Solution+VG MVG+Level+Make+Up+Test+MaANVCO08Geometry

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VG/MVG Level Test MaANVCO08 Geometry NV-College

VG/MVG_Level Make up Chapter TestGeometry MATHEMATICS COURSE A

Warning: There are more than one versions of the test.

This test is designed for those students who have alreadypassed the first test with the grade G, and are aiming to a

higher grade.

Instructions

Test period 100 minutes. Monday 19 Jan 2009.

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

graphic calculator. No telephone may be used as a calculator. You are not

allowed to borrow anything from your classmates during the test.

The test Write the solution of the problem under each problem in the designated

area. In case you need extra space, you may use extra sheets available.

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 35 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

VG: Pass with distinction: 20 points (VG/MVG-test)

MVG: Pass with special distinction: 25 points (VG/MVG-test). You should show the

highest quality work in the ¤ -problems.

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

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

G 1 2 1 1 2 2 10

VG 1 1 2 2 3 2 2 2 4 4 23

MVG ¤ ¤ ¤¤

Name:

Student number:

Grade

© [email protected] ☺Free to use for educational purposes. Not for sale! page 1 of 8





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.

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)

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: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 , no difference may be

noticed in the configuration of the equilateral triangle.

°120 P

2. Investigate isosceles triangles, which have one angle that is °20 . 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 °= 20 A , then the other two angles must be equal to each

other and their measure are:

°=°

=°−

=−

== 802

160

2

20180

2

180 AC B Answer: °== 80C B

As illustrated in the figure. [1/0]

• If the any one of the side vertices is °= 20 B , then due to the fact that

the triangle is an isosceles

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

°== 20 BC

And therefore:°×−=−= 2021802180 B A

°=°−= 14040180 A

Answer: °== 20 BC , °=140 A

As illustrated in the figure. [0/1]

• P

°35

A B

C

°35





3. Calculate the length of the diameter,3. Calculate the length of the diameter, BD , of the isosceles trapezoid illustrated below

where: cm DC 0.38= , cm AB 20= 0. , cm BC AD 0.30== . (2/2)

Suggested Solutions:We may draw a vertical line from A to theside . This is the height of the

trapezoid. Due to the symmetry of the

problem

DC

cm DM 0.92

0.200.38=

−=

The triangle AMD is a right triangle.Therefore, using the Pythagoras theorem,we may calculate the height of thetrapezoid.

cmhh 6.28819819930 222 ==⇔=−=

x is the hypotenuse of the right triangle, where BND

cmh BN 6.28== , cm ND 0.290.200.9 =+=

Therefore, using Pythagoras theorem, wemay calculate x .

cm x ND BN x 7.4016601660296.2822222 ≈=⇔=+=+=

Answer: cmcm x 7.401660 ≈=

4. How many percent of the figure below isshaded? (1/2)

Suggested Solutions: Answer:

%5.78≈stotal

shaded

A

A

The area of the square is

( ) 2242 r r Atot ==

The area of the shaded part is the areaof a circle of radius r , and that is:

2r Ashaded ⋅= π

Therefore %5.78785.044 2

2

≈==⋅

⋅=

π π

r

r

A

A

stotal

shaded

A B

C D

cm0.38

cm0.20

cm0.30cm0.30

x

A B

C D M N cm0.38

cm0.20

cm0.30cm0.30

r 2





5. A gang secede to steal the 3.00 meter high and 5.50 ton Buddha status from the temple

Chori-naram at Bangkok. In order to smuggle it out of the country, they decide on to

smelt the golden status and make 500 identical golden statuses in proportion to the

original. How high are the new status and how much do they each weight? (1/3)

5. A gang secede to steal the 3.00 meter high and 5.50 ton Buddha status from the temple

Chori-naram at Bangkok. In order to smuggle it out of the country, they decide on to

smelt the golden status and make 500 identical golden statuses in proportion to the

original. How high are the new status and how much do they each weight? (1/3)

Suggested Solutions:Suggested Solutions:Answer: Each new status is going to weightAnswer: Each new status is going to weight kg

kgm 0.11

500

5500== and they

are each tall.cmhnew 8.37≈

cmmhh

V

V

h

h

h

h

V

V new

new

orig

new

orig

new

orig

new

orig

new 8.3737798.03500

1

500

1

3

3

1

3

13

13

≈=×⎟ ⎠

⎞⎜⎝

⎛ =⇔⎟

⎠

⎞⎜⎝

⎛ =⇔

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ =⇔

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ =

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

a. The angles shown in the figure are called

exterior angles. What is the sum of all exterior

angles of an octagon? Why? (

0/2)

. If the sum of the degree measures of the interior

c. Calculate the area of the regular octagon

/2)

Suggested Solution:ny sides a polygon has, due to the fact that in

b

angles of a polygon is °9405 , how many sides

does the polygon have? (0/2)

illustrated above, if the length of each side of it is cm00.1 . (0

a) No matter how mastarting from a point, A and going round to B , ,C …we always ma

complete circle (turn) and therefore, the total sum of the exteriorangles of any polygon (and therefore, an octagon) is always °360 .

b) The total sum of angles of a polygon of number of sides n is (180

ke a

)2−⋅ n .

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

these triangles contribute °180 to the system, making the total sum of ( ) °⋅− 1802n .

Therefore, if the degree measures of the interior angles of a polygon is°9405 :

( ) ( ) 3522333180

9405294051802 =+=⇔=

°=−⇔°=°− nnn Answer: 35=n

A B

C

D

E F

G

H





c) We may draw two vertical segmentsc) 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 cm00.1

and the area:2

00. 100.100.1 cm Asq=⋅=

• Four rectangles of the sides cm00.1

and2

2, and:

2

Re4 222

200.14 cm A ct =⋅×=

2

2

2

1121 2222 =⇔=⇔=⇔=+ x x x x x

• Four rectangles of the sides cm00.1 and Four triangles of the base

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:cm00.12

2200.200.12200.1 cm A ⋅+=+⋅+= Answer: 2283.42200.2 cmcm A ≈⋅+=

Second Method:Each interior angle, α , every exterior angle a regular octagon may be

calculated as:

°=°−°===⇔°=°

=°×

= 45135180...1358

1080

8

1806C B Aα

There are eight such angles, therefore, sum of all exterior angles of aregular octagon is:

°=°⋅=+++++++= 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

x

2

2

2

2

0.145sin0.10.145sin =⋅=⋅=⇔=

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

221 +=

/⋅/+ , and

therefore area of ( )( ) 2

tan 20.120.10.1 cmcmcm A glerec+=+=

A B

C

D

E F

G

H





aa• two trapezoid of parallel sides cm0.1• two trapezoid of parallel sides cm0.1= , cmcmb 212

221 +=

/⋅/+= ,

and the height cmh2

2= therefore each of area:

( ) ( ) ( ) ( ) 222 122

12224

120.10.12

2

2

1

2cmcmcmbah Atrapezoid +=+=++⎟

⎟ ⎠ ⎞

⎜⎜⎝ ⎛

/⋅=+⋅=

Therefore, the total area of the regular octagon of sides is:cm0.1

( ) ( ) ( ) 22

tan 122122

12122 cmcm A A A trapezoid glercoctagon

+=+/

⋅/++=+=

Answer: 2283.42200.2 cmcm A ≈⋅+=





At the aspect assessment of your work with problems 6 and 7 your teacher will consider

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

carried through the task• how well you have explained your work and motivated your conclusions

• how well you have written your solutions.

7. In the accompanying diagram of the rectangle ABCD , the

diagonal AC is drawn, cm DE 0.14= , AC ⊥ DE and

°= . (2/4/¤) ∠ 0.62 DAC m

a. What is the area of the rectangle ABCD ?

b. What is the perimeter of the rectangle ABCD ?

Suggested Solutions: Answer: ;2

473 cm Area ≈ cmP 4.91≈

cmcm DE

AE AE

DE 44.7

62tan

0.14

62tan62tan =

°=

°=⇔=°

( ) cmcm DE

CE CE

DE 3.26

28tan

0.14

28tan28tan6290tan =

°=

°=⇔=°=°−

( ) DE EC AE hb

Area ⋅+=⋅

⋅=2

2

( ) 28.4720.1428tan

0.14

62tan

0.14cm Area ≈⎟

⎠

⎞⎜⎝

⎛

°+

°=

To calculate the perimeter of the rectangle, we must calculate the length of the sides AD and

AB . Using the right triangle AEDΔ :

cmcm DE

AD AD

DE 86.15

62sin

0.14

62sin62sin =

°=

°=⇔=° .

Similarly in the right triangle : DEC Δ

cmcm DE

DC DC

DE 82.29

28sin

0.14

28sin28sin =

°=

°=⇔=° .

The perimeter of the rectangle, P , is therefore:

( ) cmcmcmcm

DC ADP 4.9135.9128sin

0.14

62sin

0.1422 ≈=⎟

⎠

⎞⎜⎝

⎛ +⋅=+⋅=

A

C

D

E

B

14

°62





8. The area of the right triangle illustrated in the figure below is equal to one third of the

area of the half-circle illustrated below. Find the angles of the triangle. (2/4/¤)

8. The area of the right triangle illustrated in the figure below is equal to one third of the

area of the half-circle illustrated below. Find the angles of the triangle. (2/4/¤)

Suggested Solutions:Suggested Solutions:

Data:Data:circlehalf triangle

A A−

=3

1

Problem: , ,°= 90 A ?= B ?=C

The triangle is a right-triangle and

. This is due to the fact that the

hypotenuse of the triangle is the diameter of the circle.

°= 90 A

We may draw a vertical line from thevertex A to the base of the triangle AB and name it .h

We may realize that due to the factthat two sides of the triangle AOC Δ a

the radius of the circle, i.e.:

re

r OB AO = xOAB ABO ≡

= , it is an isosceles triangle and

therefore: ∠=∠ . Therefore x AOC 2=∠ . Th fac

that AOC ∠ exterior angle to the triangle

is is due to the t

2= is an x AOC Δ .

aming the radius of the circleN r , we may express the area relationshipas:

circlehalf triangle A A −=3

1

⎟ ⎠

⎞⎜⎝

⎛ ⋅/

=⋅/

2

2

1

3

1

2

1r BC h π

2

3

12 /⋅=/⋅ r r h π

r h ⋅= π 6

1

In the triangle we may express the height of the triangle as ADO

( ) ( ) ( ) xr hh

xh

AOC 2sin2nsin ⋅=⇔==∠ r OA

si⇔

( ) ( ) °≈°=⇔°=⇔⎟ ⎠ ⎞⎜

⎝ ⎛ =⇔=⇔/⋅=⋅/⇔⋅= − 168.156.312

6sin2

62sin

612sin

61 1

B x x xr xr r h π π π π

Answer: °≈°≈°= 74,16,90 C B A

°= 90 A

OC

B

°= 90 A

OC

Bh

D x

x

x2

Solution+VG MVG+Level+Make+Up+Test+MaANVCO08Geometry

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