Name of School:
ROUTINE INFORMATION Hillview Secondary School
Student Surname and Name:
Maphumulo Mthokozisi
Student Number:
207503878
Grade:
10, 11, 12
Subject:
Mathematics : EDMA 604
Topic:
Assignment 3: investigative task lesson. Geometric Figures: cyclic quadrilateral , isosceles triangles and their properties
Content /Concept Area:
Euclidian geometry, the cyclic quadrilateral and usefulness of geometry properties in theoretical development and geometry calculations involving algebra. Author, Year , Title , Page No. and grade
CAPS page number:
1) Mcprice and CamBell (2012) : page( 202) 2) Aird and van Duyan (2012) Chapter 9, page 296. : - Cyclic Geometry. See references: for full details Duration of Lesson: Date
45 minutes 15 August 2018
Contents: This is a brief lesson aimed at equipping learners understanding of geometry theories and their development of skills for being able deal with determination and explanation of angles of geometric figures, their length and areas through understanding of simpler concepts and theories used in geometry. In this case we are using cyclic quadrilateral, triangles and leave learners to stipulate reasons, from their average effort for the cause of scientific discovery they observe from how they interact with their task. By the end of the lesson learners should know and be able to:Explain properties of geometric using their CAPS books Construct angles to form geometric figure and interpret coordinate in a geometric plane Investigate relationship between sides of geometric figures to their angles and co-ordinates
Examine the effect that properties of geometric figure has on their sides, angles and shape. Determine geometric relationship that may develop from understanding of geometric figure. Demonstrate ability to understand conditions for corresponding angles, parallel lines and application of mid-point theorem. Analyse all theories underpinning cyclic quadrilateral and develop skills to draw circle geometry, produce perpendicular bisectors of a chord of a circle. Table of Contents: Section A: - Introduction of a Lesson To introduce this lesson questioned students to describe all properties of a circle. Following then huge protractor was used to draw a circle and the properties discussed were drawn so that students observe their label from how they were discussed. The grid pattern sheet is provided to learners who are sited according to their groups of 6- 8 members per group. Learners also got instructions which is designed a question paper to lead their group task. Briefly a cyclic quadrilateral is a circle drawn inside a circle. All vertices of cyclic quadrilateral must lie in the circumference. In total there are four (4) sides of a cyclic quadrilateral but also three or even more cyclic quadrilaterals may exist in one circle all at once. The investigative lesson aimed raising learner’s skills for being able to apply properties of geometric figures to theoretical development. Central to this task development was to enable learners to be able to use their independent effort of practical thinking. They are being trained to develop conjectures and theories supported by their understanding of geometric properties. In this case relationships were discovered but also mutual learning took place as learners sought to do even much advanced correlations from how they understand geometric relationship. Noteworthy is the paper developed by group C on which learners draw a parallelogram to develop understanding likely to conclude that angles at centre is twice the angle at the circumference. These learners did not produce isosceles triangle as one may have expected. Instead they took locate co-ordinate of a parallelogram at centre to stand against the Cartesian plane. They applied the Sine rule and cosine rule to prove that the angle at centre is twice the angle at the circumference. The second group demonstrated that the opposite angles of cyclic quadrilaterals are supplementary. They did very much demonstrated ability to interpreted how well they understood the relationship a geometric figure produced inside a circle has to a circle , their deductive reasoning were well illustrated. This task is detected directly from an investigative task, so as to suit limited time for a single period in which all students dealt with their theoretical development. Section B
Development: In this practical activity students are not aware that I am inspecting their participation and their capability against van Hiele’s level of development. I have constructed questions to lead learners towards constructing proofs that will help them to earn better understanding of definitions. This help students when they make proofs. In this task specifically I have noticed a student make use of a given grid pattern to draw a circle with a radius of 6 units. He then used grid pattern to also determine angles. But most interestingly is their skills for being able to derive understanding of properties of some cyclic quadrilateral and other geometric shapes for developing proofs in line with cyclic quadrilateral theorems in use. We cannot afford to select all groups, as these lesson was done across three different classes, we have chosen only four (4) groups. I took I group in each class to compile this report. The lesson was done from grade 10, 11, and 12.
Group 1 Teacher’s task Student task Teacher’s task Given a cyclic quadrilateral drawn Facilitate learning through help with inside of a circle. Learner’s first need to figure out what
conditions
satisfy
clarification of questions asked? Demonstrate to learners how they
the
need to handle protractor and a
quadrilateral to stand inside of a
compass used to draw. Ensure learners understand the terms
circle. Secondly learners must able to produce a line from at least two vertices of a circle where the vertices of a quadrilateral also lie against the circumference so that
they share common points. Where possible learners indicate those vertices. Ask questions for demonstration, respond
from
key
questions,
participate through discussion and write down solutions. Task Division:-
and concepts in line with the investigative lesson. Therefore foster
revision
of
properties of geometry and their applications. Did not intervene with learners choice
of
their
knowledge
generation, their solutions, proofs and discoveries were absolutely independent. Eliminate errors
Task
Group 1- G:11
Group 2- G:12
Cyclic
Cyclic
Quadrilateral
and
Outcome
Group
3
– Group 4 : G 12
G:10 Quad Properties of a Tan circle :
: parallelogram
Theorem
Euclidian
Opposite angles drawn through Geometry of cyclic quad the
use
are
vectors
supplementary
scalars.
Chord
vs
of Algebra and Outcome :-
Outcome Angles at centre is
twice
the
angle at circle Achievement 60% 80% Criteria of work Learners show Learners description
88% 60% did The topic seem From
the
ability to follow more and above to have excited beginning they instruction. Line
what they were learners because needed to gain
produced supposed
to they
were much
from
know since they requested
circumference
also introduced something they of
to a centre and vectors
more
to understanding what
is
and have been doing tangent.
also centre was scalers which I even
in
their Fortunately my
named after O 1 did not expect. previous
calculus is very
and
O
expected.
2
as Their discovery classes. is
But good. As far as I
meaningful their interesting am
because
they argument
sought to use made Cosine and Sine listing rule
to numerous
concerned
was this group was about very
much
of struck from not knowing
determine
the properties
value
the cyclic quad and vertices, secant,
of
angle at centre. naming That angle was triangles
of deflations
key like
of tangent and a with difference
also correlate to effect
to between parallel
the
of lines
theory presence
which
and
states right angles , corresponding
that the angles radius. They got angles. So far at
centre
is very good mark impression was
twice the angle from at
made when I
the elaborating
circumference.
on brief them on
what
is what conditions
difference
satisfy
for
a
between
point to lie at
isosceles, scaled tangent against and equilateral a triangle
radius.
For
and example
added that these impression was help when one about the fact must
make
a that tangent is
proof that angle perpendicular to at
centre
is radius while the
twice the angle line from centre at
to chord to pass
circumference
midpoint
must
where a line is bisect the chord produced about and also remain circumference
perpendicular.
and centre to Booklet for key form a isosceles geometric triangle
about figures and their
radius.
properties
was
also given at the end Criteria marking
of See rubrics
See rubrics
See rubrics
of
lesson. See rubrics
the
Section C – observations. Report based on what were Learner’s feedback and discoveries during the lesson commencement. The groups are selected randomly. The 1st group. – Grade 11
The first group seem to have done very much impression from being able to identify the name cyclic quadrilateral. The group seem to have understood a concept of cyclic quadrilateral because all vertices or sides of a quad lied inside the circle. The centre of the circle was drawn neatly but there were not so much complex details to confuse what they are to do. The instruments were used appropriately to draw a circle The points are made to lie in the circumference of a circle. The radius is joined from centre to the circumference. Each student shares knowledge to discussions of how to go about figure out if opposite angles of a cyclic quadrilateral are supplementary. Finally s brief summary report of their brief discussions came with the following conjecture: < O2 = 2< AED – angle at centre is twice the angle at circumference.
<O1 = 2< ABD - angle at centre is twice the angle at circumference. <O1 + <O2 = 3600 – Angles around the point, sum of complete revolution = 3600. But <O1 + <O2 = 2< AED + 2< ABD <O1 + <O2 = 2(< AED + < ABD) 3600 = 2(< AED + < ABD) 360 = 2
2(¿ AED +¿ ABD ) 2
1800 = < AED + < ABD If < AED + < ABD = 1800. The 2nd Group report The pure applied Maths and Physics Class: This class gave lot more impressions, because it was not informed what sort of relationship needs to be discovered. Instead it was given instructions to draw and circle and cyclic quadrilateral with points fixed in the circle circumference. Two distinctive impressions were made. The first one was when student chose. But I should blame myself if any mistake happen for giving hem grid pattern, to use it with geometry problems. This group drew a cyclic quadrilateral, and discovered that the angle at centre is twice the angle at circumference. But they sought to extend their proof through introducing vectors and scalers which came as a new thing to myself. They introduced a set of vectors but to convert them into parallelogram later on into parallelogram linked to a circle to form part of cyclic quadrilateral. Learners made impression for pulling out the co-ordinates points of a quadrilateral figure. Thereafter they generate angle and sides of parallelogram and begin to estimate their size using Cosine rule and Sine rule. The most interesting thing is the fact that their methods worked successfully such that you will see evidence that the angles at centre are twice as much the angle at the circumference.
In the circle above, learners used both the Euclidian geometry and their vector laws to generate a proof leading to conclusion that the angles at centre of circle is twice the angle at the circumference. The cyclic quadrilateral HGBF is produced about the radius = 8 units. Then they rely on grid pattern to read out the co-ordinates of vertices of the small vector plane E, A, C and D. Where E( -1 : 2) , A ( 0: 0) , C( 5: -1) , and D ( 4: 1). EA // DC â&#x20AC;&#x201C; EACD is a parallelogram. From then they used a coloured pen to join AE to the circumference of a circle. Similarly AD was joined to meet circle circumference both sides. Likewise DC was also joined to meet the circle at the circumference of the circle. Learners eliminated the use of protractor and ruler for measuring angles, instead they began to work out the value of angles and sides using Cosine rule and Sine rule as follows. AD represents resultant, d2 = (x2 â&#x20AC;&#x201C; x1)2 + (y2 â&#x20AC;&#x201C; y1)
(0:0)
and (4: 1).
(x1 : y1) and ( x2 : y2) AD2 = (4-0)2 + (1- 0)2 = 16 + 1 = 17
AD =
√ 17
90 90
2x
2x
EA2 = (x2 – x1)2 + (y2 – y1) :
(-1: 2)
and (0: 0)
(x1 : y1) and ( x2 : y2) EA2 = (x2 – x1)2 + (y2 – y1)2 = [(0 – (-1)]2 + (0 – 2)2 = (1)2 + (2)2 = 1 + 4 = 5 EA =
√ 5 units
EA = DC = equal.
√5
since EACD is a parallelogram – opposite sides of a parallelogram are
Check: DC2 = (x2 – x1)2 + (y2 – y1)2: D (4: 1) and C (5: -1) . (x1 : y1) and ( x2 : y2) DC2 = (5– 4)2 + (-1 – 1)2 = (1)2 + (-2)2 = 1 + 4 = 5 DC2 = 5, therefore DC = DC =
√5
√ 5 units
Similarly ED = AC = ED2 = (x2 – x1)2 + (y2 – y1)
Where E (-1: 2) and D (4: 1)
= [(4 – (-1)]2 + [(1 – 2)]2
(x1 : y1) and (x2 : y2)
= (4+1)2 + (-1)2 = 52 + (-1)2 = 25 + 1 = 26 ED2 = 26 ED =
√ 26
AC = (x2 – x1)2 + (y2 – y1)2
A (0: 0)
and C (5: -1)
(x1 : y1) and ( x2 : y2) = (5– 0)2 + (-1 – 0)2
= 52 + (-1)2 = 25 + 1
AC2 = 26 AC =
√ 26 units.
In Cos < CAD = b2 + c2 – a2 /2bc 26 Cos < CAD = √ ¿ ¿ ¿
2
+(
transformation of a theory of Cosine rule
√ 17 )2 - (√ 5)
2
26 ]/ 2 √ ¿ x ¿
= (26 + 17 – 5) / 2 x 21. 02 = 38/42.04 = 0.9 < CAD = Cos-1 (0.9xcvsd) = 260 .
√ 17 ))
That also means ADE = 260 , since ADE is a parallelogram or EA // DC – alternate angles are equal. sin< EAD EA
=
sin26 √5
sin EAD √ 26
=
sin< EAD ED
Now cross multiply the entire equation to solve for Sin < EAD
√ 26 Sin 260 = √ 5 Sin < EAD √ 5 sin < EAD √5 Sin < EAD =
=
√ 26 sin 26 √5
√26 ( 0.4 ) = 5.1 ( 0.4 ) = 2.04/ 2.2= 0.9 2.2 √5
Sin < EAD = 0.9 < EAD = Sin-1( 0.9) = 640, 2 If < EAD = 640, 2 , then it follows that <ADC = 640, 2, since EA // DC alternate <’s are equal.
Then they began to lay down their extremely rough but correct argument. There were two angles lying in the circumference lysing against the diameter drawn from separate points. However their argument was that those angles are equal because of common straight line that created supplementary angles equal to angles at the circumference. In fact they are laying correct argument because straight line was produced to meet the circle at the circumference. Then they created two different isosceles triangle. These set of isosceles triangles were supported by two equal radii. Which also created diameter. It does not matter if those angles were standing across different diameter, they remained equal from the circumference because they share equal diameter that subtends right angle equal to 900. What they did, is that they produced recall that radii are equal and they could result to isosceles with two equal angles standing at the base.
They used that knowledge together with the fact that the exterior angles of triangle = two interior opposite angles. They produced those two angles to stand along the centre of the circle O . This logic proved that the angle at centre is twice the angle at the circumference. Yet at the same time it also confirmed that straight line subtends right angle. In other hand it also make me to discover that any angles subtended by equal chords from the common circle, no matter if chords stands across different positions, those angles are also equal. That was a shocking discovery for me because I never expected transformation geometry at high school level. I could not even imagine how they integrated their understanding of vector model to Euclidian geometry. Have a look at the summary of their table, below. Missing things 1 rubrics Summary of Questions asked? Activity References