2d applications of pythagoras - part 1 slideshow 41, mathematics mr. richard sasaki, room 307
TRANSCRIPT
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2D Applications of Pythagoras- Part 1
Slideshow 41, MathematicsMr. Richard Sasaki, Room 307
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ObjectivesObjectives
• Review polygon properties and interior angles
• Review circle properties• 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. 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.
7𝑐𝑚7𝑐𝑚
3𝑐𝑚
𝑥𝑐𝑚𝑎2+𝑏2=𝑐232+72=𝑐2
9+49=𝑐258=𝑐2
𝑐=√58𝑥=√58
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Answers (Question 1)Answers (Question 1)
60o
3𝑐𝑚
30o𝑥𝑐𝑚
𝑦 𝑐𝑚
First calculate , and then after.𝑎2+𝑏2=𝑐2⇒¿¿
32
94+𝑥2=9⇒ 𝑥2=
364−94
𝑥=√ 274¿ √27√4¿ 3√3
2
3√32
𝑐𝑚
𝑎2+𝑏2=𝑐2⇒( 3√32 )2
+𝑏2=(¿¿)2
3√3⇒ 274 +𝑏2=27
⇒𝑏2=1084−274⇒𝑏=√ 814¿ 92
𝑦=¿32+92=¿6𝑐𝑚
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12
Answers (Question 2)Answers (Question 2)
𝑥
1 We know an interior angles in a hexagon add up to .720𝑜
() 720𝑜
6=¿120o
120o 30oLet’s calculate first.
1
6 0o
As we have a 30-60-90 triangle, .√32
So .
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Answers (Question 3)Answers (Question 3)
𝑥
1
We know an interior angles in an octagon add up to .1080𝑜
() 1080𝑜
8=¿135o
45o
135o
1
1
𝑎
𝑎
Let’s calculate .We have a 45-45-90 triangle.
45𝑜
45𝑜1
√21
45𝑜
45𝑜
1
√211√2
𝑎=1
√2=¿√22
𝑥=¿√22
+1+ √22
¿1+√2
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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.
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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.
6𝑐𝑚
8𝑐𝑚
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.
𝑥𝑐𝑚
𝑎2+𝑏2=𝑐2
𝑥2+42=62
𝑥2+16=36𝑥2=20𝑥=2√5
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AnswersAnswers
𝑥=√112
𝑥=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√5𝑐𝑚
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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 like…or…
32
How did we get and ?3=¿5−22=¿7−5
Let’s find the hypotenuse.𝑎2+𝑏2=𝑐2⇒22+32=𝑐2
⇒13=𝑐2⇒𝑐=√13
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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,
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FormulaeFormulaeSo simply for any and …
𝐴𝐵=√ (𝑥2− 𝑥1 )2+(𝑦 2− 𝑦1)2
Or written for the change in , and the change in , …
𝐴𝐵=√ (∆𝑥 )2+¿¿One 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.