Transcript
Page 1: 2D Applications of Pythagoras - Part 1 Slideshow 41, Mathematics Mr. Richard Sasaki, Room 307

2D Applications of Pythagoras- Part 1

Slideshow 41, MathematicsMr. Richard Sasaki, Room 307

Page 2: 2D Applications of Pythagoras - Part 1 Slideshow 41, Mathematics Mr. Richard Sasaki, Room 307

ObjectivesObjectives

β€’ Review polygon properties and interior angles

β€’ Review circle propertiesβ€’ Be able to apply the Pythagorean Theorem

to other polygons, circles and graphs

Page 3: 2D Applications of Pythagoras - Part 1 Slideshow 41, Mathematics Mr. Richard Sasaki, Room 307

Pythagorean TheoremPythagorean 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.We actually already did this, this is a review.ExampleFind the unknown value on the polygon below.

7π‘π‘š7π‘π‘š

3π‘π‘š

π‘₯π‘π‘šπ‘Ž2+𝑏2=𝑐232+72=𝑐2

9+49=𝑐258=𝑐2

𝑐=√58π‘₯=√58

Page 4: 2D Applications of Pythagoras - Part 1 Slideshow 41, Mathematics Mr. Richard Sasaki, Room 307

Answers (Question 1)Answers (Question 1)

60o

3π‘π‘š

30oπ‘₯π‘π‘š

𝑦 π‘π‘š

First calculate , and then after.π‘Ž2+𝑏2=𝑐2β‡’ΒΏΒΏ

32

94+π‘₯2=9β‡’   π‘₯2=

364βˆ’94

π‘₯=√ 274ΒΏ √27√4ΒΏ 3√3

2

3√32

π‘π‘š

π‘Ž2+𝑏2=𝑐2β‡’( 3√32 )2

+𝑏2=(ΒΏΒΏ)2

3√3β‡’ 274 +𝑏2=27

⇒𝑏2=1084βˆ’274⇒𝑏=√ 814ΒΏ 92

𝑦=ΒΏ32+92=ΒΏ6π‘π‘š

Page 5: 2D Applications of Pythagoras - Part 1 Slideshow 41, Mathematics Mr. Richard Sasaki, Room 307

12

Answers (Question 2)Answers (Question 2)

π‘₯

1 We know an interior angles in a hexagon add up to .720π‘œ

() 720π‘œ

6=ΒΏ120o

120o 30oLet’s calculate first.

1

6 0o

As we have a 30-60-90 triangle, .√32

So .

Page 6: 2D Applications of Pythagoras - Part 1 Slideshow 41, Mathematics Mr. Richard Sasaki, Room 307

Answers (Question 3)Answers (Question 3)

π‘₯

1

We know an interior angles in an octagon add up to .1080π‘œ

() 1080π‘œ

8=ΒΏ135o

45o

135o

1

1

π‘Ž

π‘Ž

Let’s calculate .We have a 45-45-90 triangle.

45π‘œ

45π‘œ1

√21

45π‘œ

45π‘œ

1

√211√2

π‘Ž=1

√2=¿√22

π‘₯=¿√22

+1+ √22

¿1+√2

Page 7: 2D Applications of Pythagoras - Part 1 Slideshow 41, Mathematics Mr. Richard Sasaki, Room 307

CirclesCirclesLet’s review some parts of the circle.

Centre (origin)

Radius

Tangent

Diameter

Chord

Question types with circles are simple, but it’s important that we understand the vocabulary.

Page 8: 2D Applications of Pythagoras - Part 1 Slideshow 41, Mathematics Mr. Richard Sasaki, Room 307

CirclesCirclesExampleConsider a circle with radius 6 cm. The circle has a chord 8 cm long. Find the distance between the centre and the chord.

6π‘π‘š

8π‘π‘š

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.

π‘₯π‘π‘š

π‘Ž2+𝑏2=𝑐2

π‘₯2+42=62

π‘₯2+16=36π‘₯2=20π‘₯=2√5

Page 9: 2D Applications of Pythagoras - Part 1 Slideshow 41, Mathematics Mr. Richard Sasaki, Room 307

AnswersAnswers

π‘₯=√112

π‘₯=2√6

Radii touch tangents (or segments of tangents) at .

or

The length required is the shortest distance from the chord, so must be at .

2√5π‘π‘š

Page 10: 2D Applications of Pythagoras - Part 1 Slideshow 41, Mathematics Mr. Richard Sasaki, Room 307

GraphsGraphsDistances between points can be calculated, based on their coordinates. Note: The shortest distance between two points is always the of a triangle.hypotenuseExampleFind the distance between two points, and .

We should visualize the triangle like…or…

32

How did we get and ?3=ΒΏ5βˆ’22=ΒΏ7βˆ’5

Let’s find the hypotenuse.π‘Ž2+𝑏2=𝑐2β‡’22+32=𝑐2

β‡’13=𝑐2⇒𝑐=√13

Page 11: 2D Applications of Pythagoras - Part 1 Slideshow 41, Mathematics Mr. Richard Sasaki, Room 307

Answers - EasyAnswers - Easy3√5 4√2√1702 √65

√2052 5

Answers - MediumAnswers - Medium

√ 41 √218 √2053√26 2√226 6 √1716√5 √794 2√557Distance is the greatest.

Answers - HardAnswers - Hard1. 2. We don’t know where in relation to the fountain that they are, 3. 4. We square the brackets,

Page 12: 2D Applications of Pythagoras - Part 1 Slideshow 41, Mathematics Mr. Richard Sasaki, Room 307

FormulaeFormulaeSo simply for any and …

𝐴𝐡=√ (π‘₯2βˆ’ π‘₯1 )2+(𝑦 2βˆ’ 𝑦1)2

Or written for the change in , and the change in , …

𝐴𝐡=√ (βˆ†π‘₯ )2+ΒΏΒΏOne of these will be provided in the exam, however you know how to do the questions without them anyway!Please bring a ruler, pencil and compass to the next lesson.


Top Related