Solution+VG MVG+Level+Test+MaANVCO08Geometry

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VG/MVG Level Test MaANVCO08 Geometry NV-College

Directions VG/MVG_Level TestCHAPTER TEST: Geometry MATHEMATICS COURSE A

Warning: This test is suitable for those students who have

passed the first test and have learned the geometry section in ahigh level.

Instructions

Test period 90 minutes for Part I and Part II as a whole. We recommend that you use at

the most 30 minutes to work with Part I. You are not allowed to use

calculator in the part I. Only after you hand in your solutions for part I you

may use a calculator.

Resources Formula sheet, your personalised formula booklet, ruler and protractor.

The test For most items a single answer is not enough. It is also expected• that you write down what you do

• that you explain/motivate your reasoning

• that you draw any necessary illustrations.

Try all of the problems. It can be relatively easy, even towards the end of

the test, to receive some points for partial solutions. A positive evaluation

can be given even for unfinished solutions.

Problems 4, 9 and 10 are larger problem which may take up to 30 minutes to

solve completely. It is important that you try to solve these problems.

Score and The maximum score is 36 points.

mark levels

The maximum number of points you can receive for each solution is indicated after

each problem. If a problem can give 2 ”Pass”-points and 1 ”Pass with distinction”-

point this is written [2/1]. Some problems are marked with ¤ , which means that they

more than other problems offer opportunities to show knowledge that can be related

to the criteria for ”Pass with Special Distinction” in Assessment Criteria 2000.

Lower limit for the mark on the test

G: Pass: 12 points

VG: Pass with distinction: 24 points (VG/MVG-test)

MVG: Pass with special distinction: 27 points (VG/MVG-test). You should show thehighest quality work in the ¤ -problems.

Only the marked problems in the box below will be graded.

1 2 3 4 5 6 7a 7b 8a 8b 8c 9 10 Sum

G 1 1 1 2 2 7

VG 1 1 2 2 4 2 1 2 2 2 2 4 4 29

MVG ¤ ¤ ¤ ¤¤¤

Grade

Name: ________________ Student number: _______ NVC08; Sjödalsgymnasiet

© [email protected] ☺Free to use for educational purposes. Not for sale! page 1 of 10





In each case, show how you arrived at your answer by clearly indicating all of the necessary

steps, formula substitutions, diagrams, graphs, charts, etc.

In each case, show how you arrived at your answer by clearly indicating all of the necessary

steps, formula substitutions, diagrams, graphs, charts, etc.

Part I VG-MVG level (without calculator)Part I VG-MVG level (without calculator)

You are not allowed to use calculator in this part. Write your answer to the first part on thisYou are not allowed to use calculator in this part. Write your answer to the first part on this

paper. Only after the submission of your solutions to the part one you may be allowed to use

your calculator. You may start working with part II before submission of your solutions to part I.

1. How many degrees must the equilateral triangle

be rotated about point P so that the triangle will

coincide with the original one? Give the least

possible angle. Explain. (0/1)

Suggested solutions:

°=°

1203

360 Answer: 120 °

This is due to special rotational symmetry of the equilateral triangle.Under a rotation of about its center point°120 P , no difference may be

noticed in the configuration of the equilateral triangle.

2. Investigate isosceles triangles, which have one angle that is °35 . Find the measure of

the other angles in the triangles you find. Motivate with figures or with calculations.

(1/1)

Suggested Solution:There are basically two distinguishably different solutions:• If the vertex °= 35 A , then the other two angles must be equal to each

other and their measure are:

°=°

=°−

=−

== 5.722

145

2

35180

2

180 AC B Answer: °== 5.72C B

As illustrated in the figure. [1/0]

• If the any one of the side vertices is °= 35 B , then due to the fact that

the triangle is an isosceles

triangle, two side angles mustbe the same, i.e.:

°== 35 BC

And therefore:°×−=−= 3521802180 B A

°=°−= 11070180 A

Answer: °== 35 BC , °=110 A

As illustrated in the figure. [0/1]

• P

°35

A B

C

°35





3. Jenny says: - I can compute the triangle’s area if I

know the angle between two sides of which I know the

lengths.

3. Jenny says: - I can compute the triangle’s area if I

know the angle between two sides of which I know the

lengths.

She is given the angleShe is given the angle A , and the lengths of the sides

AC and AB . Jenny solves the

problem correctly.Describe how you would do this. (0/2)

Suggested Solution:Let’s rename the sides of the triangle as

illustrated in the figure, namely as c AB ≡ , b AC ≡

We may draw an altitude from the vertex to

the segment

C

AB and name it , as the height of

the triangle.

h

Using the trigonometric relationship for the sinusof the angle A , we may find the height of the

triangle in terms of and the segment Asin AC :

A AC h AC

h A sinsin ⋅=⇔=

T t f the triangl

[0/1]

herefore, he area o e may be expressed as:

2

sinsin Acb AB A AC bh Area

⋅⋅=

⋅⋅=

⋅= Answer:

22 2

sin Acba

⋅⋅= [0/1]

4. The figure shows an isosceles right-angled triangle.

hat

our reasoning. (1/2/¤)

First method:

For the clarity of the solution, we may name the vertices of each produced

Are

Two of the sides are divided into five equal parts. W

fraction, or percentage, of the area of the triangle is

shaded?

Explain y

Suggested Solution:

triangle as illustrated in the figure. If length of each side is taken to be a5

25.12

2

55a

aa A ABC

⋅=⋅

=

JK AHI AFG ADE shaded A A A A A −+−= [1/0]

( )aaaaaaaa Ashaded ⋅−⋅+⋅−⋅= 223344

2

1

( ) 22

5149162

aa

Ashaded =−+−= [0/1]

%4040.0100

40

5.12

52

2

===/

/⋅

= a

a

A

A

ABC

shaded

[0/1/¤] Answer: %40= ABC

shaded

A

A

A B

C

D

E

F

G

H

I

J

K

aa a a a

a

a

a

a

a

A B

C

A B

C

h





Part II

This part consists of 6 problems and you may use a calculator when solving them.Please note that you may begin working on part II without a calculator.

arning: Submit your solutions to part I before using a calculator.

g field to be

to the observed point on the landing field, correct to the

ngles and are alternative angles.

[0/1]

W

ndin5. An airplane pilot observes the angle of depression of a point on a la

°25 . If the plane’s altitude at this moment is m1100 , find the distance from the pilot

nearest meter. (0/2)

Suggested Solution:mb 1100= , °25 , ?=c

A °25 ABC

Therefore: °=∠ 25 ABC

B

bc

c B

sinsin ==

b⇔

mmb

c 8.11

≈==m

B26032602

25sin

00

sin=

°

Answer: The distance from the pilot to theobserved point on the landing field is c 2603 m≈ [0/1]

eger and thwhat is a possible perimeter of the triangle? Explain. Is there any more possibilities? If

so name them. (0/4)

in its prime number factors:

[0/1]

This may be achieved without any great effort. Due to the fact that

6. If the lengths of the sides of a triangle are int eir product is 30.231 cm ,

Suggested Solution:Lets name the sides of the triangle as a , b and c:

20.231 cmcba =⋅⋅

We may first write 231

1173231 ⋅⋅=

6132 =++ , 231 is dividable by 3 . Dividing 231 by 3 yields. 773 = that

231

The following poss [0may be recognized relatively easily that is a product of 7 and 11.

ibilities exist for the perimeter of the triangle: /3] a b c abc c p + ba +=

3 7 11 2311173 =⋅⋅ 211173 =++

1 21 11 23111211 =⋅⋅ 3311211 =++

1 7 33 371 2313 =⋅⋅ 413371 =++

1 1 231 23123111 =⋅⋅ 23323111 =++

3 1 77 2317713 =⋅⋅ 817713 =⋅⋅

°25

m1100

°25 A

BC a

cmb 1100=

°25





7. In the quadrilateral illustrated below cm BC 0.4= , cm AB 0.5= , BA BC ⊥ and

CD AD ⊥

a) Find the perimeter of the quadrilateral? (1/1)

b) Find the measure of the angle CAB∠ and. (0/2)

Solutions:

To find the perimeter of the quadrilateral, wemay first calculate the length of each side.This may be achieved first by calculate thehypotenuse of the triangle . This may be done using the Pythagoras

theorem:

ACD∠

ABC Δ

cm AC 4125165422

=+=+= .

But the segment cm AC 41= is also the hypotenuse of the triangle

. Using the Pythagoras theorem we may write:CDAΔ

( ) ( ) 2222222222

5

41415414412 cm xcm xcm x xcm x x =⇔=⇔=+⇔=+

cmcmcm x 9.286.25

41≈≈= cmcm x 9.2

5

41≈=

Therefore, the perimeter of the quadrilateral is:

cmcmcmcmP 186.1759.175

4130.50.4 ≈≈=⋅++= Answer: cmcmP 186.17 ≈=

The measure of the angle andCAB∠ ACD∠ may be found using

trigonometric relationship for tangent of an angle in right triangle:

( ) °=⎟ ⎠

⎞⎜⎝

⎛ =∠⇔=∠ − 6.36

5

4tan

5

4tan 1

CABCAB Answer: °≈°=∠ 376.36CAB

( )°=

⎟ ⎠

⎞

⎜⎝

⎛ =∠⇔==∠ −

6.262

1

tan2

1

2tan

1

ACD x

x

ACD Answer:°≈°=∠

276.26 ACD

Second Method:

First calculate °≈°=∠ 376.36CAB and °≈°=∠ 276.26 ACD as above, then:

( )( ) ( )

cmCAB

AC AC

CAB 4.665981.36sin

4

sin

44sin =

°=

∠=⇔=∠

( ) ( ) ( ) cmcmcm ACD AC x AC

x ACD 9.286.256505.26sin40312.6sinsin ≈=°⋅=∠⋅=⇔=∠

cmcmcmcmP 186.1759.175

41

30.50.4 ≈≈=⋅++= Answer: cmcmP 186.17 ≈=

A

C

B

D

x

x2cm BC 0.4=

cm AB 0.5=





8. A regular octagon is illustrated in the figure below:

a. hown in the figure are calledles. What is the sum of all exterior

ior

a s

s the polygon have? (0/2)

f it

er how many sides a polygon has, due to the fact that in

starting from a point,

8. A regular octagon is illustrated in the figure below:

a. hown in the figure are calledles. What is the sum of all exterior

ior

a s

s the polygon have? (0/2)

f it

er how many sides a polygon has, due to the fact that in

starting from a point,

The angles sexterior ang

The angles sexterior ang

angles of an octagon? Why? (0/2)angles of an octagon? Why? (0/2)

b. If the sum of the degree measures of the inter

ngles of a polygon is °4140 , how many side

b. If the sum of the degree measures of the inter

ngles of a polygon is °4140 , how many side

doedoe

c. Calculate the area of the regular octagon

illustrated above, if the length of each side o

is cm00.1 . (0/2)

Suggested Solution:a) No matt

c. Calculate the area of the regular octagon

illustrated above, if the length of each side o

is cm00.1 . (0/2)

Suggested Solution:a) No matt

A and going round to B , …we always make acomplete circle (turn) and therefore, the total sum of the exterior

b) Th sides is

,C

angles of any polygon (and therefore, an octagon) is always °360 .

e total sum of angles of a polygon of number of n ( )2180 −⋅ n .

This is due to the fact that we may draw lines connecting a vertex of the polygon to the other vertices, making ( )2−n triangles. Each one of

these triangles contribute to the system, making the total sum of

.

Therefore, if the degree measures of the interior angles of a polygon is

°180

( ) °⋅− 1802n

°4140 :

( ) ( ) 2522323180

4140241401802 =+=⇔==−⇔°=°− nnn Answer: 25=n

c) We may draw two vertical segments BE ,

AF and two horizontal lines HC and GD .

These segments divide the octagon to:• One central square of the sides cm

and the area: .100.100.1 Asq=⋅=

• Four rectangles of the sides

and

00.120 cm 0

cm00.1

2

2

, and:

2

Re4 0.14 A ct ×= 222

20 cm=⋅

2

2

2

1121 2222 =⇔=⇔=⇔=+ x x x x x

• Four rectangles of the sides and Four triangles of the basecm00.1

cm2

2and the height

2

2, and area: 2

4 00.12

2

2

2

2

14 cm A

Triangle=⋅

/⋅

/×/=

Therefore, the total area of a regular octagon of sides is:cm00.1

A B

C

D

E F

G

H

A B

C

D

E F

G

H





22200.200.12200.1 cm A ⋅+=+⋅+= Answer: 22 83.42200.2 A = cmcm ≈⋅+

EaSecond Method:

ch interior angle, α , every exterior angle a regular octagon may be

calculated as:

°=°−°===⇔°=°

=°×

= 4135180...1358

1080

8

1806C B Aα

ere are eight such angles, therefore, sum of a

5

Th ll exterior angles of aegular octagon is:r

°=°⋅=+++++++= 360458 H F F E DC B Atot θ

The sides of corner right-triangles at the figure above i.e. x may be

calculated as:

cm x x22

220.145sin0.1

0.145sin =⋅=⋅=⇔=

Or as:

cm x x

2

2

2

20.145cos0.1

0.145cos =⋅=⋅=⇔=

We may calculate the total area of the octagon above as the sum of

• a rectangle of the width cm0.1 and cmcm 212

21 +=⋅/+ , and2/

therefore area of ( )( ) 2

tan 20.120.10.1 cmcmcm A glerec +=+=

• two trapezoid of parallel sides cma 0.1= , cmcmb 21221 +=⋅/+= ,2/

and the height cmh2

= therefore each

( )

2of area:

( ) ( ) ( ) 22 122

1222

4

1cmcm +=+=

the regular octagon of sides is:

221bah ⎞⎛ +⋅20.10.1 cm Atrapezoid

++⎟⎟

⎜⎜

/⋅==

222 ⎠⎝

Therefore, the total area of cm0.1

( ) ( ) ( ) 22122 cmm += tan 12

2

12122 c A A A trapezoid glercoctagon +

/⋅/++=+=

Answer: 22 83.42200.2 cmcm A ≈⋅+=





At the aspect assessment of your work

ve

have explained your work and motivated your conclusions

• how well you have written your solutions.

with problems 9 and 10 your teacher will consider

• what mathematical knowledge you have demonstrated and how well you ha

carried through the task• how well you

9. Rectangles with51

2

height

width

+= are usually called golden rectangles. Through the

years, such rectangles have been viewed as having especially attractive proportions.

Houses, parks, paintings and patterns are often constructed in this way. (2/4/¤)

putea. Com 51

2

+ and answer using five decimal

points.

b. State the formula for a computation of the area of a

golden rectangle with a height of .

c. Fibonacci sequence is generated by starting with two

terms and obtaining the rest by adding the previous

term of the Fibonacci sequence as a fraction of its

. Give the first ten terms as decimalscorrected to 5 decimal places. What do you notice?

t c te the relationship betw ts

width and height. What do you notice?

a)

h

two terms. The simplest Fibonacci sequence is

...,21,13,8,5,3,2,1,1 and this crop up in

natural objects. Form a sequence by expressing each

following term

d. Take any card: ID-Card, credi ard,… Investiga een i

Solutions:

61803.051

2=

+ [1/0]

b) hwh ++ 5151

w⋅=⇔=

22⇔ 22

6180.051+

2hhhw Area ⋅=⋅=⋅= [1/1/*]

c) Fibonacci sequence:

Fibonacci “fraction”:

,...55,34,21,13,8,5,3,2,1,1

,...89

55,

55

34,

3421138

21,

13,

8,

5,

5

3,

3

2,

2

1,

1

1 [0/1]

is:

,...79861.0,61818.0,76561.0,90561.0,53861.0,50062.0,60000.0,6667.0,5.0,0.1

The Fibonacci “fraction” approaches and converges to 79861.0 which in limit

6180.051

2=

+ [0/1]

d) Most of standard cards dimensions are of the “Golden ratio”. Why?Perhaps it is the nature of our sense of beauty. Perhaps, it has todo with the most compact packing sense of the nature. [0/1/*]





10. The figure below illustrates a conic water tank. It is filled to

of its full capacity. Calculate the height of the water %2.51

column in the container,w

h . (2/4/¤)

a) If the radius of the base of the container is denoted by r

and its height by h .

b) If the radius of the base of the container is mr 0.16= and

its height by

Suggested Solutions:

mh .100= .

Solutions:

Data: mr 16= , mh 96= , V V water ⋅= 512.0

Problem: ?=water h

W n of the radius of the cone. This may be achieved by linkingthve

e may express the height of the cone as a functio

h other by the tangent of em to eac the angle of thertex of the cone, θ .

Suggested Solutions

θ tan⋅=⇔= hr h

r θ tan [1/0]

2

3512.0

3r h ⋅⋅⋅=21

512.0512.0 r hV V water ⋅⋅⋅⋅=⋅=π

π

( )2tan

3512.0 θ

π ⋅⋅⋅⋅= hh V water

θ π 22

tan3

512.0 ⋅⋅⋅⋅= hh V water

θ π 23 tan3

⋅⋅h [0/1]

he other hand

512.0 ⋅=V water

On t θ θ tantan ⋅=⇔= water water hr h

r .

may also write the volume of the water as:We

( ) θ π

θ π π 232

tan3

n ⋅⋅=⋅ water h 2ta

33⋅⋅=⋅⋅= water water water water water hhr hV

θ π 23 tan⋅⋅= water water hV [0/1] 3

Therefore: θ π

θ π π 2323 tan

33water water 512.0tan ⋅⋅⋅=⋅⋅= hhV [0/1]

hhhhwater 0 [0/1] h water

⋅=⋅=⇔⋅= 8.512.0512.0 333

mhhwater 0.80100512.0512.0 33 =⋅=⋅=

Answer: hhhwater ⋅=⋅= 8.0512.03 ; mhwater 0.80= [1/0/*]

md 32=

mh .100=

wh

md 32=

mh .100=

wh





Second Method:

Therefore, we mayThe triangles are similar (They have identical angles).

⇔⋅=⋅⇔= water water hr hr r h

water water r h

hr r water ⋅=

hwater [0/1]

22

2

2r h ⋅⋅⋅π

3512.0

3

3

3hr h

r h

water water

water water

⋅⋅/

/⋅=⋅

/

/⋅⇔

⎪

⎪⎨

⋅⋅=π

512.0

r

V

V

water

water

⎪

⎪⎧

π π [1/1]

⎩

=

2

2

222

2

512.0512.0 r hh

r hhr h

hwater

⎠⎝

hr h wate⎜

⎛ ⋅⋅

water water r ⋅⋅=⋅⋅⇔⋅⋅=⎟ ⎞

[0/1]

hhhhhwater ⇔= 5.003

water ⋅=⋅=⋅ 80.012512. 33 [0/1]

mhhwater 80.5.03= 0.80100012 =⋅=⋅

Answer: mhhwater 0.8010080.0512.03=⋅=⋅= [1/0/*]

Third Method:laThe triangles are simi

Therefore, we mayr (They have identical angles).

⇔⋅=⋅⇔= water water water water hr r h

hr r h

h

hr r water

water ⋅= [0/1]

512.0512.0

3

3512.02

2

22

2

2

tan

=/⋅

/⋅⋅

⇔=

⋅⋅/

/

⋅/

/⋅

⇔=r h

h

r hh

r h

r h

V

V water water water water

k

water

π

π

[1/1]

hhhhhh

hwater water

water ⋅=⋅=⇔⋅=⇔= 80.0512.0512.0512.0 333

3

3

[0/1]

mhhhwater 0.8010080.080.0512.03 =×=⋅=⋅=

Answer: mhhwater 0.8010080.0512.03 =⋅=⋅ [1/0= /*]

Fourth Method (In this method, the generality of the problem is lost, and

The triangles are similar (They have identical angles).

Therefore, we may

therefore the MVG quality point is lost)

water water water water water ater hr hr h ⋅=⇔⋅=⇔⋅= 16.0100

1616 [0/1]w

water water r r h

⋅⇔= 10016100

water water hr ⋅= 16.0 [0/1]

3232 43693

4369161003

512.0 mr hV mV water water water water π π

π π

=⋅⋅=⇔=××⋅=

water water water water water =⇔=⋅⇔×=⋅

[0/1]

( ) 333232 107130256.01071316.043693 mhmhhmr h

( ) mhmhwater water 0.8000

10133

1333 =⇔⇔= [1/1] hmhm

water water 05120005120256.0

7 33 ==⇔

Answer: mhwater 0.80= [1/0]

Solution+VG MVG+Level+Test+MaANVCO08Geometry

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