Download pptx - 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

ObjectivesObjectives

• Review polygon properties and interior angles

• Review circle properties• Be able to apply the Pythagorean Theorem

to other polygons, circles and graphs

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

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

12


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 .


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

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.

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

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

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

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,

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.