Upload
epic-win
View
222
Download
0
Embed Size (px)
Citation preview
8/3/2019 Solution+VG MVG+Level+Make+Up+Test+MaANVCO08Geometry
http://slidepdf.com/reader/full/solutionvg-mvglevelmakeuptestmaanvco08geometry 1/8
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
8/3/2019 Solution+VG MVG+Level+Make+Up+Test+MaANVCO08Geometry
http://slidepdf.com/reader/full/solutionvg-mvglevelmakeuptestmaanvco08geometry 2/8
VG/MVG Level Test MaANVCO08 Geometry NV-College
© [email protected] ☺Free to use for educational purposes. Not for sale! page 2 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
8/3/2019 Solution+VG MVG+Level+Make+Up+Test+MaANVCO08Geometry
http://slidepdf.com/reader/full/solutionvg-mvglevelmakeuptestmaanvco08geometry 3/8
VG/MVG Level Test MaANVCO08 Geometry NV-College
© [email protected] ☺Free to use for educational purposes. Not for sale! page 3 of 8
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
8/3/2019 Solution+VG MVG+Level+Make+Up+Test+MaANVCO08Geometry
http://slidepdf.com/reader/full/solutionvg-mvglevelmakeuptestmaanvco08geometry 4/8
VG/MVG Level Test MaANVCO08 Geometry NV-College
© [email protected] ☺Free to use for educational purposes. Not for sale! page 4 of 8
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
8/3/2019 Solution+VG MVG+Level+Make+Up+Test+MaANVCO08Geometry
http://slidepdf.com/reader/full/solutionvg-mvglevelmakeuptestmaanvco08geometry 5/8
VG/MVG Level Test MaANVCO08 Geometry NV-College
© [email protected] ☺Free to use for educational purposes. Not for sale! page 5 of 8
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
8/3/2019 Solution+VG MVG+Level+Make+Up+Test+MaANVCO08Geometry
http://slidepdf.com/reader/full/solutionvg-mvglevelmakeuptestmaanvco08geometry 6/8
VG/MVG Level Test MaANVCO08 Geometry NV-College
© [email protected] ☺Free to use for educational purposes. Not for sale! page 6 of 8
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 ≈⋅+=
8/3/2019 Solution+VG MVG+Level+Make+Up+Test+MaANVCO08Geometry
http://slidepdf.com/reader/full/solutionvg-mvglevelmakeuptestmaanvco08geometry 7/8
VG/MVG Level Test MaANVCO08 Geometry NV-College
© [email protected] ☺Free to use for educational purposes. Not for sale! page 7 of 8
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
8/3/2019 Solution+VG MVG+Level+Make+Up+Test+MaANVCO08Geometry
http://slidepdf.com/reader/full/solutionvg-mvglevelmakeuptestmaanvco08geometry 8/8
VG/MVG Level Test MaANVCO08 Geometry NV-College
© [email protected] ☺Free to use for educational purposes. Not for sale! page 8 of 8
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