium Answers - Hard 3 5 4 2 41 218 205
3 5
4 2 41
218
205
170 2 65
3 26
2 226
6 17
16 5
794
2 557
205 2 5
Distance 𝐷𝐼 is the greatest.
Answers - Hard
2. We don’t know where in relation to the fountain that they are, 3 𝑚≤𝑥≤7 𝑚, 29
4. 𝐴𝐵= 𝑥 2 − 𝑥 ( 𝑦 2 − 𝑦 1 ) 2
We square the brackets, 𝑥 2 − 𝑥 ≡ 𝑥 1 − 𝑥 2 2

12. Formulae So simply for any 𝐴( 𝑥 1 , 𝑦 1 ) and 𝐵( 𝑥 2 , 𝑦 2 )…
𝐴𝐵= 𝑥 2 − 𝑥 ( 𝑦 2 − 𝑦 1 ) 2
Or written for the change in 𝑥, ∆𝑥 and the change in 𝑦, ∆𝑦…
𝐴𝐵= ∆𝑥 2 +( ∆𝑦) 2