1. Solve the following system of equations: 5 marks MTH-4111 Pretest B This is a disjoint system and therefore there is no solution (no points of intersection.

1. Solve the following system of equations:

322 xxy 72

072

xy

yx

5 marks

MTH-4111 Pretest B

24

4016

10144

1041

0104

07322

3272

2

2

2

2

))((

c;b;a

xx

xxx

xxx

This is a disjoint system and therefore there is no solution (no points of intersection between the line and the parabola).

2. A ship is moored to a pier by a chain. The chain hangs in the shape of a parabola part of which is submerged. The chain sinks below the water surface 5 meters from the pier and re-emerges 25 meters from the pier. The chain is attached to the top of the pier which is 2 meters above water level. The bow of the ship is an oblique line. The top of the bow is 30 meters from the pier and 8 meters above the surface of the water. The bow enters the water 40 meters from the pier. At what point does the chain attach to the bow of the ship?

x

y 30 m

8 mQUADRATIC EQUATION:

Zeros: (5,0), (25,0)

Other point: (0,2)

y = a(x – x1)(x – x2)

y = a(x – 5)(x – 25)

y = a(x2 – 30x + 125)

2 = a(02 – 30(0) + 125)

2 = 125a

1252

a

125250

12560

1252

125301252

2

2

xx

)xx(y

LINEAR EQUATION:

Zeros: (30,8), (40,0)

54

108

304080

12

12

xxyy

m

400

54

xy

5(y - 0) = -4(x – 40)

5y = -4x + 160

5y = -4x + 160

51604

x

y

0187520

03750402

4000100250602

160425250602

16041252506025125

2506025

1604

2

2

2

2

2

2

xx

xx

xxx

)x(xx

)x()xx(

xxx

7900

7500400

18751420

4

1875201

2

2

))((

acb

c;b;a

445444342

8810828868

2888820

2888820

12790020

2

21

21

21

.xOR.x

.xOR

.x

.xOR

.x

)(

ab

x

51604

x

y

44845

160761375

16044344

.

.

).(y

The chain is attached to the boat 34.44m to the right of the pier and 4.45m above the water surface.

x

y34.44

m

4.45 m

D)

A)

3. Two functions are described below.

f(x) = mx + b where m < 0

g(x) = ax2 + c where a > 0 and b = c.

Which of the following graphs represents the function operation, f – g?

f(x) = -1x + b

g(x) = 1x2 + b

f – g = (-x + b) – (1x2 + b)

= -x + b - x2 – b

= - x2 - x (A)

D)C)B)A)

4. Functions g and h are represented graphically to the right.

Identify the graph below that corresponds to g • h.h

gg(x) = -1x

h(x) = 1x

g h = (-1x) (1x)

= - x2 (D)

B)A) D)C)

5. The equations for the diagonals of a parallelogram are indicated in the diagram that is provided. A (-5, -7) and B (3, -5) are two of the vertices of the parallelogram. What is the perimeter of the parallelogram?

A(-5, -7)

B(3, -5)

5 marks

y = 2x + 3

21

23

xy

258

68

464

28

573522

22

212

212

.

)()(

))(()(

)yy()xx(DAB

C

D

Point C is on the vertical line x = 3 so its x-value is 3. (3, ? )

Vertex C is also on the diagonal y = 2x + 3.

y = 2(3) + 3

y = 6 + 3 = 9 C (3,9)

1414

)5(9

12

yyDAB

27.14

27.25.8

ThADmBCm

ThCDmABm

Perimeter of the parallelogram = 2(8.25) + 2(14)

= 16.5 + 28

= 44.5 units

6. Prove that the following quadrilateral is a rhombus.

B (3, 2)

A (-3, 4)

C (5, -4)

D (-1, -2)

10 marks

188

3544

12

12

)(

xxyy

mAC

144

1322

)()(

mBD

1BDAC

mm

),(,

)(,

yy,

xxMP

AC

0120

22

244

253

222121

),(,

)(,

)(MP

BD

0120

22

222

213

485.8

72

3636

)6()6(

))1(5())4(2(22

22

BDm

7. The vertices of an irregular quadrilateral are as follows: A (-3, 3), B (2, 5), C (6, -2) and D (-4, -1). Calculate its area.

C (6, -2)

B (2, 5)

D (-4, -1)

A (-3, 3)

Slope of diagonal BD:

166

)4(2)1(5

m

Equation of BD:

03

25

)2(1525

1

yx

xy

xyxy

Length of BD: Length of altitude from A:(-3,3) to side BD.

999.82

)121.2)(485.8(2

hbAABD

)3,3(),(

3;1;1

11

22

11

yx

CBA

BA

CByAxd

121.22

3

11

333

)1(1

3)3)(1()3(122

Area of ΔABD

00.332

)778.7)(485.8(2

hbACBD

00.42

00.33999.8

ABCDA

Length of altitude from C:(6,-2) to side BD.

)2,6(),(

3;1;1

11

22

11

yx

CBA

BA

CByAxd

778.72

11

11

326

)1(1

3)2)(1()6(122

Area of ΔCBD

Area of Quadrilateral ABCD

485.8BDm

03 yxEquation of BD:

C (6, -2)

B (2, 5)

D (-4, -1)

A (-3, 3)

1. Midpoint Formula:

8. Complete the demonstration of the following proposition using geometric analysis.The area of square ABCD is twice the area of

square EFGH.HYPOTHESIS:

.squareaisADCD

.squareaisEFGH

.,,

int,,

lyrespectiveADandCDBCAB

ofsmidpoareHandGFE

CONCLUSION:EFGHABCD AA 2

y

xA (0,0) B (a,0)

C (a,a)D (0,a)

E

F

G

H

STATEMENTS JUSTIFICATIONS

0,22

00,

20 a

Ea

E

2,

20

,2

aaF

aaaF

2,

22121 yyxx

5 marks

EFGHABCD AA 2

1. The coordinates of E and F are:

2. Length of the sides of both squares: 2. Distance Formula:

212

212 )yy()xx(D

aa

)()a(DAB

22

22

0

000

2242

4422

20

2

22

2222

22

aaa

aaaa

)a

()a

a(DEF

STATEMENTS JUSTIFICATIONS y

xA (0,0) B (a,0)

C (a,a)D (0,a)

E

F

G

H

3. Area of both squares: 3. Area of a Square Formula:

2sA 2aAABCD

2

2

2

2

a

aAABCD

4. Relation between Areas of both squares:

22

22

2 aa

AA EFGHABCD

9. Complete the demonstration of following proposition:

The following diagonals of the regular pentagon form an isosceles triangle with a vertex angle

of 36º. HYPOTHESIS:.pentagonregularaisABCDE

.diagonalsareECandAC

CONCLUSION:

36ACEmandECAC

STATEMENTS JUSTIFICATIONS

5 marks

A E

D

C

B36º

DCBC Congruent sides of a regular pentagonCongruent sides of a regular pentagon

EDAB

EDCABC Congruent angles of a regular pentagon

EDCABC Theorem 16 (SAS Congruency Theorem)

ECAC Theorem 39a

54025180 )(

Theorem 7

1085540EDCmABCm Congruent angles of a regular pentagon

DCEmDECmandBCAmBACm

Theorem 45

362

722

108180BACm Theorem 5

3672108

3636108

108

)(

)(

)DCEmBCAm(ACEm

10.ACDF and ABHG are similar trapezoids and the ratio of their areas is . Trapezoid GHEF is equivalent to parallelogram BCDE.

Given that the height of trapezoid ACDF is 12 cm and its large base is 20 cm, what is the length of the base of parallelogram BCDE?

10 marks

91

E

A

D

CB

F

HG

31

91

91 2

k

kA

A

Big

Small

cmh

h

h

h

h

Small

Small

Small

Big

Small

4

12331

12

31

cmB

B

B

B

B

Small

Small

Small

Big

Small

67.6

20331

20

31

hbABCDE hbB

AGHEF

2

12 cm

4 cm

8 cm

6.67 cm

12 cm

20 cm

E

A

D

CB

F

HG12 cm

4 cm

8 cm

6.67 cm

x20 - x

12 cm

Let x = base of a parallelogram

Long base of trapezoid GHEF = 20 - x

x.

)x.(

.)x(AGHEF

467106

46726

82

67620

x

xABCDE

12

12

E

A

D

CB

F

HG

x. 467106 x12

Trapezoid GHEF is equivalent to parallelogram BCDE.

cm.x

.x

.xx

x.x

AA GHEFBCDE

676

6710616

67106412

46710612

The base of parallelogram BCDE is 6.67 cm.

6.67 cm

11. ΔFDE is equivalent to rhombus IFHG.

ΔFDE ~ ΔFHG

Find the perimeter of ΔFDE.

10 marks

D E

H

F

G

I38° 38;10 GIHmcmFGm O

cmFGm

GOm 52

Theorem 28

IOmGOm

GIOtan

4006781305

538

.IOm.

IOm

IOmtan

Theorem 23

1218615705

538

.IGm.

IGm

IGmsin

IGmGOm

GIOsin Theorem 21

8012

40062

2

.

.

IOmIHm

Theorem 28

2

21

642

108122

2

cm.

FGmIHm

DDA bushomr

264cm

AA bushomrDEF

.IFGHoftorsecbiaisFG

2322

cm

AA bushomr

GHF

41412

2

3264

2

2

.k

k

kA

A

GHF

DEF

D E

F

H

F

G

10 cm

H

F

G

I

8.121 cm

8.121 cm

8.121 cm

cm.IGmGHmFHm 1218 Congruent sides of a rhombus.

cm.FDm

FDm.

FGmFDm

k

141410

4141

Theorem 50a

FEmcm.DEm.DEm

.

FHmFEm

GHmDEm

k

48111218

4141

Theorem 50a

D E

F

H

F

G8.121 cm

8.121 cm

11.48 cm

11.48 cmcm....

FEmDEmFDmFDEofPerimeter

137481148111414

12. A mother is sharing a some Tropicana orange juice between her 2 sons. However she only has 2 glasses which are both cylindrical yet have different shapes. She tries to assure her sons that she will share the juice equally between them. She fills one glass having a diameter of 6 cm to a height of 10 cm. The other glass has a diameter of 8 cm.

a) How high must she fill it to keep her promise?

b) The juice comes from a container that is a rectangular prism with a height of 19 cm whose base is a square with 10 cm sides. If after serving the juice, the container is empty, How high was the juice inside the container?

10 marks

8 cm

10 cm

6 cm

a) Volumeshort glass = Volumetall

glass

πr2h = πr2h

π(3)2(10) = π(4)2h

90π = 16πh

cm.h

h

62551616

1690

She must fill the glass to a height of 5.625 cm.

b) Volumerect prism = lwh

= (10)(10)h

= 100hThe volume in each glass is 90π. This makes a total of 180π poured from the rectangular prism.

Volumerect prism = 100h

180π = 100h

565.49 = 100h

h = 5.655 cm

The juice was 5.655 cm inside the container.

h2 = a2 + b2

= 7.52 + 202

= 56.25 + 400

= 456.25

h = 21.36 cm (slant height)

13. A company called Party Poppers plans to produce cardboard party hats for the New Years’ Eve celebrations. They have 2 different proposals. A cone-shaped hat with a diameter of 15 cm and a height of 20 cm and a cylindrical-shaped hat with the same diameter and height.

If they have 100 square meters of material to work with, how many hats can they make for each proposal? HINT: Remember the hats will be open at one end.

20 cm

15 cm 7.5 cm

20 cm

Lateral Areacone = πrs

= π(7.5)(21.36)

= 160.2π

= 503.3 cm2

100 m2 = 100 (10000 cm2)

= 1 000 000 cm2

# of cone-shaped hats = 1 000 000 ÷ 503.3

= 1986 hats

Lateral Areacylinder = 2πrh

= 2π(7.5)(20)

= 300π

= 942.5 cm2

Areacircle = πr2

= π(7.5)2

= 56.25π

= 176.7 cm2

20 cm

15 cm

Total area of cylindrical hats open at one end

= 942.5 + 176.7 = 1119.2 cm2

# of cylinder-shaped hats = 1 000 000 ÷ 1119.2

= 893 hats

Closed end of hat