GOMATHMEMOGR11

Page 1

1

Compiled by Chez Nell

2

Topic

Page

1.

Exponents

3

2.

Algebraic Factors

5

3.

Quadratic Equations

6

4.

Simultaneous Equations

10

5.

Inequalities

14

6.

Algebraic Fractions

16

7.

Number Patterns

18

8.

Financial Math

30

9.

Functions & Graphs

31

10.

Linear Programming

50

11.

Probability Theory

64

12.

Analytical Geometry

70

13.

Transformation Geometry

85

14.

Trigonometry

87

15.

Data Handling

16.

Grade 12 Statistical Data

108 122

17.

Volumes & Surface Area

125

18.

Circle Geometry

130

ďƒ“ Norma

Nell 2011

3 Exponents: Page 5: Exercise 1.1 1.

2 x 3 2 x  2 x  4. 2 x 1

2.

2 x 1 2 x 3 2 2 x 2  2 x 2  1 2 x 2.2 x 2

3.

4 x 18 x 1 (2 2 ) x 1 (2 3 ) x 1 2 2 x 2 2 3 x 3 2 5 x 1 1 1    5 x 5  4  x 1 5 x 1 5 x 5 16 32 (2 ) 2 2 2

4.

5x 25x 1 (5x.(52 ) x 1 5x52 x  2 53 x  2 1 1    3 x 1  3  x 3 x 3x 5.125 5.(5 ) 5.5 5 5 125

5.

7 x 2 49 x  2 7 x 2.(7 2 ) x  2 7 x 2..7 2   7 3 x2 7 3x2 7 3x2

6.

6 n1.12 n1.2 n (2.3) n 1 .(2 2.3) n1 .2 n 2 n1.3n 1.2 2 n 2.3n1.2 n 2 4 n1.32 n 1 1    4 n1 2 n 4  4  n  2 n 1 2 n 2 3 n 1 n  2 2 n  4 3 n 3 81 18 .8 (2.3 ) .(2 ) 2 .3 .2 2 .3 3

x4

7 3x2  70  1 3x2 7

7.

6 n 212 2 n14 2 n3 (2.3) n 2 (2 2.3) 2 n1 (2 2 ) 2 n3 2 n 2 3n 2 2 4 n 2 32 n12 4 n6 2 9 n2 33n3 3 3    9 n 3 3n  2  5  3n 1 n 1 n 3 3n 1 2 n 1 n 9 n 3 2 n  2 n 32 8 9 3 (2 ) (3 ) 3 2 3 3 2 3 2

Page 8: Exercise 1.2 1.

x 2 x 1  2 x 3 2 x.2  2 x.2 3 2 (2  18 )    2 x2  2 x 2 x..2 2  2 x 2 x ( 14  1)

2.

3 x  3 x  2 3 x  3 x.32 3 x (1  32 ) 10 10 3 15   x 1  4    1 4 2 3 x 1  3 x 3 x 31  3 x 3 ( 3  1) 3

3.

5 2 x  4.5  x 5 2 5  x  4.5  x 5  x (25  4) 21    5  x  2.5  x 1 5  x  2.5  x 51 5  x (1  10) 11

4.

3n.34  6.3n.31 3n (81  18) 63   1 63 7.3n.32 3n (63)

5

n 2 n.2 5  3.2 n.2 2 2 (32  34 ) 125 1 25     n 3 n 4 40 32 5.2 2 2 (40)

17 8 5 4

17 4 17   8 5 10

4 Page 12: Exercise 1.3

3 x яА╜ 81

1.1

3 ЁЭСЛ = 34 ЁЭСе=4 x3 = 27

1.2

ЁЭСе 3 = 33 ЁЭСе=3

5x

1.3

яАн4

3

= 80

4

ЁЭСе тИТ3 = 24 ЁЭСе = 24 ЁЭСе=

3 4

тИТ

1 8

2 x яА╜ 16 x яАн1

1.4

3.4 2 x яА╜ 48

1.7

2ЁЭСе = 24ЁЭСетИТ4

22ЁЭСе = 24

ЁЭСе = 4ЁЭСе тИТ 4

ЁЭСе=2

4

ЁЭСе=3

5.42 x яА╜ 40

1.5

24ЁЭСе = 23 ЁЭСе=

3 4

1.8. 2 xяАл1 яАл 2 x яА╜ 12 2ЁЭСе . 2 + 2ЁЭСе = 12 2ЁЭСе 2 + 1 = 12 2 ЁЭСе = 22 ЁЭСе=2

1.6

1 (2 )

x xяАн2

яА╜

1 23

ЁЭСе 2 тИТ 2ЁЭСе тИТ 3 = 0 ЁЭСетИТ3 ЁЭСе+1 = 0 ЁЭСе = 3 ЁЭСЬЁЭСЯ ЁЭСе = тИТ1

1.9 32 xяАл1 яАн 2.32 x яА╜ 45 32ЁЭСе . 3 + 2. 32ЁЭСе = 45 3ЁЭСе 3 + 2 = 45 3 ЁЭСе = 32 ЁЭСе=2

5 Surds Page 16: Exercise 1: 20 = 2 5

1.1

3

1.4

3

24 = 2 3

18 = 3 2

1.2 1.5

3

3

135 = 3 5

2+3 2=4 2

2.1

2 + 18

2.3

= 2+3 2

= 15 2

245 + 6 5

3

2.6

3

54 − 16 3

3

=3 2−2 2

=13 5

=

2 8 + 4 32 − 3 50

3

2 12− 75+2 3 3

2.8.

= 4 2 + 16 2 − 15 2

= −1

=5 2 3.1

2( 2  6)  2  2 3

3.2

3( 6  3)  18  3  3 2  3

3.3

2 5 (3 5  2 2 )  30  4 10

3.4

3 2 (2 8  18 )  24  18  6

3.5

( 2  1)( 2  1))  2  1  1

3.6

5

96 = 2 3

3 8 + 5 50 − 4 32

=7 5 +6 5

2.7

5

= 6 2 + 25 2 − 16 2

=4 2 2.5

1.6

3+3 3−2 3 = 2 3

2.2 2.4

245 = 7 5

1.3

3 2 + 5 3 2 − 5 = 18 − 5 = 13

=

2 3−5 3+2 3 3

6 2

3+ 2

3.7

= 5+2 6

3.8 ( 6  3 )( 6  3 )  6  3  3 4.1

3 3

×

3 3

= 3

2 10

=1

4.2

5 2 10

4.3

6 18 3 12

=

18 2 3 × 3 6 3

4.4

4 3 3 12

=

4 3 6 3

4.5

2 2− 3

× 2+

4.6 4.7

×

7 3+ 2 2− 3 2+ 3

×

2 3

=

2+ 3 3

=

3− 2 3− 2

2− 3 3

× 2−

4.8

7−3 2 7− 2

4.9

2 5+ 3 5 3−3 5

4+2 3 1 21−7 2 7

= =

=

5 3+3 5 3+3 5

×5

=7− 2

7−4 3 1

7+ 2 7+ 2

×

= 6

1−2 14 5

=

45+13 15 30

Algebraic Factors: Page 22: Revision Exercise 2.1. 1.1

2 = 2( x  16)

1.2

= ( x  y)( p 2  q 2 ) = ( x  y)( p  q)( p  q)

1.3

=

= 2( x  4)( x  4)

(a  b)(a  b)  (a  b) (a  b)(a  b  1)

[(2 x  y )  ( x  2 y )][(2 x  y )  ( x  2 y )] 1.4 = (2 x  y  x  2 y )(2 x  y  x  2 y ) (3x  y )( x  3 y ) [2( x  y )  3( x  y )][2( x  y )  3( x  y )] 1.5 = (2 x  2 y  3x  3 y )(2 x  2 y  3x  3 y ) (5 x  y )( x  5 y ) 1.6 = ( x  3)( x  4)

7 1.7 =

2( x 2  12 x  35) 2( x  5)( x  7)

3(3x 2  14 x  15) 1.8 = 3(3x  5)( x  3) Quadratic Equations: Page 25:Exercise 3.1.

1. ( x  5)( x  2)  0

2. (a  6)(a  1)  0

𝑥 = 5 𝑜𝑟 𝑥 = 2

𝑎 = −6 𝑜𝑟 1

3. x( x  1)  0

4. ( x  2)( x  3)( x  5)  0

𝑥 = 0 𝑜𝑟 𝑥 = 1

5. x(2 x  5)(3x  2)  0 5

𝑥 = 0 𝑜𝑟 𝑥 = 2 𝑜𝑟 𝑥 =

2 3

7. x 2  5x  6  0 𝑥+2 𝑥+3 =0 𝑥 = −2 𝑜𝑟 − 3

9. x( x  1)  6 𝑥2 − 𝑥 − 6 = 0 𝑥−3 𝑥+2 =0 𝑥 = 3 𝑜𝑟 𝑥 = −2

11. x 2  2 x  15  0 𝑥−3 𝑥+5 =0 𝑥 = 3 𝑜𝑟 − 5

𝑥 = 2 𝑜𝑟 𝑥 = −3 𝑜𝑟 𝑥 = 5

6. y 2  3 y  10  0 𝑦+2 𝑦−5 =0 𝑦 = −2 𝑜𝑟 5

8. x 2  7 x  6  0 𝑥−1 𝑥−6 =0 𝑥 = 1 𝑜𝑟 𝑥 = 6

10. ( x  3)( x  2)  12 𝑥 2 + 5𝑥 − 6 = 0 𝑥+6 𝑥−1 = 0 𝑥 = −6 𝑜𝑟 𝑥 = 1

8

12. x( x яАн 16) яА╜ 3(24 яАн 5x)

13. (2 x яАн 5)(3x яАл 2) яА╜ 2(3x яАн 11)

ЁЭСе 2 тИТ 16ЁЭСе = 72 тИТ 15ЁЭСе

6ЁЭСе 2 тИТ 11ЁЭСе тИТ 10 = 6ЁЭСе тИТ 22

ЁЭСе 2 тИТ ЁЭСе тИТ 72 = 0

6ЁЭСе 2 тИТ 17ЁЭСе + 12 = 0

ЁЭСетИТ9 ЁЭСе+8 =0

2ЁЭСе тИТ 3 3ЁЭСе тИТ 4 = 0 3

ЁЭСе = 9 ЁЭСЬЁЭСЯ тИТ 8

ЁЭСе = 2 ЁЭСЬЁЭСЯ

Completing the square: Page 29: Exercise 3.2. 1.

x 2 яАл 2 x яАн 24 яА╜ 0 ЁЭСе 2 + 2ЁЭСе + 1

2

= 24 + 1

ЁЭСе+1

2

= 25

2

ЁЭСе + 1 = ┬▒5 ЁЭСе = 4 ЁЭСЬЁЭСЯ тИТ 6 2.

x 2 яАл 9 x яАн 36 яА╜ 0 ЁЭСе 2 + 9ЁЭСе + ЁЭСе+

9 2 2

= 36 +

9 2 2

=

225 4

9

ЁЭСе+2 =┬▒ ЁЭСе=

9 2 2

225 4

тИТ9┬▒15 2

ЁЭСе = тИТ12 ЁЭСЬЁЭСЯ 3 3.

x 2 яАл 8 x яАл 15 яА╜ 0

ЁЭСе 2 + 8ЁЭСе + 4

2

= тИТ15 + 4

ЁЭСе+4

2

2

=1

ЁЭСе = тИТ4 ┬▒ 1 ЁЭСе = тИТ5 ЁЭСЬЁЭСЯ тИТ 3 ЁЭСе = 4 ЁЭСЬЁЭСЯ тИТ 6

4 3

9

x 2 яАн 7 x яАл 12 яА╜ 0

4.

7 2 2

ЁЭСе 2 тИТ 7ЁЭСе + 7 2

ЁЭСетИТ2

= тИТ12 +

1

=4

7 2

ЁЭСетИТ =┬▒

1 2

7┬▒1 2

ЁЭСе=

ЁЭСе = 4 ЁЭСЬЁЭСЯ 3

2x 2 яАн 7x яАл 6

5. 7

ЁЭСе2 тИТ 2 ЁЭСе +

7 2 4

7 2 4 7 2 1 = 16 4

= тИТ3 +

ЁЭСетИТ

7

1

ЁЭСе тИТ 4 = ┬▒4 7┬▒1 4

ЁЭСе=

ЁЭСе = 2 ЁЭСЬЁЭСЯ 6. ЁЭСе2 тИТ

3 2

2 x 2 яАн 11 x яАн 6 11 ЁЭСе 2

+

11 2 4

11 2 4

=3+

ЁЭСетИТ

11 2 4

=

11 4

=┬▒

ЁЭСетИТ

ЁЭСе=

169 16 13 4

11┬▒13 4 1

ЁЭСе = 6 ЁЭСЬЁЭСЯ тИТ 2 7.

2 x 2 яАн 3x яАн 8 яА╜ 0 3

ЁЭСе2 тИТ 2 ЁЭСе +

3 2 4 3 2

ЁЭСетИТ4

=4+

3 2 4

73

= ┬▒ 16

3

ЁЭСетИТ4 =┬▒ ЁЭСе=

73 4

3┬▒ 73 4

ЁЭСе = 2,89 ЁЭСЬЁЭСЯ тИТ 1,39

7 2 2

10 Page 32: Exercise 3.3. Quadratic Formula

x 2  4x  3  0

1.

𝑥=

−𝑏± 𝑏 2 −4𝑎𝑐 2𝑎

𝑥=

−(−4)± 4 2 −4 1 (3) 2

𝑥 = 3 𝑜𝑟 1

2 x 2  x  10

2.

2𝑥 2 − 𝑥 − 10 = 0 𝑥=

1± 1−4 2 (−10) 4

𝑥 = 2,5 𝑜𝑟 − 2

3x 2  x  2  0

3. 𝑥=

1± 1−4 3 (−2) 6

𝑥=− 4. 𝑥=

2 3

𝑜𝑟 1

x 2  6x  4  0 −6± 36−4 1 (+4) 2

𝑥 = −0,75 𝑜𝑟 − 5,24 5. 𝑥=

2x 2  7 x  4  0 7± 49−4 2 (4) 4

𝑥 = 0,72 𝑜𝑟 2,78 6. 𝑥=

2𝑥 2 + 6𝑥 + 3 = 0 −6± 36−4 2 (3) 4

𝑥 = −0,63 𝑜𝑟 − 2,37

11 Simultaneous Equations: Page34: Exercise 4.1. 1.

y  x  2 and x 2  2 xy  4  0 𝑦 =𝑥+2 𝑥 2 + 2𝑥 𝑥 + 2 − 4 = 0 𝑥 2 + 2𝑥 2 + 4𝑥 − 4 = 0 3𝑥 2 + 4𝑥 − 4 = 0 3𝑥 − 2 𝑥 + 2 = 0 2

𝑥 = −2 𝑜𝑟 𝑥 = 3 2

𝑦 = 0 𝑜𝑟 𝑦 = 2 3 2.

x  y  2 and x 2  y 2  20 𝑥 =𝑦+2 𝑦+2

2

+ 𝑦 2 = 20

𝑦 2 + 4𝑦 + 4 + 𝑦 2 − 20 = 0 𝑦 2 + 2𝑦 − 8 = 0 𝑦+4 𝑦−2 =0 𝑦 = −4 𝑜𝑟 𝑦 = 2 𝑥 = −2 𝑜𝑟 𝑥 = 4

3.

x  y  4 and xy  4 𝑦 =4−𝑥 𝑥 4−𝑥 =4

𝑥 2 − 4𝑥 + 4 = 0 (𝑥 − 2)2 = 0 𝑥=2 ∴𝑦=2

12

x  y  3 and xy  4

4.

𝑦 =3−𝑥 𝑥 3−𝑥 +4 = 0 𝑥 2 − 3𝑥 − 4 = 0 𝑥−4 𝑥+1 =0 𝑥 = 4 𝑜𝑟 𝑥 = −1 𝑦 = −1 𝑜𝑟 𝑦 = 4 5.

x  2 y  1 and x 2  2 xy  2 x  4 y  0

𝑥 = 2𝑦 + 1 (2𝑦 + 1)2 + 2𝑦(2𝑦 + 1) − 2(2𝑦 + 1 − 4𝑦 = 0 4𝑦 2 + 4𝑦 + 1 + 4𝑦 2 + 2𝑦 − 4𝑦 − 2 − 4𝑦 = 0 8𝑦 2 − 2𝑦 − 1 = 0 2𝑦 − 1 4𝑦 + 1 = 0 1

1

𝑦 = 2 𝑜𝑟 − 4 𝑥 = 2 𝑜𝑟 6.

1 2

x  y  3 and x 2  y 2  89 𝑥 =𝑦−3 (𝑦 − 3)2 + 𝑦 2 − 89 = 0 𝑦 2 − 6𝑦 + 9 + 𝑦 2 − 89 = 0 𝑦 2 − 3𝑦 − 40 = 0 𝑦−8 𝑦+5 =0 𝑦 = 8 𝑜𝑟 𝑦 = −5 𝑥 = 5 𝑜𝑟 𝑥 = −8

13

2 x  y  1 and x 2  2 yx  2 x  y 2  0

7.

đ?‘Ś = 2đ?‘Ľ − 1 đ?‘Ľ 2 − 2đ?‘Ľ 2đ?‘Ľ − 1 + 2đ?‘Ľ − 2đ?‘Ľ − 1

2

=0

đ?‘Ľ 2 − 4đ?‘Ľ 2 + 2đ?‘Ľ + 2đ?‘Ľ − 4đ?‘Ľ 2 + 4đ?‘Ľ − 1 = 0 7đ?‘Ľ 2 − 8đ?‘Ľ + 1 = 0 7đ?‘Ľ − 1 đ?‘Ľ − 1 = 0 1

đ?‘Ľ = 7 đ?‘œđ?‘&#x; đ?‘Ľ = 1 5

đ?‘Ś = − 7 đ?‘œđ?‘&#x; đ?‘Ś = 1 Maths Modeling Page 38: Exercise 4.2. 1.

The sum of two numbers is 54 and their difference is 6. Find the numbers. Let one number be x and the other y

x  y  54

x y  6 x y6

x  y  54 y  6  y  54 2 y  48

The numbers are 24 and 30

y  24 x  30 2.

The sum of two numbers is 35 and their difference is 19. Find the numbers. Let one number be x and the other y

x  y  35

x  y  35 y  19  y  35 2 y  16 y8 x  27 The numbers are 8 and 27

x  y  19 x  y  19

14 3.

In a two digit number, the sum of the digits is 12 and their difference is 4. Find the number if the tens digit is larger than the units digit. Let one digit be x the other is y

x y  4

x  y  12

x y4

y  4  y  12 2y  8 The number is 84 y4 x8 . 4.

The length of a rectangle is twice the breadth, while the perimeter is 6m. Find the length and breadth of the rectangle. { Hint: P  2(l  b) .} Let the length be x and the breadth y

P  2x  2 y 6  2( 2 y )  2 y 6  6y

x  2y

y 1 x2 The length is 2 and breadth 1 5.

A number consisting of two digits has the following properties. When the number is added to twice the tens digit the answer is 33. If the digits are reversed, the number obtained exceeds the original number by 63. What is the original number? Let x represent the tens digit thus tens digit is 10x Let y represent the units digit thus units digit is

10 x  y  2 x  33 y  33  12 x

10 y  x  10 x  y  63 9 y  9 x  63 y  x7

33  12 x  x  7  13 x  26 x2 y9

The original number is 29

y

15

6.

Samantha and Warren cycle towards each other along a straight road. They start off 70km apart. Samantha cycles at 15km/h and warren at 20km/h. How far will Samantha have cycled when they meet? dis tan ce  Speed  Time

Samantha

Warren

15x

Distance = 70 km

20x

15 km/h

Speed

20 km/h

x

Time: Let time be x

x

15 x  20 x  70 35 x  70 x2 Samantha has cycled 30 km. Quadratic Inequalities: Page 42: Exercise 5.1.

x 2  8 x  15  0 .

1.

đ?’™+đ?&#x;‘ đ?’™+đ?&#x;“ > 0 Solution: đ?’™ < −3 đ?‘œđ?‘&#x; đ?‘Ľ > −5 đ?&#x;’đ?’™đ?&#x;? − đ?&#x;— < 0

2.

đ?&#x;?đ?’™ + đ?&#x;‘ đ?&#x;?đ?’™ − đ?&#x;‘ = đ?&#x;Ž đ?&#x;‘

đ?&#x;‘

đ?’”đ?’?đ?’?đ?’–đ?’•đ?’Šđ?’?đ?’?: − đ?&#x;? < đ?‘Ľ < đ?&#x;?

x2 0 x5

3.

đ?’™ < −2 đ?‘œđ?‘&#x; đ?‘Ľ > 5

2x  1 0 x4

4. đ?&#x;?

−đ?&#x;? ≤ đ?’™ < đ?&#x;’

16

 2x  5 0 3 x

5.

𝟓

𝒙 < − 𝟐 𝒐𝒓 𝒙 > 𝟑

4 x x3

6.

𝟒

𝒙

𝒙−𝟑

− >0 𝟏

𝟒−𝒙(𝒙−𝟑) 𝒙−𝟑

>0

−𝒙𝟐 +𝟑𝒙+𝟒 𝒙−𝟑

>0

𝒙−𝟒 (𝒙+𝟏) 𝒙−𝟑

<0

𝒙 < −1 𝑈 3 < 𝑥 < 4 7.

x 𝒙 𝟏

9 x 𝟗

− ≤𝟎 𝒙

𝒙𝟐 −𝟗 𝒙

≤𝟎

𝒙−𝟑 (𝒙+𝟑) 𝒙

≤𝟎

𝒙 ≤ −𝟑 𝑼 𝟎 < 𝑥 ≤ 3 8.

2 3  x2 x3

𝟐 𝟑 − 𝒙−𝟑 𝒙+𝟐

≥𝟎

𝟐𝒙−𝟔−𝟑𝒙−𝟗 𝒙+𝟐 (𝒙−𝟑)

≥𝟎

−𝒙−𝟏𝟓 𝒙+𝟐 (𝒙−𝟑)

≥𝟎

𝒙 ≤ −𝟏𝟓 𝑼 − 𝟐 < 𝑥 < 3

17

1 2  x5 x7

9.

𝟏 𝒙−𝟓

−

𝟐 𝒙+𝟕

𝒙+𝟕−𝟐𝒙+𝟏𝟎 𝒙−𝟓 (𝒙+𝟕) −𝒙+𝟏𝟕 𝒙−𝟓 (𝒙+𝟕)

>0 >0

>0

𝒙 < −7 𝑈 5 < 𝑥 < 17 Algebraic Fractions:

Page 48: Exercise 6.1.

1.

6x x  12 2

2.

4x 2 y 3 y 2  2x 8x 3 y

3.

2 x  4 2( x  2) x  2   4 4 2

4.

xy  y y( x  1)   x 1 y y

5.

8 x 2  4 x 4 x(2 x  1)   2x  1 4x 4x

6.

x 2  1 ( x  1)( x  1) x  1   ( x  1) 2 ( x  1)( x  1) x  1

7.

x 2  x  12 ( x  4)( x  3) x  4   x 2  x  12 ( x  4)( x  3) x  4

8.

a b ( a  b)   1 b  a  ( a  b)

9.

a2 (a  2) 1   a  a  2 (a  2)(a  1) a  1 2

x 2  x  12 ( x  4)( x  3) 10.   ( x  3) or  x  3 4 x  ( x  4)

18

11.

2 1 1  2  2 x  3x  2 x  x  2 x  1 2 1 1   ( x  1)( x  2) ( x  2)( x  1) ( x  1)( x  1) 2( x  1)  1( x  1)  1( x  2) ( x  1)( x  1)( x  2) 2x  2  x  1  x  2 ( x  1)( x  1)( x  2) 5 ( x  1)( x  1)( x  2)

12.    

ab  a 2 b 2  ab a(b  a) b(b  a) b(b  a) X  X  2 2 2 b a a (b  a)(b  a) a2 a(b  a)

2

x x2 13.  x  y y2  x2 . 

x x2  ( x  y ) ( x  y )( y  x)

x( x  y )  x 2 ( x  y )( x  y )

x 2  xy  x 2 ( x  y )( x  y )  xy  ( x  y )( x  y ) 

7x 3x  2x  2 y 5 y  5x 7x 3x   2( x  y ) 5( x  y ) 5(7 x)  2(3 x)  10( x  y ) 35 x  6 x  10( x  y ) 41x  10( x  y )

14.

19 Number Patterns Page 54: Exercise 7.1. 1.1 A.

1 ; 3 ; 6 ; 10; 𝟏𝟓; 𝟐𝟏

B.

2𝑎 = 1

𝑎=

1 2

1 2

2=3 1 2

+𝑏

𝑎 + 𝑏 + 𝑐 = 𝑇1 1 2

=𝑏

1 2

+ +𝑐 =1 𝑐=0

𝑇𝑛 =

1 2 𝑛 2

+

1 𝑛 2

1

C. 𝑇10 = 2 10

1

2

+ 2 10

= 55 1.2 3

3

3

3

15 18 21 24 27 A.

15 ; 30 ; 48 ; 69; 𝟗𝟑 ; 𝟏𝟐𝟎; ….

B.

2𝑎 = 3

15 = 3

3 2

21 2

𝑎=

3 2

+𝑏

𝑎 + 𝑏 + 𝑐 = 𝑇1 3 2

=𝑏

+

21 2

+ 𝑐 = 15

𝑐=3 𝑇𝑛 = C.

3 2 𝑛 2

+

21 𝑛 2

+3

3

5

𝑇10 = 2 (10)2 − 2 10 + 16 = 258

1.3

3 5

3 8 11

A.

−12 ; −7 ; 1 ; 12; … …

B.

2𝑎 = 3

5=3

3

1 2

𝑎=2

3 2

+𝑏

=𝑏

𝑎 + 𝑏 + 𝑐 = 𝑇1 3 2

1

+ 2 + 𝑐 = −12

𝑐 = −14 𝑇𝑛 = C.

3 2 𝑛 2 3

+

1 𝑛 2

− 14 1

𝑇10 = 2 (10)2 + 2 10 − 14 = 141

20

1.4

2 4

A. B.

6

27; 31 ; 37; … … 2𝑎 = 2 4=3 1 +𝑏 𝑎=1 1=𝑏

𝑎 + 𝑏 + 𝑐 = 𝑇1 1 + 1 + 𝑐 = 27 𝑐 = 25

𝑇𝑛 = 𝑛2 + 𝑛 + 25 C.

𝑇10 = (10)2 + 10 + 25 = 135

1.5

2 5

A. B.

7

12; 17 ; 24; … … 2𝑎 = 2 5=3 1 +𝑏 𝑎=1 2=𝑏

𝑎 + 𝑏 + 𝑐 = 𝑇1 1 + 2 + 𝑐 = 12 𝑐=9

𝑇𝑛 = 𝑛2 + 2𝑛 + 9 C.

𝑇10 = (10)2 + 2 10 + 9 = 129

1.6

3 9

12

A.

−13; −4 ; 8; … …

B.

2𝑎 = 3

9=3

3

9 2

𝑎=2

3 2

+𝑏

=𝑏

𝑎 + 𝑏 + 𝑐 = 𝑇1 3 2

9

+ 2 + 𝑐 = −13

𝑐 = −19 𝑇𝑛 = C.

3 2

+

9 𝑛 2

− 19

𝑇10 = (10)2 +

1.7

9 2

10 − 19 = 176

4 5

A. B.

3 2 𝑛 2

9

25; 30 ; 39; … … 2𝑎 = 4 5=3 2 +𝑏 𝑎=2 −1 = 𝑏 𝑇𝑛 = 2𝑛2 − 𝑛 + 24

C.

𝑇10 = 2(10)2 − 10 + 24 = 214

𝑎 + 𝑏 + 𝑐 = 𝑇1 2 − 1 + 𝑐 = 25 𝑐 = 24

21 1.8

4 5

A. B.

9

7; 12 ; 21; ‌ ‌ 2� = 4 �=2

5=3 2 +đ?‘? −1 = đ?‘?

đ?‘Ž + đ?‘? + đ?‘? = đ?‘‡1 2−1+đ?‘? =7 đ?‘?=6

đ?‘‡đ?‘› = 2đ?‘›2 − đ?‘› + 6 C.

đ?‘‡10 = 2(10)2 − 10 + 6 = 196

Page 56:Exercise 7.2 Write down the next 3 numbers in each of these sequences. Explain the rule in words.

1.1

5; 11; 17; 23 _____29; 35; 41____________________________________

Add 6 to each term to get the following term 1.2

1;4 ; 9; 16; 25______36; 49; 64__________________________________________

__The set of perfect squares__________ 1.3

10; 5; 0; -5 _____-10; -15; -20_____________________________________

________The sequence is decreasing by a constant of -5____________ Question 2: 2.1

1

Look at the pattern below Draw the next pattern.

2

3

4

5

22

2.2

The table below shows the shape number and the sequence of total number of black dots in each shape. Complete the table.

Shape number 1 2 3 4 8 10 N

Number of black dots 2 6 10 14 30 38

4n – 2

Hint: ‘n’ represents a general equation for the nth term of a sequence

2.3

Draw a Graph representing the data in the table in question 2.2 y 38 36 34 32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2

O

•

•

• • • • 1

2

3

4

5

6

7

8

9

10

X

23 Question 3: Look at the sequence and the table below: 2; 6; 12; 20; 30; ………….

Terms 1st 2nd 3rd 4th 5th 6th 7th 8th

3.1

Value of term 2 6 12 20 30 42 56 72

4 6 8 10 12 14 16

2 2 2 2 2 2

How were the answers in the third column obtained?

Subtract the preceding term from the anteceding one: i.e T2 – T1 etc

3.2

What type of pattern do the answers in 3.1 form: Give reasons:

A pattern that has a difference that is increasing by constant difference of 2 .i e. the second difference is 2.

3.3

What do you notice about the fourth column?

Same value . i. e. a constant difference 4. Use question 3. as a reference to complete the table below for the sequence: 5; 11; 20; 32 ; 47; 65; 86;……………

Terms st

1 2nd 3rd 4th 5th 6th 7th

Value of term 5 11 20 32 47 65 86

1st Difference 6 9 12 15 18 21

2nd Difference

3 3 3 3 3

24 Question 5: You are given a sequence where the first term is ‘a’ and the constant ratio is ‘r’. The first for terms are written down in general form as : a ; ar ; ar2 ; ar3 5.1

What is the 5th term of this sequence?

ar 4 5.2

What is the 20th term of the sequence?

ar 19 _______________________________________________________________ 5.2

Now find an equation that will give the nth term of this sequence: i.e. Tn  ..........

Tn  ar n1

Page 59:Exercise 7.3

Look at each of the following sequences and : A. Find the next 3 terms. B. Find an equation that will give the nth term of the sequence.. C. Find the 20th term. 2 2 2 2 2 2 1. A.

4 6

8

10 12

0 ; 2 ; 6 ; 12; 20 ; 30 ; 42

B 2đ?‘Ž = 2 đ?‘Ž=1

2 = 3 1 +đ?‘? đ?‘? = −1

đ?‘Ž+đ?‘?+đ?‘? = 0 1−1+đ?‘? =0 đ?‘?=0

đ?‘‡đ?‘› = đ?‘›2 − đ?‘› C. đ?‘‡20 = 202 − 20 = 380.

25

2. -3

-2

-2

-2

-2

-2

-5

-7

-9

-11

-13

A. 2 ; -1 ; -6 ; -13; -22 ; -33 ;-46; …

B. 2𝑎 = −2 𝑎 = −1

−3 = 3 −1 + 𝑏

𝑏=0

𝑎+𝑏+𝑐 = 2 −1 + 0 + 𝑐 = 2 𝑐=3

𝑇𝑛 = −𝑛2 + 3 C. 𝑇20 = −(20)2 + 3 20 = −340

3.

-6 A.

-6

5; -1 ; -7; -13 ; -19 ; -25;..

B. 𝑻𝒏 = 𝒂 + 𝒏 − 𝟏 𝒅 𝑻𝒏 = 𝟓 + 𝒏 − 𝟏 (−𝟔) 𝑻𝒏 = −𝟔𝒏 + 𝟏𝟏 C. 𝑻𝟐𝟎 = −𝟔 𝟐𝟎 + 𝟏𝟏 = −𝟏𝟎𝟗 4.

18 18 -12

6

24

18 18 42 60

A. 8; -4 ; 2.; 26 ; 68; 128;-----;

2𝑎 = 18

−12 = 3 9 + 𝑏

𝑎=9

𝑏 = −39

𝑎+𝑏+𝑐 = 8 9 − 39 + 𝑐 = 8 𝑐 = 38

𝑇𝑛 = 9𝑛2 − 39𝑛 + 38 𝑇20 = 9(20)2 − 39 20 + 38 = 2858

26 5. 2

2

5 7

2 9

2 11

13

A. 3 ; 8 ; 15; 24 ; 35 ; 48;…

B. 2𝑎 = 2

5=3 1 +𝑏

𝑎=1

𝑏=2

𝑎+𝑏+𝑐 =3 1+2+𝑐 =3 𝑐=0

2

𝑇𝑛 = 𝑛 + 2𝑛 C. 𝑇𝑛 = (20)2 + 2 20 = 440 Page 63; Exercise 7.4. 1. 𝑇10 = 𝑎 + 9𝑑 = 33 𝑇6 = 𝑎 + 5𝑑 = 17 4𝑑 = 16 𝒅=𝟒 𝑎 = 17 − 5 4 𝒂 = −𝟑 2. 𝑥 ; 2𝑥 + 1 ; 11 2.1

2𝑥 + 1 − 𝑥 = 11 − 2𝑥 − 1 3𝑥 = 9 𝑥=3

𝑇1 = 3 𝑇2 = 7

𝑇3 = 11

27 �30 = 3 + 29 4 = 11

2.2 3. đ?‘Ž = −3

�3 = 3

đ?‘‡3 = đ?‘Ž + 2đ?‘‘ = 3 −3 + 2đ?‘‘ = 3 đ?‘‘=3 3.1

đ?‘‡25 = −3 + 24(3) = 69

3.2

đ?‘‡đ?‘› = đ?‘Ž + (đ?‘› − 1)đ?‘‘

57 = −3 + (đ?‘› − 1)3 57 = −3 + 3đ?‘› − 3 3đ?‘› = 63 đ?‘› = 21 đ?’•đ?’‰đ?’† đ?&#x;?đ?&#x;?đ?’”đ?’• đ?’Šđ?’” đ?&#x;“đ?&#x;• Geometric Progressions: Page 67: Exercise 7.5. 1. 2 ; 6 ; 18 ; 54 ; ‌. is a geometric sequence. 1.1 Continue the sequence to the 6th term. 162 ; 486.

1.2

Find the 10th term. �10 = 2 3 10 = 118098

2. Determine the 2nd and 3rd terms in the following sequence given that đ?‘‡1 = 5 đ?‘Žđ?‘›đ?‘‘ đ?‘‡4 = 40 . đ?‘‡4 = đ?‘Žđ?‘&#x; 3 = 40 5(đ?‘&#x;)3 = 40 đ?‘&#x;3 = 8 đ?‘&#x;=2 đ?‘‡2 = 10 đ?‘Žđ?‘›đ?‘‘ đ?‘‡3 = 20

28 3. In the following geometric sequences: 3.1 𝑎 = 2 𝑎𝑛𝑑 𝑟 = 3, 𝑓𝑖𝑛𝑑 𝑡𝑕𝑒 5𝑡𝑕 𝑡𝑒𝑟𝑚. 𝑇5 = 𝑎𝑟 4 = 2(3)4 = 162

3.2

1

𝑎 = 1 𝑎𝑛𝑑 𝑟 = 2 , 𝑓𝑖𝑛𝑑 𝑡𝑕𝑒 6𝑡𝑕 𝑡𝑒𝑟𝑚. 1

1

𝑇6 = 𝑎𝑟 5 = 1(2)5 = 32

3.3

1

𝑎 = 11 𝑎𝑛𝑑 𝑟 = − 3 , 𝑓𝑖𝑛𝑑 𝑡𝑕𝑒 4𝑡𝑕 𝑡𝑒𝑟𝑚.

𝑇4 = 𝑎𝑟 3 = 11 −

1 3 3

=−

11 27

4. In a geometric progression of which: 4.1 𝑡𝑕𝑒 6𝑡𝑕 𝑡𝑒𝑟𝑚 𝑖𝑠 96 𝑎𝑛𝑑 𝑎 = 3, 𝑓𝑖𝑛𝑑 𝑟. 𝑇6 = 𝑎𝑟 5 = 96 3𝑟 5 = 96 𝑟 5 = 32 𝑟=2

4.2

𝑡𝑕𝑒 5𝑡𝑕 𝑡𝑒𝑟𝑚 𝑖𝑠 4

𝑇5 = 𝑎𝑟 =

7 81

7 81

𝑎𝑛𝑑 𝑎 = 7. 𝑓𝑖𝑛𝑑 𝑟.

7

7𝑟 4 = 81 1

𝑟 4 = 81 1

𝑟=3

4.3

𝑡𝑕𝑒 7𝑡𝑕 𝑡𝑒𝑟𝑚 𝑖𝑠 192 𝑎𝑛𝑑 𝑡𝑕𝑒 2𝑛𝑑 𝑖𝑠 6, 𝑓𝑖𝑛𝑑 𝑡𝑕𝑒 𝑓𝑖𝑟𝑠𝑡 3 𝑡𝑒𝑟𝑚𝑠. 𝑇7 = 𝑎𝑟 6 = 192 𝑇2 = 𝑎𝑟 = 6 𝑟 5 = 32 𝑟=2 𝑎=3 𝐺𝑃: 3 ; 6 ; 12; ….

4.4

6

𝑡𝑕𝑒 6𝑡𝑕 𝑡𝑒𝑟𝑚 𝑖𝑠 − 32 𝑎𝑛𝑑 𝑡𝑕𝑒 9𝑡𝑕 𝑡𝑒𝑟𝑚 𝑖𝑠 6

𝑇9 = 𝑎𝑟 8 = 256 𝑇6 = 𝑎𝑟 5 = −

6 32

6 , 𝑓𝑖𝑛𝑑 256

𝑡𝑕𝑒 𝑓𝑖𝑟𝑠𝑡 3 𝑡𝑒𝑟𝑚𝑠

29 5. In a geometric progression with first 3 terms: đ?‘˜ − 4; đ?‘˜ + 2; 3đ?‘˜ + 1 5.1 find 2 possible values of k. đ?‘˜+2 đ?‘˜âˆ’4

=

3đ?‘˜+1 đ?‘˜+2 2

(đ?‘˜ + 2) = đ?‘˜ − 4 (3đ?‘˜ + 1) đ?‘˜ 2 + 4đ?‘˜ + 4 = 3đ?‘˜ 2 − 11đ?‘˜ − 4 2đ?‘˜ 2 − 15đ?‘˜ − 8 = 0 2đ?‘˜ + 1 đ?‘˜ − 8 = 0 1

đ?‘˜ = 8 đ?‘œđ?‘&#x; đ?‘˜ = − 2

5.2

find 2 possible values for first term: đ?‘“đ?‘œđ?‘&#x; đ?‘˜ = 8 đ?‘Ąđ?‘•đ?‘’đ?‘› đ?‘‡1 = 4 1 2

đ?‘“đ?‘œđ?‘&#x; đ?‘˜ = − đ?‘Ąđ?‘•đ?‘’đ?‘› đ?‘‡1 = −4,5

5.3 �10 = 4

Find the 10th term. 5 9 2

= 15258,79 1 9

đ?‘‡10 = −4,5 − 2

= 8,79

Page 71: Exercise 7.6 1.

Consider the series of numbers below. You must assume that the number of terms listed will be sufficient to conclude a rule for each sequence. Series A: 1 ; 4 ; 9 ; 16 ; 25 ; 36 ; 49 ; ------------------Series B: 1.1

1;

1 1 1 1 1 ; ; ; ; ;-------------------2 3 4 5 6

State your observations concerning the behaviour of series A:

It is the set of perfect squares 1.2

State your observations concerning the behaviour of series B

It is the set of fractions where the denominator is the set of natural numbers. 1.3

What will the tenth term ( 10th Term) in series A be?

100 1.4

What will the 20th term be in series B?

1 20

30 1.5 If ‘k’ represents the position of a term ( eg. Position 1 will be represents the position of a term ( eg. Position 1 will be k = 1) Write down the general rule , in an equation form , for series A and B, respectively. Series A:

Tk= k2

Series B:

Tk =

đ?&#x;? đ?’Œ

Find the sum ( addition) of the first 4 terms in series A.

30 Find the sum of the first 3 terms in series B. đ?&#x;?đ?&#x;? đ?&#x;”

2.1 Complete the table.

2.2

Terms

Value of term

1st

-5

2nd

2

7

3rd

11

9

2

4th

22

11

2

5th

35

13

2

What can you conclude about the second difference? It is constant

2.3

What can you deduce about all the values that you calculated in the column denoted “first difference�? They increase by 2

2.4 Determine the 10th term in the sequence ( the number occupying position ten of the sequence) 130

31 Financial Mathematics: Page 80: Exercise 8.1. 1. 𝐴 = 𝑃(1 + 𝑖)𝑛 4

𝐴 = 1000 1.115

+ 2000 1.115

3

+ 4000 1.115

2

+ 8000 1.115

1

𝐴 = 𝑅18210,90 10

2. 𝐴 = 6500 1.11

+ 7400 1.11

7

− 5800 1.11 5 𝐴 = 𝑅24046.48

3. 𝐴 = 21000 1.075 3 (1.0825)4 = 𝑅35822,53

4.1 𝐴 = 8000(1.14)14 − 15000(1.14)3 − 15000 = 𝑅12867.63 4.2 Yes. She has fulfilled her commitment to her sons as promised.

5

5. 𝐴 = 52000 1.105

3

1.12

4

1.14

= 𝑅203276,98

Page 84. Exercise 8.2 1. 𝐴 = 3000000 1 +

0.06 12×8 12

− 500000 1 +

0.06 12×5 12

+ 120000 1 +

𝑨 = 𝑹𝟒𝟑𝟏𝟏𝟔𝟎𝟒, 𝟕𝟐 . 2.1 2.1

𝐴 = 100000 1 + 𝑟 = 100

15

691888 ,66 100000

3.1 𝐴 = 200000 1 +

3.2

𝑟 = 100

20

0.10 24 4

0.15 108 12

= 𝑹𝟔𝟗𝟏𝟖𝟖𝟖, 𝟔𝟔

− 1 = 𝟏𝟑. 𝟖%

0.08 12×14 12

1241420 .79 200000

4. 𝐴 = 150000 1 +

1+

1+

0.12 4×6 4

= 𝑹𝟏𝟐𝟒𝟏𝟒𝟐𝟎, 𝟕𝟗

− 1 = 𝟗, 𝟔%

0.12 2×2 2

1+

0.15 5×4 4

1+

0.20 3×12 12

5.1

𝐴 = 1250000 1 − 9 × 0.08 = 𝑹𝟑𝟓𝟎𝟎𝟎𝟎.

5.2

𝐴 = 1250000(1 − 0.08)9 = 𝑹𝟓𝟗𝟎𝟐𝟎𝟏, 𝟕𝟎

= 𝑹𝟕𝟏𝟔𝟗𝟕𝟖, 𝟎𝟕

0.06 12×3 12

32 6.

0.085 12×10 + 15000 12 0.085 12×2 = 𝑹𝟓𝟗𝟔𝟔𝟎, 𝟗𝟔 12

𝐴 = 20000 1 + 5000 1 +

1+

0.085 12×8 12

7. Use either : Nominal to Effective: 𝑱𝒆𝒇𝒇 = 𝟏𝟎𝟎

𝟏+

Effective to Nominal: 𝑱𝒎 = 𝟏𝟎𝟎𝒎 OR 𝟏 +

𝒊𝒎 𝒎 𝒎

𝒎

− 16000 1 +

𝑱𝒎 𝒎 𝒎

0.085 12×4 12

+

−𝟏

𝟏 + 𝑱𝒆𝒇𝒇 − 𝟏

=𝟏+𝒊

𝐍𝐁 ∶ 𝒊𝒎 𝒊𝒔 𝒕𝒉𝒆 𝒏𝒐𝒎𝒊𝒏𝒂𝒍 𝒓𝒂𝒕𝒆 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅 𝒎 𝒕𝒊𝒎𝒆𝒔 𝒊𝒏 𝒐𝒏𝒆 𝒚𝒆𝒂𝒓

8.

7.1 100

1+

7.2 100

1+

0.155 12 12 0.324 2

− 1 = 35,02%

2

𝐽𝑚 = 100(12)

12

− 1 = 16,6%

1 + 0.325 − 1 = 28,5%

9. 𝐴 = 1500000 1 +

0.09 12×10 12

− 500000 1 +

0.09 12×8 12

+ 550000 1 +

= 𝑅3439847,94 9.

Functions & Graphs

Page 92: Exercise 9.1 1.

𝐴𝑡 𝑦 = 0: 𝑥 2 − 3𝑥 − 4 = 0 𝑥−4 𝑥+1 = 0 𝑥 = 4 𝑜𝑟 𝑥 = −1

𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = −4 𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑃𝑜𝑖𝑛𝑡 = (1,5 ; −6,25)

2

f(x) = x2 -3x-4

-10

-5

5

-2

-4

-6

10

0.09 12×4 12

33

𝐴𝑡 𝑦 = 0: 𝑥 2 − 4𝑥 − 5 = 0 𝑥−5 𝑥+1 = 0 𝑥 = 5 𝑜𝑟 𝑥 = −1

2.

𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = −5 𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑃𝑜𝑖𝑛𝑡 = (1,5 ; −6,25) 2

f(x) = x2 - 2x - 5 -10

-5

5

10

-2

-4

-6

𝐴𝑡 𝑦 = 0: 𝑥 2 + 𝑥 − 6 = 0 𝑥+6 𝑥−1 = 0 𝑥 = −6 𝑜𝑟 𝑥 = 1

3.

𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = −6 𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑃𝑜𝑖𝑛𝑡 = ( ; −6,25)

2

f(x) = x2 + x - 6 -10

-5

5

-2

-4

-6

10

34 4.

𝐴𝑡 𝑦 = 0: 𝑥 2 + 3𝑥 − 10 = 0 𝑥−2 𝑥+5 = 0 𝑥 = −5 𝑜𝑟 𝑥 = 2

𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = −10 𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑃𝑜𝑖𝑛𝑡 = (1,5 ; −6,25)

f(x) = x2 +3 x - 10 -15

-10

-5

5

10

15

-2

-4

-6

-8

-10

-12

-14

5.

𝐴𝑡 𝑦 = 0: 𝑥 2 − 2𝑥 − 8 = 0 𝑥−4 𝑥+2 = 0 𝑥 = 4 𝑜𝑟 𝑥 = −2

𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = −8 𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑃𝑜𝑖𝑛𝑡 = (1,5 ; −6,25)

4

2

f(x) = x2 - 2x - 8 -15

-10

-5

5

10

15

-2

-4

-6

-8

-10

𝐴𝑡 𝑦 = 0: 𝑥 2 + 4𝑥 − 12 = 0 𝑥−2 𝑥+6 = 0 𝑥 = 2 𝑜𝑟 𝑥 = −6

6.

𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = −12 𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑃𝑜𝑖𝑛𝑡 = (−2; −16)

2

-20

-15

-10

-5

f(x) = x2 + 4x - 12 5

-2

-4

-6

-8

-10

-12

-14

-16

10

15

20

35 7.

𝐴𝑡 𝑦 = 0: 2𝑥 2 − 7𝑥 + 6 = 0 2𝑥 − 3 𝑥 − 2 = 0 𝑥 = 1,5 𝑜𝑟 𝑥 = 2

𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = 6 𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑃𝑜𝑖𝑛𝑡 = (1.75 ; −0,125)

8

6

4

f(x) = 2x2 - 7x + 6

2

-10

8.

-5

5

𝐴𝑡 𝑦 = 0: 2𝑥 2 + 5𝑥 − 3 = 0 2𝑥 − 1 𝑥 + 3 = 0 𝑥 = 0,5 𝑜𝑟 𝑥 = −3

𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = −3 𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑃𝑜𝑖𝑛𝑡 = (1,5 ; −6,25)

f(x) = 2x2 + 5x - 3

2

-10

10

-5

5

10

-2

-4

-6

Page 101: Exercise 9.2 1. −2𝑥 2 − 𝑥 + 1

1 2

−2 𝑥 + 4

9

+ 16

1

𝑝𝑞 − 4 ;

9 8

𝑥 = −1 𝑜𝑟 𝑥 = 1,5

2

-10

-5

5

-2

f(x) = -(2x-1)(x+1) -4

-6

10

36

2. 𝑥 2 − 2𝑥 − 3 𝑥−1 𝑥 = 3 𝑜𝑟 𝑥 = −1

2

− 4 𝑝𝑞 1 ; −4

4

3

2

f(x) = x 2 -2x - 3

1

-8

-6

-4

-2

2

4

6

8

10

12

-1

-2

-3

-4

𝑥 2 − 6𝑥 + 8

3.

𝑥−3

2

−1 𝑝𝑞 3 ; −1

𝑥 = 2 𝑜𝑟 𝑥 = 4 7

6

5

4

3

2

f(x) = x 2 -6x+8

1

-6

-4

-2

2

4

6

8

10

12

-1

5 2

𝑥 2 + 5𝑥 + 6

4.

𝑥−2

1

−4

𝑝𝑞

5 2

1

; −4

𝑥 = −2 𝑜𝑟 𝑥 = −3 7

6

5

4

f(x) = x 2 + 5x + 6

3

2

1

-10

-8

-6

-4

-2

2 -1

4

6

8

14

37

5.

2𝑥 2 − 4𝑥 + 6 2 𝑥 − 1 2 + 2 𝑇𝑕𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑕𝑎𝑠 𝑛𝑜 𝑟𝑜𝑜𝑡𝑠

𝑝𝑞 1 ; 2

16

14

12

10

8

f(x) =2 x 2 - 4x + 6

6

4

2

-15

-10

-5

−𝑥 2 + 2𝑥 + 3 𝑥 = −1 𝑜𝑟 𝑥 = 3 f x  = -3x +2x+1

6.

5

− 𝑥−1

2

−4

10

15

20

25

𝑝𝑞 1 ; −4

2

4

3

2

gx  = -x 2+2x+3

1

-8

-6

-4

-2

2

4

6

8

-1

-2

7.

1 2

−3𝑥 2 − 2𝑥 + 1

−3 𝑥 + 3

4

+3

1

𝑝𝑞 − 3 ;

4 3

1

𝑥 = −1 𝑜𝑟 𝑥 = 3 f x  = -3x 2+2x+1

2

1

-8

-6

-4

-2

2

-1

-2

-3

-4

gx  = -3x 2-2x +1

4

6

8

38 Page 108: Exercise 9.3 1 2

9

−2 𝑥 + 4

1.1

1

+ 16 𝑝 = − 4 1

1

𝑝′ = − 4 − 4 = −4 4 1 2

9

.𝑦 = −2 𝑥 + 4 4

+ 16

𝒚 = −𝟐𝒙𝟐 − 𝟏𝟕𝒙 −

𝟓𝟔𝟗 𝟏𝟔

𝑥−1

1.2

2

−4

𝑝=1 𝑝′ = 1 − 4 = −3

2

𝒚= 𝑥+3

−4

𝒚 = 𝒙𝟐 + 𝟔𝒙 + 𝟓 2.1

𝑦 = 𝑥−3

2

−1

𝑝=3 𝑝′ = 3 + 5 = 8

𝑦 = 𝑥−8

2

−1

𝑦 = 𝑥 2 − 16𝑥 + 63 𝑦 =− 𝑥−1

2.2

2

−4 𝑝 =1 𝑝′ = 1 + 5 = 6

𝑦 =− 𝑥−6

2

−4

𝑦 = −𝑥 2 + 12𝑥 − 40 Page 110: Exercise 9.4 1.1 𝑦 = 𝑥 − 1

2

−4

𝑞 = −4 𝑞 ′ = −4 + 3 = −1

𝑦 = 𝑥−1 2 −1 𝒚 = 𝒙𝟐 − 𝟐𝒙

1.2.

𝑦 = 𝑥−3

2

−1

𝑞 = −1 𝑞 ′ = −1 + 3 = 2

𝑦 = 𝑥−3

2

+2

𝒚 = 𝒙𝟐 − 𝟔𝒙 + 𝟏𝟏

39 2.1

𝑦 =2 𝑥−1

2

+2

𝑞=2 𝑞 ′ = 2 − 3 = −1

𝑦 =2 𝑥−1

2

−1

𝒚 = 𝟐𝒙𝟐 − 𝟒𝒙 + 𝟏 2.2

𝑦 =− 𝑥−1

2

−4

𝑞 = −4 𝑞 ′ = −4 − 3 = −7

𝑦 =− 𝑥−1

2

−7

𝒚 = −𝒙𝟐 + 𝟐𝒙 − 𝟖 Page 112: Exercise 9.5 1. 𝑦 = 𝑎(𝑥 − 𝑝)2 + 𝑞 𝑦 = 𝑎(𝑥 − 2)2 + 10 2 = 𝑎(0 − 2)2 + 10 4𝑎 = −8 𝑎 = −2 𝑦 = −2(𝑥 − 2)2 + 10 2. 𝑦 = 𝑎(𝑥 − 𝑝)2 + 𝑞 𝑦 = 𝑎(𝑥 + 1)2 + 5 13 = 𝑎(1 + 1)2 + 5 4𝑎 = 8 𝑎=2 𝑦 = 2(𝑥 + 1)2 + 5 3. 𝑦 = 𝑎(𝑥 − 𝑝)2 + 𝑞 𝑦 = 𝑎(𝑥 + 4)2 − 1 2 = 𝑎(−3 + 4)2 − 1 𝑎=3 𝑦 = 3(𝑥 + 4)2 − 1

40 4. 𝑦 = 𝑎 𝑥 − 𝑥1 (𝑥 − 𝑥2 ) 𝑦 = 𝑎 𝑥 − 1 (𝑥 + 3) −4 = 𝑎 −1 − 1 (−1 + 3) −4𝑎 = −4 𝑎=1 𝑦 = 𝑥 − 1 (𝑥 + 3) 𝒚 = 𝒙𝟐 + 𝟐𝒙 − 𝟑 5. 𝑦 = 𝑎 𝑥 − 𝑥1 (𝑥 − 𝑥2 ) 𝑦 = 𝑎 𝑥 − 2 (𝑥 + 4) −14 = 𝑎 3 − 2 (3 + 4) 7𝑎 = −14 𝑎 = −2 𝑦 = −2 𝑥 − 2 (𝑥 + 4) 𝒚 = 𝟐𝒙𝟐 − 𝟒𝒙 + 𝟏𝟔

6. 𝑦 = 𝑎 𝑥 − 𝑥1 (𝑥 − 𝑥2 ) 𝑦 = 𝑎 𝑥 − 1 (𝑥 − 5) −5 = 𝑎 0 − 1 (0 − 5) 5𝑎 = −5 𝑎 = −1 𝑦 = − 𝑥 − 1 (𝑥 − 5) 𝒚 = 𝒙𝟐 − 𝟔𝒙 + 𝟓

41 Page 115: Exercise 9.6 1.1 f x  = -3x 2+2x+1

12

gx  = -x 2+x+12

10

8

6

hx  = -3x+12 4

2

-10

-5

5

10

15

1.2 𝟎 ; 𝟏𝟐

𝒂𝒏𝒅 (𝟒 ; 𝟎)

2.1

𝑦 = (𝑥 + 1)2 − 9

2.2

𝑡𝑢𝑟𝑛𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 (−1 ; −9)

2.3

𝑥 = −4 𝑜𝑟 𝑥 = 2

2.4

𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = (0 ; −8)

2.5 gx  = x 2+2x -8 2

-15

-10

-5

5

-2

-4

-6

-8

10

42

2.6 𝑦 = (𝑥 + 1)2 − 9 𝑝 = −1 𝑝′ = −1 − 5 = −6 𝒚 = (𝒙 + 𝟔)𝟐 − 𝟗 or 𝒚 = 𝒙𝟐 + 𝟏𝟐𝒙+25

𝑦 = −𝑥 2 + 4𝑥 + 5

3.1

𝑦 = −(𝑥 − 2)2 + 9

3.2 10

8

6

4

gx  = -x 2+4x+5 2

-10

-5

5

10

15

-2

3.3 10

8

6

qx  = x-5

4

gx  = -x 2+4x+5

2

-10

-5

5

-2

-4

-6

-8

3.3.1

−2 ; 7

𝑎𝑛𝑑 (5 ; 0)

10

15

20

25

43 𝑞=9

3.4

𝑞′ = 9 − 4 = 5 𝒚 = −(𝒙 − 𝟐)𝟐 + 𝟓

or

𝒚 = −𝒙𝟐 + 𝟒𝒙 + 𝟏

Page 121: Exercise 9.7 1. 8

6

4

1.1

gx  = 3x+2 2

f x  = 3x

1. -15

-10

-5

5

10

15

10

15

-2

1.2

hx  = 3x-4 -4

-6

2.

1.

𝑦=0

1.1 1.2

𝑦=2 𝑦 = −4

3. 8

6

4

3.1

gx  = 4-x+3

2

f x  = 4-x

3 -15

-10

-5

5

-2

-4

-6

-8

4. 𝑦 = 0 ; 𝑦 = 3 𝑎𝑛𝑑 𝑦 = −4

3.2

hx  = 4-x-4

44 Page 124: Exercise 9.8 1./ 1.1/ 1.2 10

8

6

hx  = 2x-4

f x  = 2x

gx  = 2x+4

4

2

-15

-10

-5

5

10

15

-2

-4

𝐴𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒 𝑖𝑠 𝑦 = 0 𝑓𝑜𝑟 𝑎𝑙𝑙 3 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠.

2. 3./ 3.1 / 3.2

10

8

6

hx  = 2-x+4 

f x  = 2-x

4

gx  = 2-x-4

2

-15

-10

-5

5

-2

-4

4.

𝐴𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒 𝑓𝑜𝑟 𝑎𝑙𝑙 3 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 𝑖𝑠 𝑦 = 0.

10

15

45 5. And 5.1 10

8

6

4

qx  = 3x-4+3

2

rx  = 3x -15

-10

-5

5

10

15

-2

-4

Page 130: Exercise 9.9 1. X Y

-6 -1

-3 -2

-2 -3

-1 -6

1 6

2 3

3 2

6 1

8

6

4

s x  =

6 x

2

-15

-10

-5

5

-2

-4

-6

-8

1.1 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑜𝑓 𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒𝑠 𝑎𝑟𝑒: 𝑦 = 0 & 𝑥 = 0.

10

15

46 1.2 8

6

4

s x  =

4 x-3

2

-15

-10

-5

5

10

15

-2

-4

-6

-8

1.3

𝐴𝑦𝑚𝑝𝑡𝑜𝑡𝑒𝑠 𝑎𝑟𝑒 𝑦 = 0 𝑎𝑛𝑑 𝑥 = 3

2. x y

-6 1

-3 2

-2 3

-1 6

1 -6

2 -3

3 -2

8

6

4

s x  =

-15

-10

2

6 -x

-5

5

-2

-4

-6

-8

2.1

𝐴𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒𝑠 𝑎𝑟𝑒 𝑦 = 0 & 𝑥 = 0

10

15

6 -1

47 2.2 8

6

4

s x  =

-15

6

2

-x+3

-10

-5

5

10

15

-2

-4

-6

-8

2.3 𝐴𝑦𝑚𝑝𝑡𝑜𝑡𝑒𝑠 𝑎𝑟𝑒 𝑦 = 0 & 𝑥 = −3 3. x y

-8 -1

-4 -2

-2 -4

-1 -8

1 8

2 4

4 2

8 1

8

6

4

s x  =

8 x

2

-15

-10

-5

5

-2

-4

-6

-8

3.1 𝐴𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒𝑠 𝑎𝑟𝑒 𝑦 = 0 𝑎𝑛𝑑 𝑥 = 0.

10

15

48 3.2 8

6

4

2

-15

-10

s x  =

8 x+2

-1

-5

5

-2

-4

-6

-8

3.3

𝐴𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒𝑠 𝑎𝑟𝑒 𝑥 = −2 𝑎𝑛𝑑 𝑦 = −1

Page 138: Exercise 9.10 1. t x  = sin x 

y

1

0

-1

1.1 1.2 1.3 1.4

360° 𝑦𝜖[−1 ; 1] 1 45 and 225

180

360

x ux  = cos x 

10

15

49

2.

y 2

ux  = cos x 

1

t x  = 2ďƒ—sinx 

180ď‚°

0ď‚°

360ď‚°

-1

-2

2.1 2.2 2.3 2.4

360ď‚° đ?‘Śđ?œ–[−2 ; 2] 2 26,5ď‚° or 206,5ď‚°

3.

y

ux  = cos x 

1

t x  = sin2ďƒ—x 

0ď‚°

-1

3.1 3.2

180ď‚° đ?‘Śđ?œ–[−1 ; 1]

3.3

1

3.4

90ď‚° ; 150ď‚° & 270ď‚°

180ď‚°

360ď‚°

x

x

50

4. y 2

ux  = cos 2ďƒ—x 

1

t x  = -sin x 

180ď‚°

0ď‚°

-1

-2

4.1

360ď‚°

4.2

đ?‘Śđ?œ–[−1 ; 1]

4.3

1

4.4

90ď‚° ; 210ď‚° &

330ď‚°

360ď‚°

x

51 11.Probability Theory Page 143: Exercise11.1 1.

A bag contains 4 yellow counters and 8 white counters. Calculate the Probability that:

1.1

The first counter drawn at random is:

1.1.1

yellow: Answer =

1.1.2

1.2

1.3

1 3 2 white. Answer = 3 For a second draw you get white if first counter was white if:

2 3

1.2.1

was replaced Answer =

1.2.2

was not replaced before second draw was made. Answer =

7 11

On drawing two(2) counters, the first is yellow and the second is white, if the first counter: 1.3.1 1.3.2

was replaced. Answer =

1 2 2   3 3 9

Was not replaced before the second draw. Answer =

1 8 8   3 11 33

2.1 Answer: James choice in pants is independent of his choice of shirt as his choice of pants does not in any way affect which shirt he chooses. 2.2 Answer: White shirt(WS)

White T-shirt(WT)

Black shirt (BS)

Green T-shirt (GT)

Grey pants(G)

G,WS

G,WT

G,BS

G,GT

Black pants(B)

B,WS

B,WT

B,BS

B,GT

Blue jeans(BJ)

BJ,WS

BJ,WT

BJ,BS

BJ, GT

2.2.1

Answer: 12

2.2.2

Answer:

2.2.3

1 4 6 1 Answer: or 12 2

52 2.3

Draw a tree diagram to show the probability of each choice that James has. WS

(G,WS)

1 12

1 4 1

WT

(G,WT)

1 4

1

1

1 12

4

4

3

1 BS

(G,BS)

GT

(G,GT)

12 1 12

WS

(G,WS)

1

1 12

4

1

1

3

4

WT

(G,WT)

1

BS

4

1

1 12

4

GT

1 (G,BS) (G,GT)

12 1 12 1

WS 1

(G,WS)

12

1

3

4

WT

1

(G,WT)

4 1 1

1 BS

(G,BS)

GT

(G,GT)

4

4

1 12

12 1 12

2.3.1

What is the probability that James chooses his black pants? Answer =

2.3.2

What is the probability that James chooses his black shirt? Answer =

2.3.3

1 4

What is the probability that James is dressed all in black? Answer =

2.3.4

1 3

1 12

What is the probability that James is not wearing any black at all?

Answer =

6 1 OR 12 2

53

Page 150: Exercise 11.2: 1. 1.1 1.2 1.3 1.4

2.

Are these two events Dependent or Independent? Answer : independent Answer : Dependent Answer : Dependent Answer : independent

Jack has a bag of 5 red, 7 blue, 8 white and 6 green marbles. He represents this sample space using the venn diagram below:

2.1.1 Nick chooses First. He chooses red, then white, then green.

P(R, then W, then G) =

2.1.2

Vusi chooses second. He chooses green, then green, then green.

Vusi’s 1st Choice B R 4 7 W 7

5 8 6 240 or 0,015    26 25 24 15600

G 5

Vusi’s 2nd B R Choice 4 7 W 7

G 4

Vusi’s 3rd B R Choice 5 7 W 7

G 3

54 2.2 There are now only 20 marbles left in the bag: 4 red ,7 blue, 7 white and 2 green. 2.2.1

P(R, then R, then R)=

4 3 2 24     0,004 20 19 18 6840 R 2 18 7

B

18

R

7

3

18

19 7

B

19

18

G

7

R

19

4

W

2

W

2

20

B

19

G

7 20

7 20

W

2 20

G

2.2.2

P(R, then W, then W)=

4 7 6 168     0,025 20 19 18 6840

R

B R 3

6 18

19 7

W

B

19

G

7

R

19

4

W

2

20

B

W

G

19

G

55 3.1.1

Blake’s 1st choice Grey

Black

2

2 White 1

3.1.2 3.1.3

Answer: There are 4 socks in the sample space for Blake’s second choice. Answer: The events are dependent as Blake is not replacing the socks after each choice. His 2nd choice is therefore dependent on his 1st choice.

3.2 Draw a tree diagram showing Blake’s sock choices, then answer the questions that follow: First Ch oice

Secon d Ch oice

Event

Probability 2

G

W, G

2

20

4 2

2 B

W, B

W

G, W

4

W

20

2 2 5

20

1 4

2

1

2

G

G

G, G

20

4

5

2 4 G, B

B

4 20

2 5 W

B, W

2 20

1 4 4

2 B

4

G

B, G

20

1 2

4 B

B, B

20

56 3.2.1 3.2.2

Answer: There are 8 outcomes altogether Answer: Two of these outcomes have 2 socks the same.

3.2.3

Answer: The probability that Blake chooses 2 socks the same is

3.2.4

Answer: The probability that Blake goes to school with odd socks on is

16  0,8 20

Analytical Geometry

Page 5: Exercise 1.1 : 1.

Find the distance between the given pairs of points: 1.1 (2 ; 3) and (4 ; 5)

Dis tan ce  ( x1  x2 ) 2  ( y1  y 2 ) 2 Dis tan ce  (2  4) 2  (3  5) 2 Dis tan ce  8 Dis tan ce  2,82 1.2 (6 ; 1) and ( -6 ;6)

Dis tan ce  (6  6) 2  (1  6) 2 Dis tan ce  144  25 Dis tan ce  13 1.3

(3 ; -7) and (-1 ; 3)

Dis tan ce  (3  1) 2  (7  3) 2 Dis tan ce  16  100 Dis tan ce  10,77 1.4 (-4 ; 3) and (0 ; 0)

Dis tan ce  (4  0) 2  (3  0) 2 Dis tan ce  16  9 Dis tan ce  5 1.5

(-2 ; 1) and -4 ; -1)

Dis tan ce  (2  4) 2  (1  1) 2 Dis tan ce  8 Dis tan ce  2,82

4  0,2 20

57

1.6

(-3 ;-1) and (4 ; -6)

Dis tan ce  (3  4) 2  (1  6) 2 Dis tan ce  49  25 Dis tan ce  8,6 2. 2.1

Given the coordinates of the vertices of ABC , in each case ( 2.2 to 2.5) A(1 ; -3) ; B(7 ; 3); C(4 ; 6) AB  36  36  72

BC  9  9  18

Perimeter = 22,2

AC  9  81  90 AB 2  BC 2  AC 2 Right angled Scalene triangle

2.2

A(5 ;1) ; B(1 ; 3) ; C(1 ; -2)

AB  4 2  (2) 2  20 BC  0  25  5

Perimeter = 14,5

AC  16  9  5 Isosceles Triangle Not Right angled 2.3

A(-2 ; -3) ; B(-4 ; 1) ; C(4 ; 5) AB  4  16  20

BC  64  16  80

Perimeter = 23,4

AC  36  64  100 Scalene Triangle

AB 2  BC 2  AC 2 Right angled 2.4

A(0 ; 0) B( 3 ; 1) ; C( 3 ; -1)

AB  3  1  2 BC  0  4  2

Perimeter = 6

AC  3  1  2 Equilateral Triangle Not Right angled

58

3. 3.1

Show that: A(-3 ; 2) , B(3 ;6), C(9 ;-2) and D(3 ; -6) are vertices of a parallelogram.

AB  36  16  52 DC  36  16  52 AD  36  64  10 BC  36  64  10

AB = DC

AD = BC

ABCD is a parallelogram ( Both prs opp sides equal) 3.2

(6 ;-4) , (5 ;3) (-2 ; 2) and (-1 ; -5) are vertices of a square. AB  1  49  50

DC  1  49  50

AD  49  1  50 BC  49  1  50 AB = DC = AD = AB ABCD is a square ( All sides are equal) Mid-points of lines: 4.

Calculate the coordinates of the midpoints of the line joining the following points: 4.1 (-3 ;1) and (1 ; 5)  31 x  1 2 Mid point ( -1 ; 3) 1 5 y 3 2 4.2 (-2 ; 3) and (6 ; 3)

26 2 2 3 3 y 3 2

x

Mid – point (2 ; 3)

59

4.3 (4 ; -1) and (-1 ; 3)

4 1 3  2 2 1 3 y 1 2

x

Mid point ( 1,5 ; 1)

4.4 (0 ;0 ) and (3 ; -8)

03 3  2 2 08 y  4 2

x

Mid Point ( 1,5 ; -4)

4.5 ( 3;1) and (3 3;1) 33 3 2 3 2 Mid point ( 2 3 ; 0) 11 y 0 2

x

5.

Determine the values of x and y if: 5.1 (-3 ; 2) is the mid-point of the line joining (-1 ; 5) and (x ; y). 1 x 5 y  3 2 2 2  1  x  6 5 y  4 x  5 y  1

5.2

(-1 ; y) is the mid-point of the line joining (0 ; -2) and x ; 8) 0 x  1 28 2 y x  2 2 y3

5.3

(x ; y) is the centre of a circle on diameter AB where A(-2 ; -1) and B(-1 ; 9).

 2 1 x 2 2 x  3 x

3 2

1 9 y 2 2y  8 y4

60

5.4

(x ; 3) is the centre of a circle with diameter MN. M (5 ; -2) and N(-7 ; y) 57 x 2 x  1

6.

2 y 3 2 2 y  6 y8

Calculate the lengths of the medians of ABC in which the coordinates of the vertices are as follows: A(-3 ;1), B(-5 ; -3) and C(1 ; -5). (NB: a median is the line from a vertex drawn to the mid-point of the side opposite the vertex) Mid Point BC = (-2 ; -4) Length = 1  25  5,1 Mid Point AC = (-1 ; -2) Length = 16  1  4,1 Mid Point AB = (-4 ; -1) Length = 25  16  41 =6,4 Page 170:The gradients and Inclinations of straight lines:

7.

Calculate the gradients of the lines joining the following points: 1 7.1 (-3 ; 2) and (1 ; 1) m   4 5 7.2 (4 ; 3 ) and (-1 ; 8) m    1 5 8 7.3 (-3 ; -5) and (1 : 3) m   2 4

8.

Write down the gradients of the lines perpendicular to the lines in 7. m4 m 1 1 m 2

9.

Calculate the inclinations of the line AB in each of the following cases. 9.1 A(-3 ; 2) and B(-5 ; 0) 2 m  1 2

Inclinatio n  45 9.2

A(-2 ; 1 ) and B(1 ; -2) 3 m  1 3

Inclinatio n  135

61

9.3

A( 3 ; 1) and B((2 3 ; -2) 3 m  1,732 3

Inclinatio n  120  9.4

A(-1 ;2) and B(1 ; -1) 3 m  1,5 2

Inclinatio n  123,7  9.5

A(-5 ; 2) and B(3 ; -1) 3 m    0,375 8

Inclinatio n  159,4  10.

Calculate the gradients of lines with inclinations of: 10.1 45º m  1 10.2 60º m  1,732 10.3 150º m  0,577 10.4 110º m  2,75 11. Calculate the gradients of the following lines and state whether they are A. Parallel B. Perpendicular C. Neither. 11.1 A(0 ; -1) , B(-4 ; -2) , C( -3 ; 1) and D ( 1 ; 2) 1 1 mAB   4 4 mBC 

1 4

Lines are parallel

11.2

A(6 ; -10) , B(0 ; 4) , C( 6 ; 0) and D ( -4 ; -3)  14 7 mAB   6 3 Neither 3 mCD  10

11.3

A(-3 ; 5) , B(5 ; -1) , C( -2 ; -1) and D ( 1 ; 3) 6 3 mAB   8 4 m  m  1 1 2 4 mCD  3 Lines are perpendicular

62

11.4

A(-2 ; -4) , B(3 ; 1) , C( 5 ; -1) and D ( -2 ; -8) 5 mAB   1 5 Lines are parallel 7 mCD   1 7

12. Show that the following points are collinear: ( lie on the same line) 12.1 A(-2 ; -6) , B(2 ; -4) , C( 4 ; -3) 2 1 mAB   4 2 A; B & C are collinear. 1 mBC  2 12.2

A(-5 ; 5) , B(1 ; 1) , C( 4 ; -1) 4 2 mAB   6 3 A ; B ; C are collinear. 2 mBC  3 Equations of straight lines:

13.

Determine the equaition of a line where the gradients and a point on the line are given as follows: 1 13.1 ; (2;3) 2 y  y1  m( x  x1 ) y3 y

13.2

3 ; (3;1) 2

1  x  2 2

1 x4 2

y  y1  m( x  x1 ) 3  x  3 2 3 9 y 1 x  2 2 3 11 y  x 2 2 y 1

63

13.3

 2; (1;3)

y  y1  m( x  x1 ) y  3  2 x  1 y  3  2 x  2 y  2 x  1 14.

Lines passing through the following points: 14.1 (-2 ; 4) and (2 ; 2) y m x

2 1  4 2 y  y1  m( x  x1 )

m

y4 

1 x  2 2

1 y   x3 2

14.2

(-1 ; 1) and (1 ; 5) 4 2 2 y  y1  m( x  x1 )

m

y  5  2 x  1 y  2x  3

14.3

(-3 ; -2) and (-1 ; -1) 1 m 2 y  y1  m( x  x1 )

1  x  3 2 1 1 y  x 2 2 y2

14.4

(3 ; -3) and (3 ; -6) 3  0 x3

m

64

15.

A line with the slope of 3 and intersecting the y – axes at 2 y  3x  2

16.

parallel to y  3x  2 and passing through (3 ; 1) y  3x  c 1  3(3)  c c  8 y  3x  8

17.

Through (-2 -1) and perpendicular to 3 y  2 x  6 2 3 y  2 x  6 m   3 2 3 y   x2 m 3 2

3 xc 2 3  1  (  2)  c 2 c2 3 y  x2 2 y

18.

Through (-1 ; 3 ) and an inclination of 120º m 3 y  y1  m( x  x1 ) y  3   3x  3 y  1,732 x  4,732

19.

A(-2 ; 1), B(3 ; 3) and C(6 ; -3) are the vertices of a triangle . Determine: 19.1

The coordinates of M, the mid-point of AC. M(2 ; -1)

19.2

the gradient of AC. m

4 1  8 2

65

19.3

the equation of the perpendicular bisector of AC.

m2 y  y1  m( x  x1 ) y  1  2( x  2) y  2x  5 19.4

the equation of the median BM

mBM  4 y  3  4( x  3) y  4x  9 19.5

the equation of the altitude from B to AC.

mAltitude  2 y  3  2( x  3) y  2x  3

Circle centre the origin: 20.

Determine the equation of a circle with centre origin and: 20.1

radius = 3 cm

x2  y2  r 2 r3 r2  9 x2  y2  9 20.2

radius = 3 2 cm x2  y2  r 2 r3 3 r 2  27 x 2  y 2  27

66

20.3

Passing through point (-2 ; 3)

x2  y2  r 2 (2) 2  (3) 2  r 2 4  9  r2 x 2  y 2  13

20.4

Passing through point ( -4 ; -2)

x2  y2  r 2 ( 4) 2  ( 2 ) 2  r 2 16  4  r 2 x 2  y 2  20 21

A(-3 ; 4) is a point on a circle with centre at the origin: 21.1 Determine the equation of the circle.

x2  y2  r 2 (3) 2  (4) 2  r 2 9  16  r 2 x 2  y 2  25 21.2

Determine the coordinates of B if AB is a diameter. (3 ; -4) By symmetry

21.3 Show that the point C(0 ; 5) lies on the circle.

x2  y2  r 2 (0) 2  (5) 2 r 2  25 Thus (0 ; 5) lies on the circle 21.4

Prove that ACˆ B is a right angle. 1 mAC  3 9 m 2 BC    3 3 m1  m2  1 ACˆ B  90 

67

Tangents to circle centre origin: Determine: 22.1

The gradient of OB. 4 3

mOB 

22.2

The equation of AC, the tangent. mTangent  

3 4

y  y1  m( x  x1 ) 3 y  4   ( x  3) 4 3 25 y  x 4 4

22.3

The equation of the circle centre O.

x2  y2  r 2 9  16  r 2 r 2  25 x 2  y 2  25 Page 24: Exercise 1.2 1.1

the length of AD.

AD  [8  (6)]2  [3  (2)]2 AD  29 AD  5,4 1.2

the mid-point of DC

  6 1  2  4 ;   2   2  5    ; 3   2 

68

1.3

The gradient of BC

m 1.4

y 1  (4) 5   x 11 2

The length of BC

BC  [1  (1)]2  [1  (4)]2 BC  29 BC  5,39 1.5

the inclination of BC

5 2 tan x  2,5

mBC  

x  111,8  1.6

the equation of BC

y  y1  m( x  x1 ) 5 y  1   ( x  1) 2 y  1  2,5 x  2,5 y  2,5 x  1,5 or 2 y  3x  2 1.7

The perimeter of ABCD

AB  49  4

AD  BC  29

AB  53

2 AD  2 29

2 AB  2 53 Perimeter  2 53  2 29  25,33 1.8

State what shape is represented in the diagram Paralellogram

69 QUESTION 2: 2.1

Show that A(-5 ; -3); B(-1 ; 0) and C(3 ; 3) lie on the same straight line.

mAB 

3 4

mBC 

3 4

A ; B & C are collinear points

2.2

P(13 ; t) , Q(7 ; 2) and R(4 ; 1) are points in a Cartesian plane. If P , Q and R are collinear, then determine the value of t.

t2 mPQ  6 1 mQR  3

t2 1  6 3 3t  6  6 3t  12 t4

QUESTION 3: 3.

Three points A 5;2 ; B(2;3) and C  3;2 in a Cartesian plane are given. 3.1 Calculate the distance AB.

AB  [5  2]2  [2  3]2 AB  50 AB  7,1 3.2 Calculate the gradient of AC.

mAC 

4  2 2

3.3 Calculate the value of t if the point D(t ; t- 3) is such that AC // BD.

mAC  2 mBD 

t6 t2

t6 t2 t  6  4  2t 2

3t  10 t  3 13 3.4 Calculate the mid-point of BC.

 23 3 2 mid  ptBC   ;  2   2  1 1 mid  po int BC    ;   2 2

70 3.5 Determine, by calculation , whether the quadrilateral is a parallelogram.

mAB 

1 7

mCD 

2 13 6

7 18

AB is not parallel to CD it is not a parallelogram. 3.6

Give the equation of BC.

mBC 

5 1 5

y  y1  m( x  x1 ) y  3  1( x  2) y  x 1 3.7

What is the size of the angle of inclination of BC with the positive x – axes. Tan x = 1 x = 45º

3.8

Calculate the size of ACˆ B .

mAC  2 tan ACˆ x  116.6  ACˆ B  116.6  45 ACˆ B  71,6

71 Trigonometry: Page 30: Exercise 2.1

X -15 (pyth)

Question:

X

X

-8 17

If sin x  

8 and 90 < x < 270 17

x  15( pyth )

Find without the use of a calculator: 1.1.1

cos x 

 15 17 2

1.1.2

 8   15  sin x + cos x =      17   17  2

2

2

64 225  289 289 289  289 1 

_X -_-3 (pyth)

1.2

1.3

3 If cos A  , and A  [180 ;360 ] Determine then value of: 5

4 3 7   5 5 5

1.2.1

sin A  cos A 

1.2.2

tan A cos A 

1.2.3

cosAsinA =

1.2.4

sin A  4 5 4    cos A 5 3 3

4 3 4   3 5 5

 3  4 12   4 5 25

If cos 40  a determine the following in terms of a:

sin 40 1 a2 = cos 40 a

1.3.1

tan 40 =

1.3.2

cos 220 = -cos 40º =-a

1.3.3

sin 40

sin 2 40  1  cos 2 40  1  a 2 sin 40  1  a 2

_--4

_X

_X _5

72

2.

Using reduction function:

Page 35: Exercise 2.2:

2.1

cos(180   x). sin(90   x). tan(180   x)  cos x  cos x  tan x  cos x  (1) cos(360   x). sin 270  cos x sin x   1 cos x  sin x sin(180   x) s tan(360   x) sin(90   x)

2.2

sin(180   x) cos(90   x) tan(360   x) cos x  sin x 1  tan x

( sin x)( tan x)(cos x) (sin x)( sin x)( tan x)

2.3

sin(360   x) sin(90   x)(cos 90   x) 

tan(180  x)( sin x) cos(180  x)

( sin x)(cos x)(sin x) sin x cos x    cos x ( tan x)( sin x)( cos x) 1 sin x

2.4

   cos(360   ). sin(90   ). tan(180   ) (cos  )(cos  )(  tan  ) 2   cos    (tan  )(  1 ) tan(180   ). sin 270

3.

Using Special Angles: Specific angle sizes given :

Page 37: Exercise 2.3:

 3 3 3   4  2 

1  2

3.1

3sin30º tan45º cos30º = 3 1 

3.2

  sin30º cos30º tan60º =    2  2  1  4

 1  3  3 

3

 3 3  1 2  4    2  2   2 3 1 2 3.3

4sin60º + tan45º + 2cos30º =  2 3  1  3

  1 3 3

3 2

73 2

2

3.4

3 1  1  1 4sin 45º - 3 sin 30º = 4   3   2   1 4 4  2  2

3.5

2  3 3 1 9  1   cos 0º +cos 30º + sin 45º = 1    2    2  =1  4  2  4    

3.6

cos 30º tan 2 45º + tan 2 30º + tan 0º

2

2

2

2

2

2

2

3 2  1    0  (1)   2  3

3 1  2 3 3 32  6 

3.7

1

1 2 sin 60   cos 30  tan 2 30   sin 45 tan 2 60  2 3 3 

2

3 1 3  1  2 1          3 3 2 2 2  3 2

1

1 3  1 2 12 18  3  12 

3.8

1 1   cos 40   2 cos 60 . cos 40 . tan 330 1 3  = 1     3 sin 210 sin 50 . sin 270  cos 40 .(1) 2

3.9

       sin 45 (  cos 45 )  (  cos 30 ) sin 315 . cos 135  cos 210      tan 45 .(1) tan 135 . sin 270

  

1

(

2

1

)  (

2  1.(1)

1 3 2

3 ) 2

 3 2

74

3.10

   sin 315 . cos 20 . sin 240    tan 135 . sin 70 cos 180

    sin 45 . cos 20 .( sin 60 )    tan 45 . cos 20 ( 1)

1

2

  1  . cos 20 ( 1)

3 2 2

2.4. Reduction Function: Page 40: Exercise 2.4

4.1

sin 230 = -sin 50º

4.2

cos150 = - cos 30º

4.3

tan125 = - tan 65º

4.4

cos230 = - cos 50º

4.5

tan320 = -tan 40º

4.6

sin145 = sin 35º

4.7

cos340 = cos 20º

4.8

sin290 = -sin 70º

4.9

tan 225 = tan 45º

4.10 sin 100 = sin 80º 4.11 cos105 = -cos 75º 4.12

tan185 = tan 5º

2.5. Calculator Work: Page 41: Exercise 2.5 5.1 3 sin 120  2,60

5.2

2 tan135 cos 240 

  cos 20  . 

4

3 2

75 5.3

2 tan125 cos 150  sin 139 

 3,77

5.4 If x  25,7  and y  137,4  calculate the value of the following:

sin 2 x cos 2 y  0,15 3 tan y

Solution of trig equations: Page 42: Exercise 2.6

2 sin x  0,545 6.1

sin x  0,2725 KeyL  15,8 x  195,8 or 344,2º

6.2

2 cos x  0,147 3 cos x  0,2205 KeyL  77,3

x  77,3 or 282,7º

3 tan x  6,605

6.3

tan x  2.2016.... KeyL  65.6 x  114.4 or x  294.4

76

1 cos x  0,245  0 2 cos x  0,490 6.4

KeyL  60.7 x  60.7 or x  299.3

tan x  8,213  0

6.5

tan x  8,213 KeyL  83.1 x  96.9 or x  276.9

6.6

3 sin x  0,369 2 sin x  0.246 KeyL  14.2 x  14.2 or x  165.8

 2 sin x  0,546  0

6.7

sin x  0.273 KeyL  15.8 x  195.8 or x  344.2

77

2 tan x  8,442  0 3 tan x  12.663 6.8

KeyL  85.5 x  94.5 or x  274.5

4 sin x  3,208  0 sin x  0.802 6.9

KeyL  53.3 x  53.3 or x  126.7

2 sin( x  20  )  1,636 sin( x  20)  0.818

6.10

Key ( x  20)  54.9 x  20  180  54.9 x  20  234.9 x  254.9 or x  20  360  54.9 x  20  305.1 x  325.1

2 cos( x  30  )  0,262 3 cos( x  20)  0.393 Key ( x  30)  66.9 6.11

x  30  66.9 x  96.9 or x  30  360  66.9 x  30  293.1 x  323.1

78

2 sin( x  25)  0,345 sin( x  25)  0,1725

6.12

KeyL ( x  25)  9,9  x  25  9,9  x  34,9 

( x  25)  170,1

or

x  195,1

3 tan( x  75 )  6,147 tan( x  75 )  2,049

6.13

KeyL ( x  75 )  64 

( x  75 )  116  x  41

( x  75 )  296 

or

x  221

 2 cos( x  15)  1,605 cos( x  15)  0,8025

6.14

KeyL ( x  15)  36,6  x  15  143,4  x  126,4 

or

x  15  216,6  x  201,6 

2 sin( x  30)  0,445  0 3 sin( x  30)  0,6675

6.15

KeyL ( x  30)  41,9  x  30  41,9  x  71,9 

6.16

or

x  30  138,1 x  168,1

3 tan( x  54)  21,213 4 tan( x  54)  28,284 KeyL ( x  54)  88

x  54  88 x  142 

or

x  54  266  x  322 

79 2.7.

General Solution in trig equations

Page 48: Exercise 2.7 7.1

tan 2 x  2,6 Key (2 x)  69 ď Ż

2 x  (180 ď Ż  69 ď Ż )  k .180 ď Ż x  55,5ď Ż  k .90 ď Ż

2 x  (360 ď Ż  69 ď Ż )  k .180 ď Ż

OR

x  145,5ď Ż  k .90 ď Ż

2 cos x  0,66  0 7.2

cos x  0,33 KeyL  70,7 ď Ż

đ?‘Ľ = 109,3° + đ?‘˜. 360° OR đ?‘Ľ = 250,7° + đ?‘˜. 360° 7.2.2 đ?‘Ľ = 109,3° đ?‘œđ?‘&#x; 250,7° đ?‘œđ?‘&#x; − 109,3°

7.3.1

1 sin x  0,825 2 1 KeyL ( x)  55,6 ď Ż 2

1 x  55,6 ď Ż  k .360 ď Ż 2 or ď Ż ď Ż x  111,2  k .720 7.3.2

1 x  (180 ď Ż  55,6 ď Ż )  k .360 ď Ż 2 1 x  124.4 ď Ż  k .360 ď Ż 2 x  248,8ď Ż  k .720 ď Ż

Find the value(s) of x above if đ?’™ ∈ [−đ?&#x;?đ?&#x;•đ?&#x;Ž°; đ?&#x;?đ?&#x;•đ?&#x;Ž°] in 7.3.1 đ?‘Ľ = 111.2° đ?‘œđ?‘&#x; 248,8°

7.4 Find the general solution for the following:

tan 3x  4,302 KeyL (3x)  76,9 ď Ż 3x  103,1ď Ż  k .180 ď Ż x  34,4 ď Ż  k .60 ď Ż

or

3x  283,1ď Ż  k .180 ď Ż x  94,4 ď Ż  k .60 ď Ż

80 7.5

Find the general solution for the following:

2 cos(2 x  60  )  0,684 cos(2 x  60  )  0,342 KeyL (2 x  60  )  70 

2 x  60   70   k .360  2 x  130   k .360 

2 x  60   290   k .360  or

x  65  k .180  7.6

2 x  350   k .360  x  175  k .180 

Find the specific solutions for 7.6 if x  [360 ;360 ]

If k = 0 then x = 65º or 175º If k = 1 then x = 245º or 355º If k = -1 then x = -115º or -5º If k = -2 then x = -295º or -185 7.7

Find the general solution for the following:

4 sin(3x  120  )  2,812 sin(3x  120  )  0,703 KeyL (3x  120  )  44,7 

3x  120   224,7   k .360  3x  344,7   k .360 

3x  120   315,3  k .360  or 3 x  435,3  k .360 

x  114,9   k .120 

7.8

x  145,1  k .120 

Find the specific solutions for 7.8 if x  [180 ;360 ]

If k = 0 then x = 114,9º or 145,1º If k = 1 then x = 234,9º or 265,1º If k = 2 then x = 354,9º If k = -1 then x = -5,1º or 25,1º If k = -2 then x = -125,1º or -94,9º 2. 8.

Use the fundamental identities to simplify the following:

81

Page 52: Exercise 2.8 sin x sin x sin x cos x     cos x tan x sin x 1 sin x cos x

8.1

8.2

1-sin2x = cos2x

8.3

1-cos2x = sin2x

8.4

sin 2 x  1  cos 2 x = sin 2 x  (1  cos 2 x)  sin 2 x  sin 2 x  0

8.5

cos x cos x )  sin x(sin x  ) sin x tan x cos x  sin x cos 2 x      1  sin x   1  sin x  sin x  1 2 2   sin x  sin x  cos x   sin x  

8.6

sin x cos 2 x  sin 3 x = sin x(cos 2 x  sin 2 x)  sin x

sin x(sin x 

2.9.

Proving Fundamental Identities:

Page 53: Exercise 2.9:

9.1

cos x 1 1   2 tan x cos x sin x

LHS 

cos x 1  sin x cos 2 x cos x

cos 2 x 1  sin x cos 2 x 1  sin x  RHS 

82

(1  cos 2 x) 

1 2

 cos 2 x

tan x 9.2

LHS 

2

sin x cos 2 x  1 sin 2 x

 cos 2 x  RHS 1  1  cos x   sin x  sin x  tan x  cos x LHS  cos x   sin x sin x 9.3

cos 2 x sin x   sin x 1 cos 2 x  sin 2 x sin x 1  sin x  RHS

sin 3 x  cos 3 y  (sin x  cos x)(1  sin x cos x) 9.4

LHS  (sin x  cos x)(sin 2 x  sin x cos x  cos 2 x)  (sin x  cos x(1  sin x cos x) (sin x  cos x) 2  1  2 sin x cos x

9.5

LHS  sin 2 x  2 sin x cos x  cos 2 x  1  2 sin x cos x (1  sin 2 x)(1  tan 2 x)  1 LHS  cos 2 x(1 

9.6

.  cos 2 x  sin 2 x

1  RHS

sin 2 x cos 2 x

)

83

1 cos x

cos x(1  tan 2 x 

 sin 2 x  LHS  cos x 1   cos 2 x    cos x 

9.7

sin 2 x cos x

cos 2 x  sin 2 x cos x 1  cos x 

2.10

If sin 20 = k , then represent the following in terms of k:

Page 57: Exercise 2.10 10.5.1

sin 160 = sin20º = k

1 10.5.2

cos70 = sin20 = k

10.5.3

cos 20 =

10.5.4

tan 20 =

k

1 k 2

sin 20  cos 20 

2.11

20 00

k

1 k 2

1 k 2

Solution of Triangles:

Page 59: Exercise 2.11 Calculate the area of: 11.1.1

ABC in which BC = 7cm; AC = 6cm and Cˆ  27,6 . AreaABC 

1 (7)(6) sin 27,6  = 9,73 cm2 2

84 11.1.2

EFG in which EG = 29cm ; EF = 54cm and Eˆ  61,4 AreaEFG 

11.1.3

FGH in which Hˆ  61,4 ; GH = 9,5cm and FH = 2,3cm AreaFGH 

11.2

1 (29)(54) sin 61,4   687,5 cm2 2

1 (9,5)(2,3) sin 61,4   9,6 cm2 2

Calculate the area of a parallelogram in which two adjacent sides measure 100mm and 120mm and the angle between them is 65º.

AreaABC 

1 (100)(120) sin 65  5437,8 2

Area of parallelogram = 2 x 5437,8 = 10875,7 cm2 11.3 If the area of XYZ is 3000m2 and x  80m and y  150m , calculate two possible sizes of Zˆ

1 (80)(150) sin Zˆ 2 sin Zˆ  0,5 Zˆ  30  3000 

2.12.

Sine Rule:

Page 62: Exercise 2.12: Solve the following triangles: 12.1.1

ABC in which Aˆ  30 ; Bˆ  45 and a  2

2 AC  sin 30 sin 45

AC 

2 sin 45  2,83 sin 30

85

2 sin 105 Cˆ  105 ( L sum triangle) AB   3,86 sin 30

12.1.2

PQR in which Pˆ  115 ; Qˆ  20  and q  15,3

15,3 QR 15,3 sin 115  QR   40,54 sin 20 sin 115 sin 20 15,3 sin 45 Rˆ  45 ( L sum triangle) PQ   31,6 sin 20

12.1.3

XYZ in which Yˆ  64  ; Zˆ  21 and x  30

30 XZ 30 sin 64  XZ   27,1 sin 95 sin 64 sin 95 30 sin 21 Xˆ  95 ( L sum triangle) XY   10,8 sin 95 12.1.4

ABC in which ; Bˆ  50,1 ; Cˆ  72,3 and AC  5,34

5,34 AB 5,34 sin 72,3  AB   6,63 sin 50,1 sin 72,3 sin 50,1 5,34 sin 57,6 Aˆ  57,6 ( L sum triangle) BC   5,9 sin 50,1

In PQR , Pˆ  123 ; Qˆ  20  and PQ = 57,3m. Calculate the length

12.2 of RQ.

Rˆ  37  ( L sum triangle)

57,3 QR 57,3 sin 123  QR   79,9 sin 37 sin 123 sin 37

86 In KMS , Kˆ  30,7  ; Sˆ  19,1 and KM = 4,2m. Calculate the

12.3

length of KS.

Mˆ  130,2 ( L sum triangle)

4,2 KS 4,2 sin 130,2  KS   9,8 sin 19,1 sin 130,2 sin 19,1

Cosine Rule:

2.13. Page 65: Exercise 2.13:

Solve the following triangles:

ABC in which Aˆ  60 ; AB = 5cm and AC = 8cm

13.1

7 5  sin 60 sin C

BC 2  49 BC  7 13.2

2

2

3 2 sin 135  0,6 5

Pˆ  36,9   Rˆ  8,1 ( Lsum )

GHK in which GH = 8cm ; HK = 9cm and GK = 10cm

8 2  10 2  9 2 cos G  2(8)(10) 9 8  cos G  0,51875 sin 58,8 sin K Gˆ  58,8 13.4

sin P 

5 3 2  sin 135 sin P

PR 2  25 PR  5

13.3

Cˆ  38,2   Bˆ  81,8 ( Lsum )

PQR in which Qˆ  135 ;QR = 3 2 and PQ

PR  (3 2 )  1  2(3 2 )(1) cos 135 2

5 sin 60  0,6186 7

sin C 

BC 2  5 2  8 2  2(5)(8) cos 60 

sin K 

8 sin 58,8  0,7603 9

Kˆ  49,5  Hˆ  71,7  ( Lsum )

LMN in which LM = 7cm ; MN = 13cm and NL = 8cm

87

8 2  7 2  13 2 2(8)(7) 13 7  cos L  0,5 sin 120 sin N Lˆ  120 

sin L 

cos L 

7 sin 120  0,466 13

Lˆ  27,8  Mˆ  32,2  ( Lsum )

A

13.5

20 m

D

120

ABCD is a trapezium in which: 30

ˆ B  120 AD = 20 m. BC = 30 m, DA

B

30 m

and ABˆ D  30 . AD // BC. Show that : 13.5.1

BD = 20 3m

BD 2  20 2  20 2  2(20)(20) cos 120  BD 2  1200 BD  1200  12  100 BD  20 3 13.5.2

DC = 10 3m

DC 2  (20 3 ) 2  30 2  2(20 3 )(30) cos 30  DC 2  300 DC  300  3  100 DC  10 3

13.6

In the following sketch AC  BC  x and Cˆ  y 

C y x

A

x

B

C

88 13.5.3 Show that the distance AB  x 2(1  cos y) and hence,

AB 2  x 2  x 2  2( x)( x) cos y  AB 2  2 x 2  2 x 2 cos y  AB 2  2 x 2 (1  cos y  ) AB  x 2(1  cos y  ) 13.5.4 Calculate AB if x  150m and y  112 

AB  x 2(1  cos y  ) AB  150 2(1  cos 112  ) AB  248,7

2.14

Two and 3 dimensional Problems in trigonometry:

Page 69: Exercise 2.14 14.1

In the diagram QP = 10,28 cm PR = 5,73 cm and Qˆ  32 

Calculate Pˆ

sin R 

10,28 sin 32  0,9507 5,73

5,73 10,28 ˆ R  71,9   sin 32 sin R Pˆ  76,1 ( Lsum )

14.2 In the diagram AB = 5cm , AC = 4cm and BC = 6cm A

4cm 5cm

C 6cm B

89

Calculate all the angles of ABC

52  4 2  6 2 2(5)(4) 6 5  cos A  0,125 sin 82,8 sin C Aˆ  82,8 cos A 

sin C 

5 sin 82,8  0,82676 6

Cˆ  55,8  Bˆ  41,4  ( Lsum )

14.3 In the figure PQR is right agled at Q: RS = 10cm, Rˆ  40  and PSˆQ  68 Calculate: 14.3.1 PS

10 PS  sin 28 sin 40 10 sin 40 PS  sin 28 PS  13,69

14,3,2

PQ 13,69 PQ  13,69 sin 68

sin 68 

PQ  12,69

14.4 In the diagram A B & C lie on the same horizontal plane . HC is a vertical

H

height equal to 100m. 100m

Calculate:  ACB=104,5

26,5

14.4.1

AB

C

A 21,8 P B

90

BC 

100 100  250 and AC   200,6 tan 21,8 tan 26,5

AB 2  200,62  2502  2200,6250 cos 104,5 AB 2  127853,47 AB  357,6 11.4.1 CP

AreaABC 

1 (200,6)250  sin 104,5 2

Area  24276,3m 2 1 (357,6)(CP ) 2 2(24276,3) CP  357,6 Area 

CP  135,8 14.5

In the figure B and D are points in the same horizontal plane as C, the foot of a vertical tower AC.

A

ˆ C  90   AC = BD = x and BD

x

C 

B

Show that: BC  x

1 sin 2 

 2 cos 

D

91

CD 

x tan 

BC 2  BD 2  CD 2  2( BD )(CD ) cos BDˆ C 2

 x   x  BC  x     2( x)  cos   tan    tan   2

2

 x 2 cos 2    2 x 2 cos 2     BC 2  x 2    sin 2    sin        cos 2  2 cos 2   BC 2  x 2  1    sin   sin 2  

sin 2   cos 2   2 cos 2  sin 2 

 1  2 cos 2    BC 2  x 2   sin 2     BC  x

3

1 sin  2

 2 cos 2 

Data Handling:

Page 76: Exercise 3.1: 10A 97766666655421

10B 0 1

89 0344446789

10C 0 1

257 2224444444

92

64322

1.1

2 3

0034789 0

2 3

01146

Calculate the mean for each class. Answers (10A = 17,5); (10B = 18,4); (10C = 14,6)

1.2

Calculate the mode for each class. 10A = 16 ; 10B = 14 ; 10C = 14

1.3

Calculate the median for each class. 10A = 16 ; 10B = 17,5 ; 10C = 14

1.4

Calculate the range for each class. 10A : 26-11 =15 ; 10B : 30-8 = 22 10 C : 26-2 = 24

1.5

Calculate the lower quartile (Q1) for each class. 10A = 15 ; 10B =14 ; 10C = 12

1.6

Calculate the upper quartile (Q3) for each class. 10A = 22 ; 10B =23,5 ; 10C = 20

1.7

Calculate the inter-quartile range for each class. 10A = 7 ; 10B =9,5 ; 10C = 8

1.8

Calculate the semi-quartile range for each class

10A = 3,5 ; 10B =9,54,75 ; 10C = 4

Page 90: Exercise 3.2: 1. The number of points scored by four (4) Formula One racing drivers over a number of races is in the table below A B C D

1 1 1 2 1.1

1 2 1 2

1 6 2 2

2 8 2 4

6 8 4 4

6 8 4 6

8 8 6 6

8 8 6 8

Calculate the mean for each of the drivers. A : Mean = 6.1 B : Mean = 7.5 C: Mean = 4.7 D: Mean = 6

1.2

List the Five Number Summary for each driver. A: {1; 1,5; 8; 9; 10} B: { 1; 7 ; 8 ; 9 ; 10} C: { 1; 2 ; 4 ; 8 ; 10}

8 8 8 8

8 10 8 10

10 10 10 10

10 10 10

10 -

93

D: { 2; 3 ; 6 ; 9 ; 10} 1.3

Calculate the difference between the mean and median for each driver. A: mean – median = -1,9 B: mean – median = -0,5 C: mean – median = 0,7 D: mean – median = 0

1.4

Draw a Box and Whisker plot for each driver.

A: 1

1,5

8

1

7

8

1

2

4

2

3

9

10

B: 9

10

C: 8

10

9

10

D:

1.5

6

Discus each drivers distribution of scores in terms of the spread about the

median and mean. A: Mean – Median < 0 the data is negatively skewed(skewed to the left) B: Mean –Median < 0 the data is negatively skewed(skewed to the left) C: Mean – Median > 0 the data is positively skewed(skewed to the right)

94

D Mean – Median = 0 Data is symmetric 1.6

Compare the performance results for each driver by using the information Obtained. above.

2. The following set of data records the number of chocolates sold by a convenience store over a period of 44 days.

2.1

2.2

9

30

23

13

26

29

38

16

13

27

42

36

50

19

37

11

31

51

36

14

22

23

42

25

23

33

24

21

33

32

22

13

6

8

29

23

45

23

17

15

12

5

46

32

Draw a Stem and Leaf Plot to organize the data.

0

5689

1

1113345679

2

1223333345679 9

3

01223366 78

4

2256

5

01

List the Five Number Summary. { 5 ; 15,5 ; 23,5 ; 33; 51}

2.3

Draw a Box and Whisker Plot.

2.4

5 2.5

15,5

23,5

33

51

Draw the cumulative frequency graph (Ogive Curve) using the Box and Whisker plot as a starting point.

95

1

2

3

4

5

5

6

7

51

15,5

23,5

33

3. The following table (grouped frequency distribution) shows the mark obtained by 220 learners in a Science exam. % 1-10 11-20 21-30 31-40 41-50 51-60 61-70 71-80 81-90 91-100 Frequency 2 6 11 22 39 59 45 20 11 5 3.1

Copy & Complete the cumulative frequency table below for this data:

Marks 1-10 11-20 21-30 31-40 41-50 51-60 61-70 71-80 81-90 91-100 Total

3.2

200 180 160

165

5.5 14.5 25.5 35.5 45.5 55.5 65.5 75.5 85.5 95.5

Frequency 2 6 11 22 39 59 45 20 11 5 220

Cumulative Frequency 2 8 19 41 80 139 184 204 215 220 1112

On a set of axes draw the cumulative frequency graph ( Ogive Curve) for the data.

220

Class midpoint

96

3.3 Determine the lower quartile (Q1). Answer = 43,5

3

3.4

Determine the median. Answer = 50,5

3.5

Determine the upper quartile (Q3). Answer = 67,5

The following is list of heights of learners in a class. Heights are in centimeters(cm) 152

153

147

151

138

181

159

149

155

153

167

180

132

157

151

142

183

168

150

145

145

đ?‘„1

đ?‘€

đ?‘„3

132 138 142 145 145(đ?&#x;?đ?&#x;’đ?&#x;”) 147 149 150 151 151 đ?&#x;?đ?&#x;“đ?&#x;? 153 153 155 157 159(đ?&#x;?đ?&#x;”đ?&#x;‘) 167 168 180 181 183

4.1 Determine: 4.1.1

median = 152

4.1.2

arithmetic mean = 155

4.1.3

standard deviation = 13,7

97

4.1.4

4.2

first and third quartiles. Q1 = 146 Q3 = 163

Draw a box-and-whisker diagram for the data set.

132

183 146

4

152

Copy and complete the table below: Interval

Frequency

Class midpoint

0  x  10 10  x  20 20  x  30 30  x  40 40  x  50 50  x  60

7 11 22 25 10 0

4,5 14,5 24,5 34,5 44,5 54,5

5.1 4

3.5

3

2.5

2

1.5

1

0.5

0

10

20

30

40

50

5.2 đ??¸đ?‘ đ?‘Ąđ?‘šđ?‘Žđ?‘Ąđ?‘’đ?‘‘ đ?‘šđ?‘’đ?‘Žđ?‘› âˆś

đ?‘“.đ?‘‹ đ?‘›

= 27,2

đ?‘†đ?‘Ąđ?‘Žđ?‘›đ?‘‘đ?‘Žđ?‘&#x;đ?‘‘ đ??ˇđ?‘’đ?‘Łđ?‘–đ?‘Žđ?‘Ąđ?‘–đ?‘œđ?‘› = 11,5

5.3

163

60

Cumulative frequency 7 18 40 65 75 75

98 4

X

3.5

Q2

3

Median is closer to 30

2.5

2

1.5

1

0.5

0

5

10

20

30

40

50

60

The ages to nearest year of 27 members of a Cricket Club are : 17 21 23 19 27 18 20 21 21 24 30 25 19 22 27 35 20 30 27 21 23

28 18

31 27

18 22

6.1 Organise the ages using a stem & leaf diagram.

1

788899

2

0011112233 4577778

3

0015

6.2 Using 5 classes, and starting at 16, construct a frequency table and histogram for the data. Min = 17 Q1 = 20 M = 22 Q3 = 27 Max = 35 Interval

Frequency

Class midpoint

15  x  20 20  x  25 25  x  30 30  x  35

6 11 6 4

17,5 22,5 27,5 32,5

6.3

12 10 8

6

Cumulative frequency 6 17 23 27

99

6.3 Use the histogram to construct a frequency polygon on the same set of axes.

12 10 8

6 4 2

0

15

20

25

30

35

40

6.4 Describe the shape of the frequency polygon. Use the shape to predict the relation ship between the median and mean. Polygon is skewed to the right or positively skewed thus x  m

Q2 is situated to the left of the shape. 6.5 6.5.1 6.5.2 6.5.3 7.

Find: Median = 22 An estimate of the mean using grouped data. 22,5 Mean – median. 0,5 The following table shows the prices , correct to nearest rand, of second-hand VW Golfs for sale in Car Finder . a magazine listing second-hand cars for sale. Prices in Rands Frequency Cumulative Frequency 0 - 19 999 8 8 20 000 – 39 999 19 27 40 000 – 59 999 45 72 60 000 – 79 999 25 97 80 000 – 99 999 23 120 100 000 – 119 999 9 129

100

120 000 – 139 999 140 000 – 159 999

11 1

140 141

7.1 Find estimates of: 7.1.1 The mean and standard deviation of the prices Mean = 69857.5 Std Dev = 29129.83 7.1.2

The median price. 40000 to 59999,5 ( 49999,5)

7.2.1 Calculate mean – median. = 19858 7.2.2 Use the answer to predict whether the distribution is symmetrical, positively skewed or negatively skewed. The data is positively skewed because x  m 7.2.3

Draw a histogram to illustrate the data.

45 40 35 30 25

20 15 10 5

0

19

20

40

60

80

100

120

140

X (1000)

7.3.2

Describe the shape of the histogram. Does the answer confirm the prediction in 7.2.

The data is skewed to the right as the median is situated more to the left side of the histogram and the mean < median.

101

Page 106: Exercise 3.3: 1. 12

11

10

9

P l a n t

8

7

6

G r o w t h i n

5

4

3

2

1

c m 1

2

3

4

5

6

Hormone growth in ml

-1

1.2

The growth of the plant depends on the amount of growth hormone

1.3

The y axis is directly proportional to the x- axis.

1.4

A linear function

2..1 5

035

6

24455678

7

2 2 2 3 4 4 (74,5) 5 5 5 5 8 9

8

0 00 002489

9

00

2.2 Mean = 73,4 Median = 74,5 Mode = 80 2.3 Q1 = 66

and Q3 = 80

2.4 2.4.1

the interquartile range.

IQR = 14

2.4.2 the semi-interquartile range. SIQR = 7 2,4,3 the range for the class. Range = 40

7

102

2.4.4

Write down the maximum and minimum scores. Min = 50 and Max = 90

2.5 100

P 75 E R C

50

E N T 25

0

90

66

0

1

2

3

S

2.6

4

C

O

5

R

E

6

74.5

7

80

8

S

Standard Deviation. 10,1

2.7

What % of scores lie within 1 standard deviation from the mean. 74%

2.8

What % of scores lie within 2 standard deviations of the mean. 94%

3.1 4

5

5

00345

6

00244577777889

7

0003455566

8

0255

9

10

103 9

0

3.2 Median = 68; Mode = 67 Mean = 67,9 3.3 Q1 = 62 and Q3 = 75 3.4 3.4.1

the interquartile range. IQR = 7

3.4.2

the semi-interquartile range. SIQR = 3,5

3.4.3

the range for the class. Range = 45

3.5 Write down the maximum and minimum scores. Max = 90 and Min = 45 3.6

Do a box and whisker diagram using the five-number summary (L;Q1;M; Q3;H)

45

90 62 68

75

5

0

10

15

3.7 đ?‘ đ?‘Ąđ?‘Žđ?‘›đ?‘‘đ?‘Žđ?‘&#x;đ?‘‘ đ?‘‘đ?‘’đ?‘Łđ?‘–đ?‘ đ?‘Ąđ?‘–đ?‘œđ?‘› đ?œŽ = 10,5 3.8 28,6% 3.9 94,3% 4. 23 48 30 46

4.1

30 45 25 45

31 50 35 44

52 47 24 53

42 20 30 45

15 43 40 54

45 60 52 35

36 40 75 28

29 37 34

104 1

5

2

034589

3

0001455 67

4

0023455556 7 8

5

02234

6

0

7

5

8 9

4.2 Find the median, mode and mean for the data Median = 40 ; Mode = 45 & 30 Mean = 39,5 4.3 Find the lower and upper quartile Q1 = 30 Q3 = 47 Calculate: 4.4 the interquartile range. Answer : IQR = 17 4.4.1

the semi-interquartile range. Answer : SIQR = 8,5

4.4.2

the range for the class. Answer : Range = 60

4.5 Write down the maximum and minimum scores. Answer: Max = 75 Min = 15 4.6

Do a box and whisker diagram using the five-number summary (L;Q1;M; Q3;H

75

15

30

4.7

0

40

Standard Deviation 10 20 30 40

47

50

60

70

80

90

100

Answer = 12,2 4.8

What % of scores lie within 1 standard deviation from the mean. {27,0 ; 47,4}

105

22  63% 35 4.9

What % of scores lie within 2 standard deviations of the mean. {16,8 ; 57,6}

32  91% 35

Grade 12: Optional Work: 5.1 Do a stem and leaf diagram for the data 2

258

3

77

4

44556

5

344668

6

0037889

7

1155568

8

14

9

028

5.2

Find the median, mode and mean for the data

Median = 60 mode = 75 mean = 60,7 5.3 Find the lower and upper quartile Q1 = 45

5.4

Q3 = 75

Calculate:

5.4.1

the interquartile range.

IQR = 30

5.4.2

the semi-interquartile range.

SIQR = 15

5.4.3

the range for the class.

Range = 76 5.4.4

Write down the maximum and minimum scores.

LOWEST = 22 5.4.5

HIGHEST = 98

106

5.4.6 5.4.7

Standard Deviation = 18.9

What % of scores lie within 1 standard deviation from the mean. 60,7 + 18,9 = 79,6 60,7 – 18,9 = 41,8 {41,8 ; 79,6}

25  71% 35 5.4.8

What % of scores lie within 2 standard deviations of the mean. 60,7 + 37,8 = 98,5 60,7 – 37,8 = 22,9

34  97% 35

5.4.9

6. . The following table represents the maths scores for the entire grade 11 maths group at Northwood School. The data is grouped due to the size of group. Class to 9 to 19 to 29 to 39 to 49

0 10 20 30 40

Frequency(f) 15 10 17 40 35

Mid-points(X) 4.5 14.5 24.5 34.5 44.5

fX 67.5 145 416.5 1380 1557.5

Cum Freq 15 25 42 82 117

107

50 to 59 60 to 69 70 to 79 80 to 89 90 to 99 100 to 109 Totals

22 20 20 15 5 1 200

6.1 6.2 6.3 6.4

54.5 64.5 74.5 84.5 94.5 104.5

1199 1290 1490 1267.5 472.5 104.5 9390

139 159 179 194 199 200

Complete the last column of the table i.e (fX) Find the modal class đ?&#x;‘đ?&#x;Ž đ??­đ??¨ đ?&#x;‘đ?&#x;— Find the median class đ?&#x;’đ?&#x;Ž đ??­đ??¨ đ?&#x;’đ?&#x;— Find the interval where Q1 and Q3 lie đ?‘¸đ?&#x;? : đ?&#x;‘đ?&#x;Ž đ?’•đ?’? đ?&#x;‘đ?&#x;— . đ?‘¸đ?&#x;‘ : đ?&#x;”đ?&#x;Ž đ?’•đ?’? đ?&#x;”đ?&#x;—

6.5

Calculate the estimated mean.

NB estimated mean =

ďƒĽ fX n

 46,95

6.6Use the grouped data to display the data on a histogram 6.7 Draw the relevant frequency polygon on the histogram.

7. 0

10

20

30

40

50

60

70

80

90

110

100

x1

f

F x1

xx

( xx)

40

1

40

35,5

1260.25

50

1

50

25,5

650.25

2

108 65

2

130

10.5

110.25

70

1

70

5.5

30.25

75

4

300

0.5

0.25

78

5

390

-2.5

6.25

79

1

79

-3.5

12.25

80

1

80

-4.5

20.25

81

2

162

-5.5

30.25

82

3

246

-6.5

42.25

86

1

86

-10.5

110.25

88

1

88

-12,5

156.25

90

1

90

-14,5

210.25

x  75,5

n  24

 (x  x ) 1

2

=

2639.25 Variance

109.97

STD DEV

10.4

Height (h) in cm

Mid points

Frequency

Fx

Cumulative Frequency

Co-ordinates

135  h < 140

137,5

2

275

2

(140 ; 2)

140  h < 145

142,5

5

712.5

7

(145 ;7)

145  h < 150

147,5

10

1475

17

(150 ;17)

109 150  h < 155

152.5

17

2592.5

34

(155 ;34)

155  h < 160

157.5

19

2992.5

53

(160 ; 53)

160  h < 165

162.5

15

2437.5

68

(165 ; 68)

165  h < 170

167.5

4

670

72

(170 ; 72)

170  h < 175

172.5

2

345

74

(175 ;74)

175  h < 180

177.5

1

177.5

75

(180 ; 75)

75

8.1

11677.5

155,7

Question 8.2 & 8.3 20 18 16 Frequency

14 12 10 8 6 4 2 0 130 £ h < 135 £ h < 140 £ h < 145 £ h < 150 £ h < 155 £ h < 160 £ h < 165 £ h < 170 £ h < 175 £ h < 180 £ h < 135 140 145 150 155 160 165 170 175 180 185

Height in cm

8.4

Modal Group = 155 – 160 Median Height =157.5 Upper Quartile =170 Lower quartile = 145

8.5

110 80 70 60 50 40 30 20 10 0 130 £ 135 £ 140 £ 145 £ 150 £ 155 £ 160 £ 165 £ 170 £ 175 £ 180 £ h< h< h< h< h< h< h< h< h< h< h< 135 140 145 150 155 160 165 170 175 180 185

Volumes & Surface Area: Exercise 5.1:Page 135 1. Calculate the volume and surface area of the following closed prisms: Prism P Q R S Length (mm)

52

47

43

39

T 36

111 Breadth (mm)

20

18

17

15

14

Height (mm)

85

77

70

64

58

Volume

88400

65142

51170

37440

29232

Surface Area

14320

11702

9862

8082

6808

Determine the following ratios correct to 2 decimals.

VolumeP  1,36 VolumeQ

2.1

2.3

VolumeR  1,37 VolumeS

2.2

VolumeQ  1,27 VolumeR

2.4

VolumeS  1,28 VolumeT

3.

SurfaceAre aP  1,22 SurfaceAre aQ

3.2

SurfaceAre aQ  1,19 SurfaceAre aR

3.3

SurfaceAre aR  1,22 SurfaceAre aS

3.4

SurfaceAre aS  1,19 SurfaceAre aT

4. Are the volumes of the prisms approximately in proportion? Give reasons for your answers. Answer: Yes , approximately, the ratios around 1,3

5.1

How much smaller in volume is prism T than prism P? Give the scale factor (not the change in volume).Answer: About

1 smaller. 3

5.2

Are the surface areas of the prisms in proportion? Give reasons for pour answers. Answer: Yes, approximately, ratios are around 1,2.

5.3

How much smaller in surface area is prism T than prism P? Give the scale factor (not the change in area). Answer: About

6.

1 smaller. 2

Determine the scale factor used : to reduce the dimensions of the prisms. Answer: 0,9 To enlarge the dimensions of the prisms. Answer 1,1

7. What is “The Golden Ratio” ? Answer: the ratio is 1,6



1 (1  5 )  1,6180 2

112 7.1 Determine which ratio of the faces comes closest to this ratio. NB: You must choose a ratio greater than 1. Answer: The height to length ratio is 1,6 in each case. 8.1 Reduce each of the dimensions of prism P by a factor of

1 , then calculate the volume 2

and surface area of the new prism X. Answers: V  26  10  42,5  11050mm3 Answers: SA  3580mm2 8.2 How much smaller in volume and in surface area is this new prism X? Give the scale factor in each case. Answers: Vol Pr ismX 

1 volume of prism P. 8

SA of X = 9.1

1 SA of prism P. 4

Using the answers to question 8 , estimate the volume and surface area of prism Y, where each dimension of prism P has been enlarged by a factor of 2. Answers: Volume of Y = 8 x volume of prism P. = 8 x 88400 = 707200mm3 Answers: SA of prism X = 4 x SA of prism P = 4 x 14320 = 57 280 mm2

9.2

Calculate the volume and surface area of prism Y, and compare your answers to your answers to question 9.1 Vol Y = 104 x 40 x 170 = 707200 SA of Y = 2(104 x 40) + (40 x 170) + (170 x 104)= 57 280mm2

10.

Copy and complete the following tables:

Dimensions of the prism is length; breadth and height. i.e. l ; b ; h Volume of prism = l  b  h Factor k =_2_____ Prism

l (cm)

b (cm)

h (cm)

Volume (cm3)

Vxk

Factor

113 A

6

4

3

72

V

k

B

12

4

3

144

Vx2

k

C

6

8

3

144

Vx2

k

D

6

4

6

144

Vx2

k

E

12

8

3

288

Vx4

k2

F

6

8

6

288

Vx4

k2

G

12

8

3

288

Vx4

k2

H

12

8

6

576

Vx8

k3

Volume of prism = l  b  h Factor k =_3_____ Prism

l (cm)

b (cm)

h (cm)

Volume (cm3)

Vxk

Factor

A

4

3

2

24

V

B

4

3

6

72

Vx3

k

C

4

9

2

72

Vx3

k

D

12

3

2

72

Vx3

k

E

12

9

2

216

Vx9

k2

F

12

3

6

216

Vx9

G

4

9

6

216

Vx9

k2

H

12

9

6

648

V x 27

K3

k2

11. A cold - drink can measures approximately 65 mm in diameter and 75mm in height. Calculate the volume of the can ( in mm2 and cm2). Volume of can = r 2 h  32,5 2  75  248873mm3

V  248,873cm 3  249cm 3 The writing on the can says that it contains 200 ml of liquid. How much air space is there in the can? ( 1ml  1cm 3 ) Air space is approximately 49cm3

What is the height of the liquid in the can? Height  60mm

114 12.1

Calculate the total surface area of the can (in cm3 ) , assuming that the can is a closed cylinder.

SA  2(ď ° r 2 )  ď °D h Surface area ď‚ť 220m2 12.2

If the metal to make the can costs 0,25 cents per square centimeter, calculate the cost of making each can. Cost of a can = 55 cents

13

he manufacturer of Lemon Twist wants to double the volume of the can , but keep the radius as it is. By which factor must the height be increased? Increase height by a factor of 2

14.1

If the radius is increased by a scale factor of 2, but the height is kept the same by which factor will the volume increase?

vol of new can ď ° ( 2r ) 2 h  4 vol of original can ď ° (r ) 2 h 14.2

By which scale factor will the area of the top of the can increase?

surface area of new lid ď ° ( 2r ) 2  4 surface area of original lid ď ° (r ) 2 14.3

By which factor will the area of the lateral surface ( the area of the curved side) increase?

surface area of new side 2ď ° (2r )h  2 surface area of original side 2ď ° (r )h

Exercise 5.2 Page 155 5.2.1 1

Volume = 3 đ?œ‹đ?‘&#x; 2 đ?‘• = đ?œ‹(4)2 Ă— 11 = 184,31đ?‘?đ?‘š3 2 2 đ?‘†đ?‘˘đ?‘&#x;đ?‘“đ?‘Žđ?‘?đ?‘’ đ??´đ?‘&#x;đ?‘’đ?‘Ž = ď °r h  r

115

=𝜋 4

137 = 147,1𝑐𝑚2

5.2.2 4

𝑉𝑜𝑙𝑢𝑚𝑒 = 3 𝜋 𝑟 3 = 2144,66𝑐𝑚3 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 = 4𝜋 𝑟 2 = 804,25𝑐𝑚2 5.2.3 𝑉𝑜𝑙𝑢𝑚𝑒 = 𝑙𝑏𝑕 = 10 × 25 × 5 = 1250𝑐𝑚3 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎 = 2 𝑕𝑙 + 2 𝑕𝑏 + 2 𝑙𝑏 = 2 5 × 25 + 2 5 × 10 + 2(25 × 10) = 850𝑐𝑚2 5.2.4 1

𝑉𝑜𝑙𝑢𝑚𝑒 = 3 (𝑎𝑟𝑒𝑎 𝑜𝑓 𝑏𝑎𝑠𝑒) × 𝑕𝑒𝑖𝑔𝑕𝑡 1

= 3 (233.5)2 × 71,2 = 1185533,4𝑐𝑚3 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎 = 4 =4

233.5 2

1 2

𝑏𝑎𝑠𝑒 × 𝑕𝑡 + 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑏𝑎𝑠𝑒

× 136,75 + 233.52

= 63861,29𝑐𝑚2 + 54522,25𝑐𝑚2 = 118383,54𝑐𝑚2

5.2.5 1

Volume of Cone = 3 𝜋 𝑟 2 𝑕 1

= 3 × 52 × 8 209,4𝑐𝑚3

116

𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 = 𝜋 𝑟 2 𝑕 = 𝜋 × 52 × 20 = 1570,8𝑐𝑚3 𝑻𝒐𝒕𝒂𝒍 𝑽𝒐𝒍𝒖𝒎𝒆 = 𝟏𝟕𝟖𝟎. 𝟐 𝒄𝒎𝟑 2 2 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑐𝑜𝑛𝑒 = r h  r

89 = 148,2𝑐𝑚2

=𝜋 5

𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 = 𝜋 𝑟 2 + 2𝜋 𝑟𝑕 = 𝜋 × 52 + 2 × 𝜋 × 5 × 20 = 706,9𝑐𝑚2 𝑻𝒐𝒕𝒂𝒍 𝑺𝒖𝒓𝒇𝒂𝒄𝒆 𝑨𝒓𝒆𝒂 = 𝟖𝟓𝟓, 𝟏𝒄𝒎𝟐 Exercise 5.2 Page 143: 1

5.2.1 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐶𝑜𝑛𝑒 = 3 × 𝜋 × 𝑟 2 1

= × 𝜋 × 42 3

= 16,76 𝑐𝑚3 1

𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎 𝑜𝑓 𝐶𝑜𝑛𝑒 = 2 𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 × 𝑠𝑙𝑎𝑛𝑡 𝑕𝑒𝑖𝑔𝑕𝑡 1

= 2 × 2 × 𝜋 × 4 × 11,7 = 147,03𝑐𝑚2 4

5.2.2 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑝𝑕𝑒𝑟𝑒 = 3 𝜋𝑟 3 4

= 3 × 𝜋 × 83 = 2144,66 𝑐𝑚3 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒 𝑜𝑓 𝑠𝑝𝑕𝑒𝑟𝑒 = 4𝜋𝑟 2 = 4 × 𝜋 × 82 = 804,25𝑐𝑚2 5.2.3 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑃𝑟𝑖𝑠𝑚 = 𝑙𝑏𝑕 = 10 × 25 × 5 = 1250𝑐𝑚3 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑝𝑟𝑖𝑠𝑚 = 2𝑕𝑙 + 2𝑕𝑏 + 2𝑙𝑏 = 2 5 25 + 2 5 10 + 2 25 (10)

117

= 850𝑐𝑚2 1

5.2.4 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑃𝑦𝑟𝑎𝑚𝑖𝑑 = 3 (𝑎𝑟𝑒𝑎 𝑜𝑓 𝑏𝑎𝑠𝑒) × 𝑕𝑒𝑖𝑔𝑕𝑡 1

= 3 × 233,52 × 71,2 = 1293994,73𝑐𝑚3 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑃𝑦𝑟𝑎𝑚𝑖𝑑 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑏𝑎𝑠𝑒 + 4(𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑖𝑑𝑒𝑠) = (233,5)2 + 4

1 × 233,5 × 136,75 2

= 118384,5𝑐𝑚2 5.2.5 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑎𝑙𝑡 𝑐𝑒𝑙𝑙𝑎𝑟 = =

1 3 1 3

𝜋𝑟 2 𝑕 + 𝜋𝑟 2 𝑕 × 𝜋 × 52 × 8 + (𝜋 × 52 × 20)

= 1780,24𝑐𝑚3 1

𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑠𝑕𝑎𝑝𝑒 = 𝜋𝑟 2 + 2𝜋𝑟𝑕 + 2 × 𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 × 𝑠𝑙𝑎𝑛𝑟 𝑕𝑒𝑖𝑔𝑕𝑡 = 𝜋 × 52 + 2𝜋 × 5 × 20 +

1 × 2𝜋 × 5 × 89 2

= 855,05𝑐𝑚2 2

1

5.2.6 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑕𝑎𝑝𝑒 = 3 𝜋𝑟 3 + 3 𝜋𝑟 2 𝑕 = 218,08𝑐𝑚3 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑠𝑕𝑎𝑝𝑒 = 2𝜋𝑟 2 +

1 × 2𝜋𝑟 × 𝑠𝑙𝑎𝑛𝑡 𝑕𝑒𝑖𝑔𝑕𝑡 2

1 2 × 𝜋 × 3.52 + × 2𝜋 × 3,5 × 10,6 2 = 98,6𝑐𝑚2 5.27

1

𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑕𝑎𝑙𝑓 𝐶𝑦𝑙𝑖𝑛𝑑𝑒𝑟 = (𝜋𝑟 2 𝑕) 2

=

1 × 𝜋 × 32 × 20 2 = 282,74𝑐𝑚3

𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 = 2 =2

1 2

1

𝜋𝑟 2 + 𝑙𝑏 + 2 2𝜋𝑟𝑕) 1 2

× 𝜋 × 32 + 6 × 20 +

1 2

2 × 𝜋 × 3 × 20

= 336,77𝑐𝑚2 5.2.8 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎 𝐵𝑖𝑔 𝐶𝑜𝑛𝑒 = 𝜋𝑟 × 𝑠𝑙𝑎𝑛𝑡 𝑕𝑒𝑖𝑔𝑕𝑡

118

= đ?œ‹ Ă— 20 Ă— 552 + 202 = 3677,139925đ?‘?đ?‘š2 đ?‘†đ?‘˘đ?‘&#x;đ?‘“đ?‘Žđ?‘?đ?‘’ đ??´đ?‘&#x;đ?‘’đ?‘Ž đ?‘œđ?‘“ đ?‘†đ?‘šđ?‘Žđ?‘™đ?‘™ đ??śđ?‘œđ?‘›đ?‘’ = đ?œ‹đ?‘&#x; Ă— đ?‘ đ?‘™đ?‘Žđ?‘›đ?‘Ą đ?‘•đ?‘’đ?‘–đ?‘”đ?‘•đ?‘Ą = đ?œ‹ Ă— 10 Ă— 252 + 102 = 845,90đ?‘?đ?‘š2 đ??żđ?‘Žđ?‘šđ?‘? đ?‘†đ?‘•đ?‘Žđ?‘‘đ?‘’ đ?‘›đ?‘’đ?‘’đ?‘‘đ?‘ = 3677,1399 − 845,90 = 2831,24đ?‘?đ?‘š2 đ?‘œđ?‘“ đ?‘šđ?‘Žđ?‘Ąđ?‘’đ?‘&#x;đ?‘–đ?‘Žđ?‘™.

Circle Geometry: 1.

O is the centre of the circle.

D

N

AB = 60mm ; OM = 40mm Calculate the radius of the circle and the

C O

Length of CD if ON = 30mm B

M

AM = MD =30 ( Midpt chord thm)

A

ON = 30 AO 2  OM 2  AM 2 (Pythagoras)

AO 2  40 2  30 2 AO 2  1600  900 AO  50 CN 2  50 2  30 2 (pythag)

CN = 40 CD = 80

2.

CD = 80 mm; AB = 60mm and AB // CD D

If the radius is 50mm find the distance between the chords

C O

B A

RTC: Distance between chords.

119

Calc: CM = MD = 40 AN = NB = 30 Radius = 50

OM 2  50 2  40 2 OM  30 On 2  50 2  30 2 OM  40 Distance between chords = 70mm

3.

CD // AB E

C

D

A

B

Prove CD = 2AB NB: A and B are centres of the circles. RTP: CD = 2AB Proof: Join A to CE perpendicular at M. Join B to ED perpendicular at N. CM = ME ( Mid pt chd thm) EN = ND ( Mid pt chd thm) Let CM = ME x and EN = ND = y CD  2 x  2 y

AB  MN ( Opp sides rectL ABNM) AB  x  y  2 AB  CD 4.

M is the mid-point of AB, O is the centre.

E

RTP: AMC  BMC A

M C

Proof: AMˆ C  BMˆ C  90 (Mid pt chd thm)

120

In ”s AMC and BMC AMˆ C  BMˆ C ( proved)

AM  BM (Mid pt chd thm) MC  MC(common)

 AMC  BMC ( SAS)

O is the centre of the circles 5 to 10. Find the sizes of x and y in each case. 5.

6. A

A x

O

O

C

C

B

B

y

RTC: x Calc:

x  62 ( L at centre = 2 L at circle)

RTC: x and y BOˆ C  116 ( L’s at a point)

x  58 (L at centre = 2 L at circle)

y  122 ( L at centre = 2 L at circle)

7.

8.

O

O

110

70 C

A

x B

C

x A B

121

RTC : x

RTC : x

Calc:

Calc:

reflex AOˆ C  250 ( L’s at a pt)

Cˆ  35 (L at centre = 2 L at circle)

x  125 (L at centre = 2 L at circle)

x  35 Alt L’s AO // BC

C

9. O 70 x 10

A

B

RTC : x Calc: Cˆ  10 Alt L’s AC // BO

Oˆ  20 L at centre = 2 L at circle) x  20 ( Alt L’s AC // BO)

10. A

Prove that Bˆ1  Aˆ  90 O

1 B

Proof: 1 C

Let Bˆ1  x  Cˆ1  x ( L’s opp equal sides: radii)

122

BOˆ D  180  2 x ( L sum Triangle BOC)  Aˆ  90  x (L at centre = 2 L at circle) Bˆ1  Aˆ  90

Find the values x ; y and z in 11 to 16.

11.

12.

D

A

D

x z

A

60

O

y

y

2 1 x

O

C

B

20 B

C

z  120 (L’s on a str line)

x  60 (L sum Triangle)

x  60 (L at centre = 2 L at circle)

DAˆ C  30 ( L at centre )

Cˆ  90 ( L in semi circle)

BAˆ C  65 ( L at centre )

y  70  ( L sum Triangle)

y  95

13. F x A 2

y

1

2

1 20

3

E

4 B

C

2

85 D

123

Calculations:

Cˆ 4  20  ( L’s in same segt) Cˆ 3  85 ( L’s in same segt)

Cˆ 2  75 ( L’s on a str line) Cˆ1  20 ( Vert opp L’s) x  20 (L’s in same segt)

y  75 (L’s in same segt)

E

14.

15. O is the centre of circle ABCD 100

A

D

A

z

y

z

O x

C

x

120

E

B

D

y C

z  80 (L’s on st line)

z  30 (Ext L Triangle)

x  100 ( Ext L cyclic quad)

x  120 ( Altb Segt thm)

y  90  (opp L’s cyclic quad)

y  90  (L in semi-circle)

16.

DE is a tangent to circle ABC.

z  62 ( Alt segt thm)

A z

y  124 ( L at centre)

B

x  28 ( L sum Isosc Tri.)

O y

x 62 D

C

E

B

124

17.

A

Let Cˆ1  x

2 1

Bˆ1  x ( alt segt thm)

B

2 1

2 1

3 2

4 E

Dˆ 1  x (L’s opp = sides) D

ˆ  x (L’s opp = sides) A 1 Cˆ 4  x (Alt segt thm)

1 F

C

Aˆ 2  x (alt segt thm)

18.

D A E

Chord AB is parallel to chord CD. Sˆ1  Sˆ 2

Prove that QS = RT

1 2 B

Proof:

C

ˆ  Sˆ ( alt L’s AB //CD) A 1  Aˆ  Sˆ 2 (given Sˆ1  Sˆ 2 )

DE = BC ( equal L’s subtend = chords)

D

19. 1

2

E

Chord AD equals chord AC and Aˆ1  Aˆ 3 1 1

2 A

2

3

C

Prove: 19.1 Proof:

B

Eˆ  Bˆ

125

AD = AC  Eˆ  Bˆ ( equal chrds subtend = angles) 19.2

AE  AB

Proof: In  AED & ABC Eˆ  Bˆ ( proved)

AD = AC (given)

Aˆ1  Aˆ 3 ( given)  AED  ABC (AAS) AE = AB.

20.

D

A

Prove: T L

K

C

20.1 AKLD is a cyclic quad Proof:

M

N

B

126

AKˆ D  ALˆ D  90

AKLD is cyclic (Conv L’s in same segt) 20.2

KL // CB. Proof: DAˆ T  TRˆ L (L’s in same segt)

DAˆ T  BCˆ D (L’s in same segt) TRˆ L  BCˆ D KL // CB ( corr L’s equal) 20.3

If AK and Dl produced cut CB at M and N respectively, prove AMND is a cyclic quad. Proof: In KLM:

Mˆ 1  Cˆ  90 ( L sum ) In ALD:

Aˆ1  ADˆ L  90 (( L sum ) But ADˆ L  Mˆ 1 AMND is cyclic (conv ext L)

21.

P 1 2

F 1 2

PA and PC are tangents to the circle at A and C. AD //PC and PD cuts the circle at B. CB is produced to meet AP at F. AB, AC and DC are joined.

A 4 3

2

1

4 3 2 B 1

1

Prove:

2 3

C

4

21.1 AC is the bisector of PAˆ D . Proof:

1 2 D

127

PAˆ C  PCˆ A (L’s opp = sides) Aˆ 2  PCˆ A ( Alt L’s CP // DA)

ˆC  A ˆ PA 2

21.2 Bˆ1  Bˆ 3 . Proof:

Bˆ 3  ADˆ C ( ext L cyclic quad ABCD)

PAˆ C  ADˆ C ( alt segt thm) Aˆ 2  PCˆ A ( Alt L’s PC // AB)

Aˆ 2  Bˆ 2 (L’s in same segt)

Bˆ1  Bˆ 3 21.3 AP = AC. Proof: AP = AC. ( Tan from same pt) 21.4

APˆ C  ABˆ D . Proof: ˆ ( corr L’s CP //AD) APˆ C  A 1

Cˆ 3  Aˆ1 ( Alt segt thm) Cˆ 3  Bˆ 2 ( L’s in same segt)

APˆ C  ABˆ D

21.5 Aˆ 4  Pˆ2 Proof: ˆ ( Alt segt thm) Aˆ 4  D 1 ˆ Pˆ2  D 1

(alt L’s CP // AD)

Eˆ 3  Bˆ 3 CEBF is cyclic ( conv L’s in same segt)

128

22. FEˆ C  ADˆ C (i.e.Eˆ 3  Dˆ 1 2 )

F

B 3 1 2

2 A

1

C

2

2 E1 3 6 4 5

Prove:

1

22.1 CEBF is a cyclic quad. Proof:

Bˆ 3  ADˆ C (ext L cyclic quad ABCD)

1 2

Eˆ 3  ADˆ C (given)

D

Eˆ 3  Bˆ 3 CEBF is cyclic (conv L’s in same segt) 22.2 CGAF ARE concyclic. Proof: EFˆC  Bˆ 2 ( L’s in same segt) Aˆ1  Bˆ 2 L’s in same segt) ˆ  EFˆC A 1

CGAF is cyclic ( conv L’s in same segt)

22.3 AC bisects BCˆ G Proof: BFˆE  Cˆ 2 ( L’s in same segt)

ACˆ G  BFˆE ( L’s in same segt) ACˆ G  Cˆ 2

AC bisects BCˆ G A

23.

1 2

D 1

129

AB is a diameter. ADP and BCP are straight lines. PQTR is a straight line.

Prove: 23.1 DQCP is a cyclic quadrilateral. Proof:

Dˆ 1  90  ( L in a semi – circle) Cˆ1  90  ( L in a semi – circle) Dˆ 2  Cˆ 2  90 (L’s on st lines) DQCP is cyclic (conv L’s in same segt) 23.2 If Qˆ 5  TBˆ C , then PT  AB Proof:

Pˆ2  Qˆ 5  90 ( L sumQCP)

Aˆ1  Bˆ  90 ( L sumQCP) But

Bˆ  Qˆ 5 (given) ˆ  Pˆ2  A 1

In TBP:

Bˆ  Cˆ 2  90 Tˆ2  90 PT  AB 23.3 DATQ is a cyclic quad. Proof:

Tˆ2  Pˆ1  90 ( proved) DATQ is cyclic ( conv ext L = int opp). 24. AOD and EOB are diameters. AF  EB A 1 E

3

3 1

2

2 O

130

Prove : 24.1 EFHD is cyclic. Fˆ  90  (given) 1

EDˆ B  90 (L in semi-circle) ˆB Fˆ1  ED

EFHD is cyclic ( conv ext L cyclic quad) 24.2

BAˆ D  DHˆ C . Eˆ1  Aˆ 2 ( L’s in same segt)

Eˆ 2  Aˆ 3 (L’s in same segt) Cˆ1  Eˆ1 2 ( Ext L cyclic quad EFHD)

Cˆ1  Aˆ 23

BAˆ D  DHˆ C 24.3 Cˆ 3  Aˆ1 . Bˆ  Cˆ (L in same segt) 3

2

Cˆ 23  90 ( L in semi-circle) Bˆ 23  90  (L in semi-circle)

Bˆ 2  Cˆ 3 But Bˆ 2  Aˆ1 ( L’s in same segt)

Cˆ 3  Aˆ1 24.4

EB bisects ABˆ C Proof: In EBC:

Cˆ 2 3 25. T

A

In the figure BOD is a diameter of the circle with centre O. BA and BC are chords of the

131

circle. BA produced and CD produced meet in T and AD produced and BC produced meet in S. Prove: 25.1 ATSC is a cyclic quad BAˆ D  90 (L in semi-circle)

Aˆ1  90  (L in semi-circle) BCˆ D  90 (L in semi-circle) SCˆ T  90 (L’s on st line)

ATSC is cyclic ( conv L’s in same segt)

ADˆ B  ATˆS ADˆ B  ACˆ B (L’s in same segt)

25.2

ATˆS  ACˆ B (ext L cyclic quad ATSC) ADˆ B  ATˆS 25.3

OA is a tangent to circle ˆ  ATˆS ( Radii) OAˆ D  D 1 OA is a tangent ( Con alt segt thm)

26. D 3 2

1 T 2 1

C

3 2 A

4

1 B W

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In the figure above, TD is a tangent to circle ABCD at D. AD // BC, AB and DC produced meet at W. TBS is a straight line. If WBˆ T  CBˆ D , Prove that: 26.1 BWTD is a cyclic quadrilateral. Bˆ1  Bˆ 3 (given)

Dˆ 1  Bˆ 3 (alt segt thm) ˆ  Bˆ D 1 1

BWTD is cyclic ( conv L’s in same segt) 26.2

TBS is a tangent to the circle at ABCD.

Aˆ  Bˆ1 2 ( Corr L’s AD // BC)

Bˆ1  Bˆ 3 ( given)  Aˆ  Bˆ 23 TBS is a tangent ( conv Alt segt thm) 26.3

TW // BC

ˆ  Tˆ ( Alt segt thm) Bˆ 2  D 2 1

TW // BC ( Alt L’s are =)

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