Download - 2D Applications of Pythagoras - Part 1 Slideshow 41, Mathematics Mr. Richard Sasaki, Room 307
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2D Applications of Pythagoras- Part 1
Slideshow 41, MathematicsMr. Richard Sasaki, Room 307
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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
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Pythagorean TheoremPythagorean Theorem
We know how to find missing edges of triangles now. Letβs 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.
7ππ7ππ
3ππ
π₯πππ2+π2=π232+72=π2
9+49=π258=π2
π=β58π₯=β58
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Answers (Question 1)Answers (Question 1)
60o
3ππ
30oπ₯ππ
π¦ ππ
First calculate , and then after.π2+π2=π2βΒΏΒΏ
32
94+π₯2=9β π₯2=
364β94
π₯=β 274ΒΏ β27β4ΒΏ 3β3
2
3β32
ππ
π2+π2=π2β( 3β32 )2
+π2=(ΒΏΒΏ)2
3β3β 274 +π2=27
βπ2=1084β274βπ=β 814ΒΏ 92
π¦=ΒΏ32+92=ΒΏ6ππ
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12
Answers (Question 2)Answers (Question 2)
π₯
1 We know an interior angles in a hexagon add up to .720π
() 720π
6=ΒΏ120o
120o 30oLetβs calculate first.
1
6 0o
As we have a 30-60-90 triangle, .β32
So .
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Answers (Question 3)Answers (Question 3)
π₯
1
We know an interior angles in an octagon add up to .1080π
() 1080π
8=ΒΏ135o
45o
135o
1
1
π
π
Letβs calculate .We have a 45-45-90 triangle.
45π
45π1
β21
45π
45π
1
β211β2
π=1
β2=ΒΏβ22
π₯=ΒΏβ22
+1+ β22
ΒΏ1+β2
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CirclesCirclesLetβs review some parts of the circle.
Centre (origin)
Radius
Tangent
Diameter
Chord
Question types with circles are simple, but itβs important that we understand the vocabulary.
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CirclesCirclesExampleConsider a circle with radius 6 cm. The circle has a chord 8 cm long. Find the distance between the centre and the chord.
6ππ
8ππ
Note: It doesnβt 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₯ππ
π2+π2=π2
π₯2+42=62
π₯2+16=36π₯2=20π₯=2β5
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AnswersAnswers
π₯=β112
π₯=2β6
Radii touch tangents (or segments of tangents) at .
or
The length required is the shortest distance from the chord, so must be at .
2β5ππ
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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 likeβ¦orβ¦
32
How did we get and ?3=ΒΏ5β22=ΒΏ7β5
Letβs find the hypotenuse.π2+π2=π2β22+32=π2
β13=π2βπ=β13
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Answers - EasyAnswers - Easy3β5 4β2β1702 β65
β2052 5
Answers - MediumAnswers - Medium
β 41 β218 β2053β26 2β226 6 β1716β5 β794 2β557Distance is the greatest.
Answers - HardAnswers - Hard1. 2. We donβt know where in relation to the fountain that they are, 3. 4. We square the brackets,
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FormulaeFormulaeSo simply for any and β¦
π΄π΅=β (π₯2β π₯1 )2+(π¦ 2β π¦1)2
Or written for the change in , and the change in , β¦
π΄π΅=β (βπ₯ )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.