1

Vocabulary:

3 5

0.75 which can be written as a ratio is called a rational number

and 2 which can't be written as ratios are called irrational numbers

or or the symbol is called a RADICAL

n the number beneath the

radical sign is the RADICAND

n is the principal or positive square root of n

n is the negative square root of n

n is the positive or negative square root of n

n is the positive or negative square root of n

What is the approximate value of these radicals? Think about how close they are to the square numbers 4, 9, 16, 25, 36, 49,64, 81

7 20

37

12 102

70 55 48

2.6 10.1 4.5 3.5

8.4 6.1 7.4 6.9

Simplifying radical terms.1. look for factors that are square terms.2. find the square root of that square term.3. put that outside the radical.

459 5

9 5

84 2

4 2

7525 3

25 3

4816 3 16 3

3 5 2 2 5 3 4 3

700 100 7 10 7100 7

16x 8x

3x 2x x x x

7m 6m m 3m m

23y 22y y 11y y

78h 64h 2h 32h 2h1345p 129p 5p 63p 5p

Adding and subtracting radicalsSimplify the terms then gather like terms.

5 7 2 5 2 7 31 5. 3 11 6 22. 2 11 8 2

3 7 5 5 11 2 2

8 3 2 11 3 5 13. 1 6 7 3 5. 64 3

7 3 3 11 4 6 6 3

5 15. 8 2 8

5 9 2 2 4 2

5 3 2 2 2 2

15 2 4 2 19 2

2 756. 5 27

2 25 3 5 9 3

2 5 3 5 3 3

10 3 15 3 5 3

5 12 2 45 3 277. 5 4 00

5 4 3 2 9 5 3 9 3 4 100 5

5 2 3 2 3 5 3 3 3 4 10 5

10 3 6 5 9 3 40 5

19 3 34 5

2 28 3 20 50 58 3. 0 6

2 4 7 3 4 5 100 5 5 9 7

2 2 7 3 2 5 10 5 5 3 7

4 7 6 5 10 5 15 7

11 7 16 5

Simplifying Radicals with Fractions Simplify the two parts of the radical and then simplify again if required.

99

100

99

100 9 11

10

3 11

10

50

27

25 2

9 3

5 2

3 3

300

3 810 3

6 2

5 3

3 2

50

10 4925 2

10 7

5 2

10 7 2

14

100 3

3 4 2

2

Solving Equations with Radicals2 there are actually two solutions.

Can you think of them both?x 4

which is written as 2x 2 or 2

The rest of the steps in solving equations just use the reverse operations that we have used before.

22x 3 9 2 add 3 on both 2x 12 sides2 divide 2 on both s6 sx ide

square root on both sx 6 ides

23x1 5

2

2

add 1 on both 3x

6 2

sides

2 multiply by 2 on both 3x 12 s side22 divide by 3 on both s4 sx ide

square root on both s d sx i e2

23x 15

4

2 multiply by 4 on bot3x 1 2 h 0 es sid 2 add 1 on both 3x 21 sides2 divide by 3 on both s7 sx ide

square root on both sx 7 ides q

22x 45 11

7

2

minus 5 on bot2x

h 4

6 s7

s ide

2

72x

o4 4

n .s2

b 2

4 o2x 3

.s8

n b 2

2 on sx 19

b.

x 19

2

4.2x

16

5

x 7

23x2 4

52. x 10

23(m 1)63.

2

m 3 2 1.5y4 11 y 2

24g2 10

35. g 3 25(h 1) 16. 9 h 3

27y 17y 1. 27 3

3

y 2 2 25n 1 138 2n. n 2

23z2 89.

5 z 10 25 3b 110. 2 b no solution

23(5z1

411.

2)7

z 2 25 2(4t 1 112 1. ) t 1

Finding the Area of Complex Shapes – see Looking For Pythagoras. Investigation 2

1. Make a rectanglearound the shape.2. Make triangles andrectangles in the outside.

Find the area of the rectangle 3 4 12

A BC

D

E

Find the area of the rectangle. 3 4 12

Find the area of the outside shapes. 1 2B 12 1 1A 0.5

2

C 2 1 2 2 2D 22

3 1E 1.52

Find the total outside area 7Subtract the area of the outside fromthe rectangle area to get the inside area. 5

Find the area of the rectangle. 5 4 20

B

C

A

DE

Find the area of the outside shapes. B 2 1 2 3 2A 3

2

1 2C 12

4 1E 22

Find the total outside area 14Subtract the area of the outside fromthe rectangle area to get the inside area. 6

3 4D 62

3

Finding the Length of Segments – see Looking For Pythagoras. Page20

Make a squarearound the segment.Counting the number of dots on each side makes it easy.

Find the area of the outside square 6 6 36Find the area of the outside square. 6 6 36

Find the area of the outside triangles. )5 14( 102

Find the area of the inside square. 36 10 26

Therefore the length of one sideof the square is 26

Find the area of the outside square 5 5 25Find the area of the outside square. 5 5 25

Find the area of the outside triangles. )3 24( 122

Find the area of the inside square. 25 12 13

Therefore the length of one sideof the square is 13

Pythagorean Theorem. See page 482 in the textbook.

In a right triangle the area of a square on the hypotenuse is equal to the sum of the area of the squares on the other two sides.

Note: The hypotenuse is the long side of the triangle. It is opposite the right angle.

2 2 2a b c The sum of the areas gives the equation:

When the hypotenuse is the unknown.

2 2 2a b c 2 2 23 5 x

2234 x

x 34

When a leg is the unknown.

2 2 2a b c 2 2 25 x 11 2 2 2x 11 5

2x 96x 96x 4 6

4

When the side lengths include variables.

2 2 2a b c 2 2 2(2x) (3x) 8 2 24x 9x 64

2x

8

3x

213x 6464

x13

8

13

2 2 2a b c 2 225p 6 7p

2 2 225p 6 49p 2 26 24p3 23

p2

3p

2

When there is more than one triangle in the diagram.

2 2 29 8 y

Choose the triangle where you know two sides. Label the third side with a variable.

2145 y

Note: We don’t need to know what y is. Solve the other triangle.

2 2 2y x 13 2 2 2

we know y 145 so 145 x 13 2 2x 13 145 2x 24x 24

x 2 6

Dot Paper Problems:

Find the perimeter of this shape.

Construct right triangles and then solve using Pythagoras.

2 2x 3 2 13x 3 2 13 2 2y 5 2 29

perimeter 4 13 29 2 5

2 2z 4 2 20 2 5

Find the perimeter for these shapes.

2 10 6 2 51 2

8 1 2 52 0

3 3 3 53 2

3 10 5 24 5

2 13 2 10 35 2 822 13 2 10 35 2 82

6 2 13 10

5 13 2 7 657 5 1

Questions Involving Recognising Pythagorean Relationships:

1. Recognising right triangles:Are the following triangles right triangles?

The sides of a right triangle will make the Pythagoras equation true. Try it for this triangle6

9

7

2 2 26 7 9 36 49 81

85 81Not true so not a right triangle.

15

20

16

72

90

54

2 2 215 16 20 481 400

Not true so nota right triangle.

2 2 212 16 20 400 400

True so it isa right triangle.

16

20

12

2 2 254 72 90 8100 8100True so it isa right triangle.

5

Questions Involving Recognising Pythagorean Relationships:

2. Recognising pythagoras triples:• These are a set of three numbers that fit the pythagorean relationship?

Are these sets of numbers pythagoras triples?

35,12,372 2 235 12 37

1369 1369

23,11,252 2 223 11 25

650 625It is a pythagorean triple Not a pythagorean triple

9,40,412 2 29 40 41

1681 1681It is a pythagorean triple

17,24,29

865 841Not a pythagorean triple

2 2 217 24 29

If you recognise the common pythagoras triples 3,4,5 and 5,12,13 it can make solving questions much quicker.

12 416 4

34

24 226 2

x 2 1

12135 0

16

12

x

24

26x

2516

x 4 254 4

0

40

x

50

60

x

512

2

13

5 560 5

x 5 65

453

40 1050 10

x 10 30

Special Right Triangles. See page 493 in the textbook.We will use two special triangles that are derived using Pythagoras.1. The 45-90-45 triangle.

Using this triangle and proportions we can solve any other45-90-45 triangle.

Use the angles to match up the sides in the two triangles

13

M

90

45 45

Use the angles to match up the sides in the two triangles.• put the unknown on the top for the proportion.

M 2

13 1

2M 13

1

M 13 2

2. The 30-60-90 triangle.

Using this triangle and proportions we can solve any other 30-60-90 triangle.

11

N

60

30

Use the angles to match up the sides in the two triangles.• put the unknown on the top for the proportion.

N 3

11 2

11 3N

2

K

F

12

45

4590

M

12

60 30

90F 1

12 2

12F

2

M 2

12 3

24F

3

Use special triangles to solve these questions:

5

14H

60K 2

5 1

F 5 2

H 1

14 2

H 7