Download Apply geometric reasoning in solving problems Example Task A

New Zealand
Association of
Mathematics
Teachers
1
Level 1 Mathematics and Statistics
91031 (1.6): Apply geometric reasoning in solving problems
Example Task A 2011
Credits: Four
You should answer ALL parts of ALL questions in this booklet.
You should show ALL your working.
If you need more space for any answer, use the page(s) provided at the back of this booklet and clearly
number the question.
Check that this booklet has pages 2–9 in the correct order and that none of these pages is blank.
YOU MUST HAND THIS BOOKLET TO YOUR TEACHER AT THE END OF THE
ALLOTTED TIME.
For Assessor’s
use only
Achievement
Apply geometric reasoning
in solving problems.
Achievement Criteria
Achievement
with Merit
Apply geometric reasoning,
using relational thinking, in
solving problems.
Achievement
with Excellence
Apply geometric reasoning,
using extended abstract
thinking, in solving
problems.
Overall Level of Performance
NZAMT 91031 (1.6): Apply geometric reasoning in solving problems Example Task A 2011 – page 1 of 14
You are advised to spend 60 minutes answering the questions in this booklet.
QUESTION ONE
(a) WXYZ is a parallelogram. Angle XYT is 66o
W
X
66°
Z
(i)
Y
T
Calculate the size of angle ZWX giving reasons for each step.
Calculation
Reasoning
Angle ZWX = ___________________________
(ii)
LMNPQR is a regular hexagon. Calculate the size of angle LRM giving reasons for
each step.
L
M
N
R
Q
Calculation
P
Reasoning
Angle LRM =
NZAMT 91031 (1.6): Apply geometric reasoning in solving problems Example Task A 2011 – page 2 of 14
(b)
In the kite below EF is 80cm long and the kite is 120cm wide.
E
Diagram not to scale
80cm
J
H
F
x
I
G
120cm
(i)
Find the angle EFJ.
Angle EFJ =
(ii)
To fly the kite, Bobby ties the ends of a string to points E to I. There is a knot in the
string at point K. The string can be pulled tight, so that point K is 10cm directly above
point J. If the total length of the string is 90cm find the length x, the distance from I to
J.
Length of IJ =
NZAMT 91031 (1.6): Apply geometric reasoning in solving problems Example Task A 2011 – page 3 of 14
(c) A farmer has a rectangular shaped field. He describes the path he walks across the field
starting at point A. First he walks on a bearing 070o for 1000m to reach point B. He then
turns at point B and walks 500m on a bearing 323o reaching point C on the north side of the
field. Calculate how far he still needs to walk to reach the corner, G of the field.
Diagram not to scale
1500m
G
N
A
NZAMT 91031 (1.6): Apply geometric reasoning in solving problems Example Task A 2011 – page 4 of 14
QUESTION TWO
(a) When a 9m long ladder is placed against a wall, the base of the ladder is 6.5m away from
the base of the wall.
9m
6.5m
(i)
Find how far up the wall the ladder reaches.
Length up the wall = ____________________ m
(ii)
If the 9m ladder is now placed against the wall at an angle of 75o to the horizontal
ground, how far up the wall will the ladder reach?
Length up the wall = ____________________ m
(b) When a new fireman places his ladder against a wall at an angle of 49o to the horizontal. The
base of the ladder is 6.5m away from the base of the wall.
Diagram not to scale
Window
49
°
6.5m
(i)
Calculate how far up the wall the ladder will reach.
Length up the wall = ___________________________
NZAMT 91031 (1.6): Apply geometric reasoning in solving problems Example Task A 2011 – page 5 of 14
(ii)
The fireman realises the top of the ladder is 2m below the bottom of the widow. He
moves the base of the ladder closer to the wall. Find the angle the ladder needs to be to
the horizontal so that it reaches the bottom of the window.
The angle needs to be =
(iii) Flick the fireman is curious about how the angle with the horizontal changes as the
ladder shifts up the wall. If a ladder of length l initially reaching height h up the wall,
shifts up the wall by 2m, by how much does the angle change? Show working to
explain your answer.
NZAMT 91031 (1.6): Apply geometric reasoning in solving problems Example Task A 2011 – page 6 of 14
QUESTION THREE:
(a)
KM is parallel to LN. IJ = JK. Angle IHN is 152o.
K
Diagram not to scale
L
J
I
M
152
°
H
N
.
Calculate the size of the angle JKI explaining the reasons for each step of your answer.
Calculation
Reasoning
Angle JKI =
(b)
AB is parallel to DE. O is the centre of the circle.
Diagram not to scale
A
D
20 °
30 °
O
C
B
E
(i)
Calculate the size of angle AOB giving reasons for each step.
Calculation
Reasoning
Angle AOB =
NZAMT 91031 (1.6): Apply geometric reasoning in solving problems Example Task A 2011 – page 7 of 14
(ii)
Calculate the size of the angle OBC giving reasons for each step.
Calculation
Reasoning
Angle OBC =
(c)
Find angle OCD in terms of x, where O is the centre.
A
B
Diagram not to scale
O
C
2x
D
E
Explain the reason for each step of your answer.
Calculation
Reasoning
Angle OCD =
NZAMT 91031 (1.6): Apply geometric reasoning in solving problems Example Task A 2011 – page 8 of 14
Extra paper for continuing your answers, if required.
Clearly number the question.
Question
number
91031 (1.6): Apply geometric reasoning in solving problems Example Task A 2011 – page 9 of 14
SAMPLE ASSESSMENT SCHEDULE a
Mathematics and Statistics 91031 (1.6): Apply geometric reasoning in solving problems
Assessment Criteria
Achievement
Apply geometric reasoning in
solving problems will involve:
Merit
Relational thinking will involve one
or more of:

using a range of methods
when solving problems

selecting and carrying out a
logical sequence of steps

demonstrating knowledge
of geometrical concepts
and terms

connecting different
concepts and
representations
and communicating
solutions that would usually
require only one or two
steps.


Excellence
Extended abstract thinking will
involve one or more of:

devising a strategy to
investigate or solve a
problem

identifying relevant
concepts in context
demonstrating
understanding of concepts

developing a chain of
logical reasoning, or proof

forming and using a model

forming a generalisation

relating findings to a
context

using correct mathematical
statements

communicating thinking
using appropriate
mathematical statements.

communicating
mathematical insight.
91031 (1.6): Apply geometric reasoning in solving problems Example Task A 2011 – page 10 of 14
Evidence Statement
One
Expected coverage
Achievement
Apply geometric
reasoning in solving
problems by:
Merit
Apply geometric
reasoning, using
relational thinking by:
Excellence
Apply geometric
reasoning, using
extended
abstract thinking
by:
:
(a)
(i)
 ZWX = 114o

Finding both
angles correctly.
 finding both angles
correctly, giving a
coherent
explanation of the
geometric
reasoning

Finding both
angles correctly.
 finding both angles
correctly, giving a
coherent
explanation of the
geometric
reasoning

setting up a more
complex
trigonometric ratio
correctly, but not
solving the
problem

solving a more
complex
problem using
trigonometric
ratios.

Solving part of the
problem correctly,
and allowing for a
consistent error.

Solving both
parts correctly,
but using
incorrect
mathematical
statements.
 Solving both
parts
correctly,
using correct
mathematica
l statements.

Correctly find one
or more horizontal
distance

finding the
bearing
correctly but
with a chain of
logic that is
not clear
(check
diagram), or
finding the
bearing of B
from O (72.5o)
with
reasoning.
 developing a
clear chain
of logic to
find the
bearing
correctly
 on a straight line supp
Co-interior ’s on //
repeated lines or
equivalent
(ii)
 LRM = 30o
Int ’s of hexagon and
base  of an isos △ or
equivalent
(b)
(i)
(ii)
 EFJ =
-1 60
cos
80
= 41.41 °
EJ = 2800
=52.91502622
EK= 2900 =53.85164807
IK = 90 – 53.85164807
= 36.14835193
2
2
IJ = (90 – 2900 ) – 10
Length of IJ =
34.74763589cm
(c)
Distance to walk to G =
861.215m
91031 (1.6): Apply geometric reasoning in solving problems Example Task A 2011 – page 11 of 14
Two
Expected coverage
Achievement
Apply geometric
reasoning in solving
problems by:
(a)
(i)
2
2
length = 9 – 6.5
= 6.22494979 m
Merit
Excellence
Apply geometric
reasoning, using
relational thinking by:
Apply geometric
reasoning, using
extended abstract
thinking by:
 Finds correct
length
Accept any rounding or
truncating. Units not
required.
(ii)
length = sin 75 °  9
= 8.69332437 m
 Finds correct
length
Accept any rounding or
truncating. Units not
required.
(b)
(i)
6.5 tan 49o = 7.477m
 finding the angle
correctly

(ii)
Ladder = 9.9m
6.5
cos
49 ° or
Either
 finding a length
correctly or finding
the angle with a
consistent error
 finding the correct
angle
 writes one
equivalent
expression for
either angle with
minor notation
errors
 writes one
equivalent
expression
correctly for either
angle
2


2
7.477 + 6.5
Angle needs to be
-1
sin  9.477 
 9.9  = 73.19o
Accept any correct
rounding
(iii)
The initial angle was
-1
sin  h 
l
then the
second angle will be
-1
sin  h + 2 
 l 

The difference is
-1
-1
 sin  h + 2  – sin  h 
 l 
l
91031 (1.6): Apply geometric reasoning in solving problems Example Task A 2011 – page 12 of 14
Correctly
expresses the
complete,
difference
equation.
Three
(a)
Expected coverage
JIM = 152o Corrsp ’s
on //
Achievement
Merit
Excellence
Apply geometric
reasoning in solving
problems by:
Apply geometric
reasoning, using
relational thinking by:
Apply geometric
reasoning, using
extended abstract
thinking by:
TWO of:
TWO of:
ONE of:
 Correct angle
found
 finding angles

correctly, giving a
coherent explanation
of the geometric
reasoning
 Correct angle
found
 finding angles

correctly, giving a
coherent explanation
of the geometric
reasoning
 Any two correct
consecutive angles
found
 finding two angles
correctly in this
sequence, giving a
coherent explanation
of the geometric
reasoning
OR
JIK = 28o ’s on
straight line are supp
JKI = 28o base ’s of
isos △ equal
Or equivalent reasoning.
(b)
(i)
 ABO = 20o base  isos
△, equal radii
 AOB = 140o ’s in a △
Or equivalent reasoning.
(ii)
 BOE = 20o alt ’s // =
 AOD = 20o alt ’s // =
 DCB = 80o ’s at centre
are twice angle at
circumference.
 OBC = 360o – 80o – 30o
– 200o = 50o
 OED = 2x corr ’s // =
 CAB = x  at centre =
2x  at circum
 EAB = 90o  in semi
circle
 EAC = 90 o – x
 EDC = 180 – (90 – x)
= 90 + x opp ’s
= in cyclic quad
 EOC = 180 – 2x ’s on
str line
 OCD = 360 – (180 -2x)
– 2x – (90 + x)
= 90 – x ’s in a
quad up to 360o
finding the
size of angle
OBC correctly,
using a
sequence of
clearly
explained and
logical steps

finding an
expression for
angle OCD
coherently,
using a clearly
explained
sequence of
steps with
geometric
justification for
most steps.
 finding expressions
for at least two of the
intermediate angles
with geometric
reasoning and clear
logic.
’s at a pt or  sum of
quad
Or equivalent reasoning.
(c)

 Any two correct
consecutive angles
found
 finding the correct
expression for
angle OCD without
a clearly explained
sequence of steps
with geometric
justification for
most steps
OR
 finding expressions
for at least two of
the intermediate 
angles with
geometric
reasoning and clear
logic.
91031 (1.6): Apply geometric reasoning in solving problems Example Task A 2011 – page 13 of 14
Judgement Statement
Achievement
Achievement with Merit
Achievement with Excellence
Minimum of:
2A
Minimum of:
2M
Minimum of:
2E
For Each Question 1 – 3
N1 = one step on 2q
N2 = 1a
A3 = 2a
A4 = 3a
M5 = 1m and 3a or 2m
M6 = 2m and 3a or 3m
E7 = 1e
E8 = 1e + 2m
91031 (1.6): Apply geometric reasoning in solving problems Example Task A 2011 – page 14 of 14