The Tangent Theorem

The Tangent Theorem A university of Natal Academy Mr Maphumulo Mthokozisi has proved The Tangent Theorem. This theorem has failed many academics so to prove. In order words it has been used without proof.

Forewords

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Steps of the research process 

Step 1: Identify the Problem. ..

There are many properties of a tangent to a circle. The tangent touches the circle once ant share common point with its circumference. In order to call a point a tangent of a circle there must be a straight line passing through its point of contact. We make a difference between tangents as secant. Secant of a circle touches the two points of circles from one circumference. Secants differ from chords, as chords maintain the extension of their lines to remain inside the borders of circle. However a secant can pass through two points of circle and extend its borders of straight line outside of a circle.

We have a serious problem in pure applied mathematics and it is universal problem concerning the tangent theorem in relation to its radius or diameter. Out of million textbooks are written to express that the tangent is perpendicular to radius , yet not even a single book ever prove the truth and essence for that scientific phenomenon. The problem is that our pedagogical system are now getting transformed beginning with adoption of cognitive science theories which situate learners / students at centre of construction of knowledge. The essence for question of this probing question took place at Hillview Secondary School during their grade 11 and 12 classroom teachings, when a young White boy came out to challenge a pure applied mathematics teacher of the truth and validity for using a scientific phenomenon to prove theories who are regarded as real. Three teachers including Sir Maphumulo Mthokozisi, Mr Toheey, Mr Gorbin and Mr Weich remain embarrassed because they had nothing to say. The fact is that even books used have no real scientific proof. However today we are going to be surprised because a careful scholar Sir Maphumulo Mthokozisi Maphumulo has overcome the complexity behind the lack of scientific proof of that theory.

In other words, we have a real proof and theory to prove that tangent is indeed perpendicular to a radius.



Step 2: Review the Literature. ...

Sir Maphumulo joined the University of Natal as Geography and Environmental Management student. He completed his first degree in Geography and Environmental Management, where he also did advance applied mathematical practical in Geographic Informative System (GIS). After that he completed honours degree in Economics. However he also became addicted in writing lengthy research where he questioned universal evolution.

He is exactly the professional who wrote a book called ‘The Scientific Philosophy of New Universal Invention’. In year 2018 he re-join the University of Natal to study geography of teaching (FET). Mr Eric Bishop motivated Sir Maphumulo Mthokozisi Maphumulo to take mathematics. This followed after he has seen how capable he was for demonstrating and illustrating using pure applied mathematics examples. Soon he got to register for FET mathematics and the school of mathematics education called University of Natal never hesitated to register him. Very soon a school dean issued letters for all pure applied mathematics students to go to make practical training in South African Schools. Sir Maphumulo was indeed very good mathematics teacher, he also passed higher grade mathematics during his school times. He was introduced to Mr Weich, Mr Gorben and Mr Toheey as his mentor while teaching during school practical.

These are White teachers and they examined Sir Maphumulo such that there was no stone remained unturned. Sir Maphumulo taught Geometry, Algebra, Function graphs of all types and Calculus. He proved almost 98% of mathematical theories however he was challenged by the Tangent theorem. Since then he step back and look for information across 2000 pure applied mathematics books and observed that there was no scientific proof over why the tangent is perpendicular to radius.

It became truth that a student complain over continuous use of tangent – radius theory was compromising the student conceptual understanding of mathematics. As a matter of fact Sir Maphumulo saw a chance to investigate factors underlying the lack of any scientific theory to prove that tangent is perpendicular to radius. Over a period of a year a perfect response and solution to that problem was captured. From now and forever we have overcome the mist that stood on front of mathematician for why the tangent is perpendicular to radius or diameter.



Step 3: Clarify the Problem. ...

The essence of this problem is with a Hillview Secondary School, a South African student doing grade 11 who felt unsatisfied for how we continue to use the tangent theorem stating that a diameter or radius is perpendicular to tangent if there is no scientific proof for that theorem. Most books also offer no solution to that query.

Therefore a student question was not fluke results, it’s a problem that needs careful scientific case study.

Mr Gorben is a mathematics literacy teacher for grade 11 and 12 in Hillview Secondary School. He was observed by Sir Maphumulo teaching area and volume of cylinder.

Who are Sir Maphumulo’s key role model of tangent theorem?

Mr Gorbin a mathematics literacy teacher for grade 11 and 12 had powerful influence due to the fact that he followed constructivist approach from how he taught areas and volume of geometric shapes such as cylinder. Secondly there is Mr Weich and Mr Toheey who teaches pre applied mathematics who introduced advanced algebra and Euclidian geometry which also attracts Sir Maphumulo. Mr. Phatha Mahlabela and Mr Bishop also set key activities which includes transformation geometry that sought to challenges. Mrs Rowlath is also teaching advanced mathematics at University of Kwa-Zulu Natal–Edgewood Campus using online software which are more or less the same as Geogebra and that is where Sir Maphumulo benefitted a lot from how secant and tangent shift within the circle. Mr Mahlabela and Dowlath are key lectures who taught Sir Maphumulo …a man who has discovered a tangent theory concerning the radius to remain perpendicular to a tangent itself.

Contributions of Mr Weich, Mr Toheey and Mr Gorban from Hillview Secondary School.

Constructivist approach is highly encouraged these days. This is to encourage teachers to start to shift away from traditional approach and adopt relational approach to teaching which is informed by conceptual understanding of learning. Teachers are informed to remain the facilitator of learning and classroom is encouraged to be a learner –centred (see Getting Practical Classroom learning Approach). Often teachers make use of interactive charts and internet resources to supplement textbook knowledge. They also make use of their lesson plan to teach scientific proofs. Group works in which students share knowledge and device their solutions are encouraged. The idea is that students are following problem based learning

in which they have to construct their own conceptual understanding of theories and learning.

However this expose the student potential and teacher’s potential to device critical reasoning.

Procedural approach was extremely difficult for learners because it informed that to cram methods and learn steps to calculate. They did this with no critical understanding of what is the meaning for doing it. The teaching approach informed by constructivist approach is too liberal and too dependent on student conceptual understanding of knowledge so it is open t criticism of learning approach used for each and every classroom activity. Teachers must respond to questions, demonstrate and also design lesson plan which seek to allow for participation of student to ensure they construct their meaning of learning. That is in line with relational approach. Their level of understanding of work is examinable and they can trace methods using models and resources to prove understanding. These theoretical understanding are the ones advocated by Piaget, Vygotsky and van Hiele . They differ slightly in meaning but they all refer to cognitive development of student.

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Step 4: Clearly Define Terms and Concepts. ...

Tangent: line which touches a circle or ellipse at just one point. Below, the line is a tangent to the circle when it goes outside and have one point to touch the circle once and share a point with the interior of a circle. Do not confuse a secant to a circle or chord to a circle to a tangent.

The essence of Sir Maphumulo’s theory is the assertion that the radius to the point of tangency is always perpendicular to the tangent line.

A cylinder: is one of the most basic curved geometric shapes, with the surface formed by the points at a fixed distance from a given line segment, known as the axis of the cylinder. The shape can be thought of as a circular prism. Both the surface and the solid shape created inside can be called a cylinder.

Circle: is a sphere usually drawn in a two dimensional plane surface so that there is a circumference to go around the fixed point about 360 0 to produce a circumference perimeter like distance that goes around the centre. We tend to forget that the line from centre to circumference is radius. Sometimes we confuse diameter to any two set of radii as if any two radii are radius … when a diameter is just a straight line that runs from circumference of a

circle to the centre of a circle and then to the circumference across the opposite side of a circle again. 

Step 5: Define the Population involved. ...

Student, teachers and future generations are placed at risk for consuming a theory which lacks basic instrumental proof. We often state that tangent is perpendicular to radius with no sound engineering proof to demonstrate that to our learners. Academics often get into trouble to forget recalling this fact when they make geometry proofs. This cost student time and lectures valuable energy because it demote their intellectual capacity to students. This is a worldwide academic problem because all universal institutions including universities and high schools consume tangent theory without scientific proof to demonstrate how it is perpendicular to a radius.

1) Constructivist theory denies instrumental approach from educators who introduce topics for the first time and prefer constructivist approach in which learners participation is planned and implemented.

2) Sir Maphumulo Got a question from a Hillview Secondary School from explaining how faithful he was to his learners to proceed using a theory which has no strong scientific proof without challenging this idea 3) Mr Mahlabela , Mr Weich , Mr Tohey , Mr Gorban and Mrs Rowlath taught many Euclidian geometry cases which involved corollaries and area using both textbook knowledge , online aids such as geogebra and interactive charts 4) Both university and secondary school kids are so curious to uncover why the tangent is perpendicular to a radius. Teachers including Sir Maphumulo were fascinated to uncover the truth.

How have we overcome this problem?

Key role 1: player is Mr Gorban and Mr Weich who taught the volume and area of a circle using constructivist approach in which he took the actual cylinder box and separate it apart in front of all students. The aim is to demonstrate how the concept of area of a circle and volume of a circle came to exist. Learners observe what makes up an area and also what makes up a volume of a cylinder. A cylinder is one of the most basic curved geometric shapes, with the surface formed by the points at a fixed distance from a given line segment,

known as the axis of the cylinder. The shape can be thought of as a circular prism. Both the surface and the solid shape created inside can be called a cylinder.

Key role 2: is Mr Phatha Mahlabela who designed a geometry investigative research program which sought to make learners discover Euclidian geometry conjectures in which several corollaries and co-ordinate geometry were used to illustrate various mathematics proofs.

Corollaries where proved to students where they used their deductive reasoning to explain reasons behind mathematical phenomenon such as formation of isosceles triangle inside the circle , the right angle along the diameter of a circle and also the fact that why chords subtends equal angles and the corollaries .

Key role 3: We cannot exclude a student who questioned Sir Maphumulo why he persist to use a theory which has no sound proof concerning that tangent is perpendicular to a radius or diameter.

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Step 6: Develop the Instrumentation Plan. ...

A) A lesson plan of Mr Gorben and Mr Weich used to illustrate the area and volume of a circle at Hillview Secondary School.

B) Similarly Mr Phathamahlabela also demonstrated that the area of a circle and volume.

C) The Euclidean Geometry and Investigative task in which protractor, ruler and scientific calculator skills are used to demonstrate to learners.

D) Learners participation in which key objective and skills to be covered are outlined in Sir Maphumulo and Mr Gorban lesson plan

E) Geogebra and Some key lessons from Mrs Rowlath

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Step 7: Collect Data. ...

1) Students, University Lectures and Secondary School Teachers represent data source

2) Books including university textbooks, academic journals and CAPS book including Ace it and Jika Imfundo are used to enhance student query.

3) Mr Mahlabela’s 2018 PGCE –FET, 2018 EDMA604E2 Mathematics Teaching 401

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Step 8: Analyze the Data.

Question of Research This research questions why the tangent is perpendicular to radius or diameter. We often use this theory to prove that the angles between chord and a tangent through one the end points of chord is equal to the angle in the alternate segment. However student’s mostly White students laugh at teachers because it tells their common sense that there has to be a scientific proof of why the tangent is perpendicular to a radius or a tangent even before that theory is used to make proofs of tangent-chord theorem. The question is around how one use an assertion whose underlying base lacks scientific proof as a proof to other theories. The fact is that we often produce a line from the tangent common point of contact to pass through to the center of a circle as if that produces a line that shall remain perpendicular to radius or circle.

Therefore the aim of this research is to prove that the tangent is perpendicular to radius or diameter. It is the first time in the history if anyone ever tried to do it so you have to pay careful attention.

Measurment Priorities : Recall that there is a theorm of Pythagoras which assert that the square of sides of the longest side of a righnt angle triangle line called hypotenuse Is equal to the sum of the squares of the two sides that are adjacent to the right angle opposite to the hypotenuse. Secndly remember that there is Cartesian plane which allow one to draw vertical lines and horisontal so that they are called y-axis and x-axis respectively. Now We need to produce these two lines so that they remain parallel along two identical circles using ruler , protractor and pencil . Remember to note the co-ordinate points of the points of a circle when you have less ability to deal with co-ordinate geometry when it is integrated to eucludian geometry . Remember that imaginary lines that run parallel to vertical lines and horizontal lines can produce either alternate or corresponding angles . I n this regard we are talking about lines parallel to y- and x- axis .

Two circles are drawn along the x-axis so that they share a common point of contact. These two circles are flipped together to share common point so that an imaginary line is drawn to run parallel to y-axis. This straight line becomes a common tangent to two identical circles.

Data Analysis :

Y –Axis

X-Axis

The two identical circle circle centered at C and also another circle centered at A as drawn above.

Note that line BE is parallel to Y –axis .

Two circles are identical , meaning they are equal in terms of radius and circumference around their fixed points.

Take a look at those two triangles which are formed by joining the line from centres of identical circles to the apex of imaginary line BE .

Required to prove that line BE is a tangent to both circles with centres C and A. Ignore any other feuteures such as co-ordinates points or function equation of these two circles .

Choose any acute angle between o – 89 ( preferable 45, 60, ) to be constructed from centre of the circle to meet any point of line E, such that line EB is either reduced on increased . Take note that the angles with same value in terms of degrees meet from same apex point along the line ED. You must construct angles from centre of cirlce from both centres of those circles to take any equal values so that they meet at any point of line EB .

Ststement :

Reason :

In Triangle EBA and Triangle EBC BA = BC

- Construction

EB =EB

- Common side

Angle EAB = angle ECB

- Construction

Therefore triangle EAB is congruent to triangle ECB

- Side , Side , Angle (SSA).

Deductive Reasoning : Evaluate the following : Line ABC is a straignt line Therefore : angle ABE + angle CBE = 1800 But angle ABE = angle CBE ∴

i.e.: 2 ¿x 2< x 2 ¿x

¿ AEB=¿ CEB = ¿ x ¿x

+ ¿x

= 180

= = 900.

180 2

= 1800

- sum of a straight line - ∆ EAB ≡

∆

ECB

That means angle ABE = angle CBE = 900.

But Line AB and Line CB are radii.

Then if Line EB shares common point of contact to both point circles with radius AB and CB. But this are identical circle , where Line EB is perpendicular to both radii given by line AB and Line CB. Therefore line EB is a tanngent to both circle with both radii namely AB and CB to remain perpendicular to EB. Therefore a tangent is perpendicular to a radius , since both common tanget line EB and radii of line AB and CB meets at right angles equal to 90 0.

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Step 5: Interpret Results.

What if the two identical circles are drawn to lie not parallel to the y-axis … however to find out those identical circles still have common tangent ? Is that common tangent between them not any of Sir Maphumulo’s features of identical circles whose common tangent reflect to the one depicted by an imaginary line to prove that tangents are perpendicular to radius . Refer to a graph below and see culcuations that Sir Maphumulo developed .

Can you test yourself here?

First step you are not required to culculate the function of these two graphs however … notice that they are both equal . You may also know that these are not circles centred at origin . These circles are centred away from origin … in a professional mthematical language we often say these are circles centred off origin .

RTP :Required To Prove?

Required to prove that circles that touche one another to share common tangent have their radii perpendicular to the common tangent. ( Tangent

⊥ Radius ).

Hint : Join these two circle radii and form a line tht go intersect with y-axis …not tht you have created a straight line with the y-intercept. (hint y = mx+ c ).

Proof : Draw a Common tangent between circles ..assume it touches common point B once .

Statement : Straight line gradient M =

Reason: y 2â&#x2C6;&#x2019; y 1 x 2â&#x2C6;&#x2019;x 1

straignt line Gradient

Given co-ordinate point of circle centres A(-4:4) and B( -8: 2)

(x1: y1) and (x2:y2) M=

2−4 −8−(−4)

−8+4 2−4 ¿ ¿

=

=-2/-4=1/2

1 2

M=

y = mx + c

¿ 1 2

For y =

1 2

x+c

x+c

B( -8: 2) -

(x : y) 1

Y= 2 x + c

2=

1 2

substitute B( -8: 2)

(-8) + c

2= -4 + c C = 2+ 4 = 6 c = 6,

therefore y =

1 2

x+c =

1 2

x+6

1

Therefore y = 2 x + 6.

Make a deravative of any circle eqution Circle equation : (x-a)2 +( y-b)2 = r2.

Hint this is a version of a distance equation given A (-4:4) a centre of one of circle ‫( ؞‬x-a)2 +( y-b)2 = r2.

A (-4:4)

(x:y)

‫( ؞‬x-a)2 +( y-b)2 = r2.

(a: b)

‫( ؞‬x-(-4))2 +( y-4)2 = r2. (x + 4)2 +( y-4)2 = r2.

But r = √ 5…. Use a distance formular to prove that D2 = (x2 – x1)2 +( y2-y1)2

A(-4:4) B ( -6: 3) (x1:y1)

(x2: y2)

= (-6 –(-4) )2+ (3 – 4)2 = (-6+4)2+ (-1)2

D2= (2)2 + (1)2= 4+1 = 5 D 2= 5 Therefore D = √ 5

because radius cannot be negetive ( leave out square

differences ).

Sir Maphumulo argued that other learners can choose to use hypotenuse or gradient paths to determine the radius of circles . it fine … but all answers return the value of 5 units . e.g : look at the grid pattern … it is allowed.

Then look distance of a radius of circle centred A(-4:4) and the point at the circumference B( -6 : 3) .

Acknowledge that gradient of these two point = rise / run = 1 divided – 2.

Teachers must help learners to discover how to do gradient by construction so as to discover the relationsip above ..

You may also use number line to expose this relationship

Using these two points together with Pythagoras theorem .. we can determine the radius of the circle Therefore r2= x2+ y2 r2= 12+ (-2)2

in this cae rise = 1 and run = -2 …. Apply your deductive

reasoning here on point B( -6 : 3) and A(-4:4) .. where A B is a radius or simple make drawing and watch Cartesian so that you take a quote of a gradient using a triangle to illustrate r2= 12+ (-2)2 r2 = 1+ 4 r2 = 5 r=√ 5 What next ?

You have to be careful this time around .. you will not construct two points or make any angle because a tangent has not been placed parallel to y-axis . It is not an imaginary line of y-axis as you may have seen in the other tangent proofs earlier .

Ask yourself what should you have done to prove that common tangent like the one shown above is perpendicular to a radius NB you know all function of ciecles and given that they both have equal radii = square root of 5 units ( r=√ 5 units ) .

Solutions The product of the gradients of perpendicular lines is equal to getive one If gradient of line AB is perpendicular to Line GH then their radients(Gradient) have a product = -1

Let G stand for gradient Therefore GAB x GGH = -1

you may look some other maths book for such

algebraic –eucludian geometry theory in your other spare time GAB x GGH = -1 1

GAB = 2

use any two points to prove that but also remember that straight 1

line gradient DE was proved to be equal to 2 We assume that GAB x GGH = -1

If so, then

GAB x GGH = -1 1 2

‫؞‬

x GGH = -1

Solve for gradient of line GGH

1 2

x GGH = -1

2 1

x 2 x GGH = -1x 1

GGH =

1

−2 1

2

= -2

Therefore gradient of line GH is GGH = -2 Does that mean we have proved a theory ?

Answer = No In order to overcome this we can adopt two approaches

It either we introduce parabola to turn around points B ( -6: 3) or determine the y – intercept of a straight line GH .

Let look at straight line GH indepth y = mx + c y = -2x+ c

since gradeint of line GH is GGH= -2

let manipulate a gradient formular to outfit our culculations m( x2 – x1) =

y 2− y 1 x 2−x 1

× x2 – x1

m( x2 – x1) = y2 – y1 change position sothat y takes the RHS since y = mx + c . m( x2 – x1) = y2 – y1 y2 – y1= m( x2 – x1) switch off figures from y2 and x2 and leave only x and y as shown below y2 – y1= m( x2 – x1) y – y1= m( x – x1) y – y1= m( x – x1)

********** use any points to determine y – intercept

What is y- intercept? i.e. a point at which a straight line intersect with y – axis .

In this case you may expect a negative y –intercept … hint look at the direction of a tangent y – y1= m( x – x1)

use points on the straight line GH

B( -6:3) ( x: y)

Therefore y – y1= m( x – x1) y – y1= m( x – x1)

B( -6:3) and let x1 = -6 and y1 = 3.

y – 3= m(x – (-6)) y -3 = m(x +6) But M = G = -2 as per our assumption and also as per our calculations perfumed above y -3 = m(x +6) Therefore: y -3 = -2(x +6) Write this in standard form or plot this on a graph and see if it does not fit to be a tangent .

Standard form of straight line is given by y = mx + c y -3 = -2(x +6) is y = -2(x +6) +3 y = -2x -12 + 3 = -2x – 9 Therefore y = -2x -9 If you draw this line against the two circles it seem to show that y –intercept = -9.

To prove if this is a tangent mark the y – intercept to be point Q and label the co-ordinate points of Q Even a grade 8 child can tell that co-ordinate of Q are (0: -9) What do you think we can do to eliminate assumption that this line GHQ is a tangent to both circles?

Assume EQ is a hypotenuse, the longest side of right angled triangle Hint use Pythagoras theorem to see if distance (EQ)2= (EB)2+ (QB)2

(EQ)2= (EB)2+ (QB)2

Assumption by theorem of Pythagoras

Look at the graph: remember how helpful the number line is. So add the distances NB there is no negative distance EQ= 6 + 9 = 15 But (EQ)2= (15)2 EB2 = (x2- x1)2+ (y2- y1)2

use co-ordinate points E (0:6)

and B (-6:3) (x 1:y1) and B (x2:y2)

EB2 = (-6- 0)2+ (3- 6)2

= (-6)2+ (-3)2= 36 + 9 = 45

EB2 = 45 EB = √ 45 Similary (BQ)2 = (x2- x1)2+ (y2- y1)2

where B( -6:3) and Q( 0: -9) (x1:y1) and B (x2:y2)

(BQ)2 = (x2- x1)2+ (y2- y1)2 (BQ)2 = (0- (-6))2+ (-9- 3)2 = (0+6)2+ (-12)2 = 36 + 144 = 180 ‫(؞‬BQ)2= 180 ‫(؞‬BQ)= 180

‫(؞‬BQ) = √ 180

(EQ)2= (EB)2+ (QB)2 It is now a time to exemine if our assumption that common tangent is perpendicular to radius We have LHS and RHS LHS = (EQ)2 and RHS = (EB)2+ (QB)2 If LHS = RHS it means that triangle EQB is a right angled triangle What is a right angle triangle? …this is a triangle with one interior angle equal to 900.

This becomes an angle that remain opposite to the hypotenuse. We then revisit the Pythagoras Theorem which argue that: In a right triangle the area of the square on the hypotenuse is equal to the sum of the areas of the squares of its remaining two sides.

Results : LHS = (EQ)2 and (RHS)2 = (EB)2+ (QB)2 If LHS = RHS it means that triangle EQB is a right angled triangle , meaning in a

∆

EQB angle EBQ is a right angle.

(EQ) = 15 But (EQ)2= (15)2 = 225 RHS = (EB)2+ (QB)2 EB = √ 45 But (EB)2= (√ 45 )2 ‫(؞‬BQ) = √ 180 NB there is no diference between BQ or QB if naming a line

‫(؞‬BQ) = √ 180 But (QB)2

= (√ 180)2

Is LHS = RHS ?

(RHS)2 = (EB)2+ (QB)2 Proof LHS = EQ)2= (15)2 = 225 Proof RHS = (EB)2+ (QB)2

= (√ 45 )2 + (√ 180 )2

)

= 45+ 180

(hint

exponents) = 225.

The LHS = EQ)2= (15)2 = 225 Therefore triangle EQB is a right angled triangle, meaning in a

∆

EQB angle

EBQ is a right angle.

Theorem of Pythagoras argue that in any right triangle the area of the square on the hypotenuse is equal to the sum of the areas of the squares of its remaining two sides.

The situation our credentials are giving born to a new proof that a radius is perpendicular to a tangent and vice versa because we used Pythagoras theorem as our prior knowledge.

Congratulations to Sir Maphumulo a UKZN PGCE student for having successfully completed this theory.

That means that tangent GBQ is perpendicular to AB .

Therefore a tangent is perpendicular to radius Therefore common tangents are perpendicular to radii.

NB Sir Maphumulo added that there would be alternative methods to prove that tangent is perpendicular to radius following his Theory other than the one he used. For example he mentioned the trigonometry, Euclidian geometry and the gradients of the straight lines.

Was that theory a theory or hypothesis?

This is a theory because it eliminated assumption when the straight line of a tangent was proved to be perpendicular to a radius and hence radii. Assumption was totally eliminated when one also test if a triangle formed was a right angle using the theorem of Pythagoras.

We can see that tangent was not planned but it was drawn using the free hand.

To calculate the y-intercept Sir Maphumulo used gradient formula as assumption and it was successful because the straight line did remain perpendicular to radius. Beside that assumption eliminated error of free hand drawing when it is drawn in the graph of grid pattern. It was necessary to know the centre points of the circle and the other things such as distance radius covers and those things are given. But you should discover yourself whether the line that touches the circle in their common point of contact is a tangent and then if it is perpendicular to radius. There are clear evidence that Sir Maphumulo adopted a culculus method to determine gradient of the straight line that maintain the common point of contact of the two identical circles. NB even if it

not identical circles you should be reminded that the radii are perpendicular to a tangent. Or common tangent at their common point of contact.

In the future editions Sir Maphumulo is expected to further illustrate what is it that he was trying to say when he mentioned that one could make use of parabola to prove that tangent is perpendicular to radius in his theoretical drawings.

The end:

Extra Section : The Statement of Sir Maphumulo’s Theory The tangent is perpendicular to radius The common tangent to circles remain perpendicular to their radii irrespective of the size of their circles . However we adopt identical circles to prove that theory . We can also make use of parabola graph together with Sir Maphumulo’s proofs to rectify that tangent is perpendicular to radii.

The underlying theories and collaborators 1) Pythagoras Theorem 2) Mr Mahlabela’s Co-ordinate conjectures and Piaget and Skimp 3) 4) 5) 6) 7) 8)

ideas Co-ordinate Geometry Straight line formular of a gradient manipulation Culculus understanding of a gradient Parallel lines and Perpendicular lines Van Hiele’s level of Thinking ( Geometry Thinking) .

Hillview Secondary School Mthematician Masters as mentioned

suchs as Mr Gorben , Mr Weich and Mr Tohey ) 9) Mrs Rowlath and His UKZN Lectures 10) Sir Maphumulo Himself and His Geogebra

11)

Key students who asked questions concerning why tangent

theory is used continuously without a proof 12) Social Science Educators such as Dr Masondo and EPDP 610-630 educators of PGCE UKZN 2018.

Fill In the missing words Sir Maphumulo has thus proved the theorem that Tangent is ……… to a radius .

No one expected that co-ordinate geometry can be used together with Eu………. Geometry in that way . It was exciting to se his two identical …… which were used to illustrate that two identical circles can share a common tangent . Straight line was manipulated a little bit using understanding of culculus . Therewas an error correction made about a straight line of a tangent following the culculstion of the gradient and y-intercept . The tangent wasline was not fixed because it was laid down on the basis of the assumption that this line is a tangent . However the co-ordinate point of gradient common contact to the identical circles were given as shown in the grid pattern made os system of squares. You should know what is given and what is missing . Yous should also be able to know what is an assumption and what is proof . The assumption is given statement and the hypothesis is statement that tangent is perpendicular to radis prior to actual discovery that radius are standing next to right angled triangle as shown. Triangle was not planned but we use prior knowledge that gradients of perpendicular lines have their products equal to -1. However by doing that you did not send and graph to restore a tiangle that remain perpendicular to radius or radii. This meant that a line drawn between the common contact of a circle was a tangent and it is perpendicular to radii . Does that means that any two circles to share common poit of contact to their tangent as common tangent are perpendicular to radius? This is true for any form of circle drawn . But does that means that any line passing the common point of contact of circles about their circumference and one fixed points are

perpendicular to radius . Answer to that is no? Whay not ? it is No because the common point of contact has to be perpendicular to radius . A line can pass to a cirlce common point of contact to found out that it is a secant . Visual skeletal of circle and lines drawn as tangent cannot be termed tangents unless there is a proof for that or something is given . Even Sir Maphumulo hiself can agree that he had a serious problem prior to his ability to prove that tangent are perpendicular to radii . Indeed we can see that he adopted some other theories and other conjecutures used in algebra. We cought him call his case a algebraic geometry or algebraic euludian geometry or agebra trigonometry .

Indeed that exose him as a person on an aeroplane not totally landed. However his theory is not hypothesis because it has been proved and his proofs are totally correct. He has ability to even list other people who influence his theories and even name the problems outlined by Hillview Children about tangent and radius theory . That expose his great endeavours about the need of his theoretical development . He used assumption to develop theory . Eventually he eliminate assumption and even errors of handwriting about a tangent . This are true meaningful systems of making a theory . A theory does not situate him at centre of knowledge instead it prioritise the student worries about geometry . It also centralise Pythagoras theorem , constructivist theorist and even his university teachers . Indeed this theory did something so perfect Can Sir Maphumuloâ&#x20AC;&#x2122;s theory be used locally , nationally or globally ?

Sir Maphumulo set people free to use this theory because he was willing to let nations gain confidence when they deal with geometry cases and culculus . He also draw many examples from geometry and algebra.

Where can we use Sir Maphumuloâ&#x20AC;&#x2122;s Theory ?

It can be used at Universities , Schools and all level of teachings . Engineers and Scientist could enjoy using it for building of rocket , aeroplanes and cars. It can be used for Culculus .

Challenges paused to Sir Maphumulo by those who cross exemined his theory Challenge Number 1.: It was mentioned that your theory can work with parabola , where and how could you set this to work ?

Response : Take for example y = x2……. this is hyperbolic function where it is very easy to test derivative of a radius . If this is drawn to the Cartesian plane , we can also derive its gradient using culculus . Following then we have ability to know the equation of a tangent to the curve . This open student engineer’s mind about the move towards deriving the tanegent gradient via curve graph function. Indeed the curve graph function hehave the same as circle . This exposes what influenced a theorist to even think about gradient of a tangent or secant to circles and curves .

Consider if you are dealing with two identical circles such as (x+8)2+ (y-2)2= 5 and (x+4) + (y -4) = 5 .

These are two circles which differ in terms of location but with similar radii.

Similar shapes of structures does not same function statistically however method for their function may remain the same . Let us redraw these two circles

so that we see how their tanget and secants respond to the culculus considerations .

These are two identical circles to shar ecommon point of contact at (-6:3) The equation of a given parabola is y = (x+6)2+ 3.. we did that with understanding that parabola must share turning point with common point of contacts of circles .

Once again we do not need to know the equation of a straight line …but we must discover it .

Verify any two points to which you can draw a sraight line to pass through circle common point of contact and the curve graph .

Unig the grid pattern or geogebra you will see that those points are (-6:3) & (8:7) You are not forced to choose which form of parabola to share point with common point of contact of two circles …even the hyperbola or exponential grap or graph of trigonometric function can do the same thing The gradient between two points chosen is M = 4/-2 = -2 y = -2x+c y = -2x+c using points (-6:3) ( x:y) 3 = -2(-6)+c 3=+12 + c 3-12 = c C = -9 Thereofre y = -2x+c = -2x – 9 y = f(x) (x+6)2+ 3= x2 + 12x +39 Derivative of y’ =f(x’) = 2x2-1 + 1(12)x1-1 +0(39x)0 y’ =f(x’) = 2x+12 The derivative gives a stupid straight line which moves very far that the gradient point of contact But why is is going to be useful later

It useful when you change the gradient into negetive form Change y’ =f(x’)= 2x+12 Into y’ =f(x’)= -2x+12 Notice that the line go to form a straight line which more or less parallel to a secant Parallel lines have equal gradients Indeed y = -2x – 9 and y’ = -2x+12 have same gradient so these are gradients of the secants Let us now approach point y’ =f(x’)= -2x+12 We move with c = 12, c= 11, c=10….,…., … c= -9 Therefore y’ =f(x’)= -2x+12 y’ =f(x’)= -2x+11 y’ =f(x’)= -2x+10 y’ =f(x’)= -2x+9 y’ =f(x’)= -2x+8 y’ =f(x’)= -2x+7 y’ =f(x’)= -2x+6 y’ =f(x’)= -2x+5 y’ =f(x’)= -2x+4 y’ =f(x’)= -2x+3 y’ =f(x’)= -2x+2 y’ =f(x’)= -2x+1 y’ =f(x’)= -2x+0 y’ =f(x’)= -2x-1 y’ =f(x’)= -2x-2 y’ =f(x’)= -2x-3 y’ =f(x’)= -2x-4 y’ =f(x’)= -2x-5 y’ =f(x’)= -2x-6 y’ =f(x’)= -2x-7 y’ =f(x’)= -2x-8 y’ =f(x’)= -2x-9 You should have noticed that the graph come close and closer to the gradient of the secant and they occupy common position.

Take notice that y = 2x + 1 form another tanget to a corcle of (x+4)2+ (x-4)2=5 Similarly we can also discover other tangents by mocing towards these directions y’ =f(x’)= -2x-10

y’ =f(x’)= -2x-11 y’ =f(x’)= -2x-12 y’ =f(x’)= -2x-13 y’ =f(x’)= -2x-14 y’ =f(x’)= -2x-15 y’ =f(x’)= -2x-16 y’ =f(x’)= -2x-17 y’ =f(x’)= -2x-18 y’ =f(x’)= -2x-19 Note that y’ =f(x’)= -2x-19 is yet another tangent of circle of (x+8)2 + (y -2)2= 5 Eventually you can see that we have discovered three tangents viz 1) y’=f(x’)= 2x + 1 2) y’ =f(x’)= -2x-9 3) y’ =f(x’)= -2x-19

This time around things are changing slightly because you have to look if lines of radii are not perpendicular . You will cilculate each gradiant and then use formula M1xM2 = -1 Therefore LHS = M1xM2 and RHS = -1 Let M1 be gradient of a line to which identical circles share their common points If so then we are talking about seant of y = -2x-9 because it goes passes (-6:3) a point where identical circles meet ( you may download circles or redraw to see).

We got many co-ordinate options as are (0:-9) , (-6:3) and (-8:7) to shoose from for a gradient of a secant that lies nearly straight to the point of common point of contact of those two circles .

Msecant = M1 =

y 2− y 1 x 2−x 1

(0 :-9) ,

(-6 : 3)

and (-8:7

( x1:y1) 3−(−9) −6−0

Msecant = M1 = M1 =

(x2 :y2)

(3+9) −6

=

(12) −6

= -2

Let M2 = straight line between radii , so join then to intersect with secant line .

You know all co-ordinates of the straight line by centres of circles incding the point of common contact of circles and the secant These are ( -8:2) , ( -6:3) and (-4:4) all the way to (0:6) M2 =

y 2− y 1 x 2−x 1

( -8:2) ,

( -6:3)

(x1:y1) (x2:y2) M2 =

3−2 −6−(−8)

=

−6 +8 1 ¿¿

1

= 2

1

M2 = 2 LHS xRHS = -1 then line of a secant is perpendicular to radii and diameter Therefore also that line is a common tangent to a circle .

LHS = M1xM2 M1= Msecant = -2 M2= gradient of tsraight line between two radii and their point of contact = = 1 2 1

LHS = M1xM2 = -2x 2 = -1 But RHS = -1

**Therefore lines are perpendicular â&#x20AC;&#x201C; the product of gradient of perpendicular lines is equal to -1 ( negetive one ).

This means that line of a ecant is perpendicular to radii or diameter of two cirlces with common point of contact about their circumference .

Therefore a secant to a curve is a tangent to those two circles Can you see how we can make use of culculus to also derive whether tangent to circle is perpendicular to radii. Indeed in three ways we have seen how Sir Maphumulo proves that the tangent to circle is perpendicular to radii.

Then if LHS = RHS then the two lines are perpendicular Your guideline would be points of a parabola of a graph function of y = f(x) = (x+6)2+ 3 Unlike Mr Mahlabela and Dr Dowlath Sir Maphumulo could not set up any geometric sequance and series but you can see that some level of sequance and series or geometric progression prevail with this some . Actually Sir Maphumulo had clear mind about need to eliminate assumption â&#x20AC;Ś hence these lines have eliminated assumption so that it becomes a theory that tangent to the common point of contact of tangent are perpendicular to radius or diameter of circles.

Challenge Number 2 to Sir Maphumulo What could happe with your tangent to circle radius relationsip if you derive any of equation of the circle ?

The answer is subject to what happened with the derivative of a parabola .. it gives you equation of a straight line that must be reflected about its gradient to form a line that remain parallel to all tangents of the circles with equal gradients

The circle derivatives are likely to emain similar to their tangents For example : one circle has an equation of (x+4)2+ (y-4)2=5 y = x3+13x2+40x+80 NB (x+4)2+ (y-4)2=5 a function of x3+12x2+y2+ 48x+8y+80 =5 is same as function of (x+4)2+ (y-4)2=5.

It extremily difficult to bring about derivative of the graph like this unless you meaning of circle completely . 5 is radius and not the function of a graph given by the whole equation itself . Functions are alwas defined with respect to y sothat you have f(x) .

So we may need to change the entire equation so that y takse the subject of a formula.

Therefore (x+4)2+ (y-4)2=5. (y-4)2= 5 - (x+4)2 This becomes y = √5 - (x+4)2 +4

and this becomes a semicircle

equation .

Then y = f(x) = √5 - (x+4)2 +4 y =f(x) = (5 - (x+4)2 )1/2 + 4 y = 51/2 - (x+4)2/0.5 + 4 = 51/2 - (x+4) + 4 = 51/2 - x-4 + 4 = 51/2 = - x + 51/2 Y’ =f(x’) = -1 Or y =f(x) = 51/2 - (x2+8x+16 )1/2 + 4 = 51/2 -(x+4) +4 = 51/2 -x-4 +4 = 51/2 -x = -x + 51/2 Y’=f(x’) = -1 is a derivative of y = √5 - (x+4)2 +4 Likewise we can take the inverse of y = -1 which is y = +1 and move closer and closer to circle tangent points as follows

Y = -1 Y=0

Y =1 Y=2 From that derivative we can see that y = -1 is a parallel line about x-axis We can move from y = -1 to y =1 and y =2 and y =3 and y = 4 Lest us close by looking at derivative of y = (x+5)2+ 2 y= x2+10x+25 +2= x2+10x +27 y=f(x) = x2+10x +27 yâ&#x20AC;&#x2122; = f(xâ&#x20AC;&#x2122;) = 2x+10 Therefore by deriving using function of a curve you will get a radient of a secant . However the derivative of a function is equivalent to the gradient of the tangent of a curve to a curve graph . That means that we can either add or subtract constant term to see where does a tangent lies against graph . It is useful to also observe how the lines move together to form tangents while also forming right angled triangle . Once that is done you may need to prove if the triangle is aright angle Challennge Number 3 Would you also use trig ratio to prove tangent theorem . Actually that is where it started because it began with co-ordinate geometry . You can simple determine the angle between points of a triangle and conclude if the triangle has a right angle . You can also use Cosine rule , Area rule or Sine rule .

Where else could we apply Sir Maphumulo Engeneering and Aiport operations . In designing of computer cases , since it involves culculus . To analyse the flow of current . To develop airline and other navigation scheme . To help in GIS and ICT with planning of roads , aiport and fleight scheme . You can see how the lines are moved using geogebra and real

culculations .. so you can use them together with exponents , hyperbola and even algebra and egeometry to design heavy mechanical structures . You can use it in geography to analyse gradient and valley .

What are hidden cost or profit of getting started with its use .

You must be computer literate and you must somehow posee mathematics and physics knolwledge . It can create chaos for complex universal structures to be built . It can be solution to eliminate traffic on roads and syncronization of robbots. In this case we may refer to a case of SA Pinetown where pedestrian cross very difficult with unplanned city roads along the line of cars . We can plan how we could twist and tur a traffic bridge to allow to be elevated above the ground so that cars flow smoothly .

What are your dreams other than universal view of your theory ?

Indeed I am looking to design a universe like earth . I have a vision and I have capital so I have no worry to test some Physics theory of motor and wind interaction to valley . I have a vision to create inanimate brain using this system . In USA and UK and China we have seen marvellous elders creating a brain to interact to environment â&#x20AC;Ś so why we as African should fail to use our Models to do so .

Have you ever Triend Anything similar to this ?

Indeed I did constructed a wind like model to depict direction of air , ponetial to rain and temperature measurement .

They say you are studying teaching â&#x20AC;Ś are you a model of classroom ?

Classroom times are not my types .. I see myself joining mechanical engineering department perhaps in USA , or UK or China where I should begin to xpand my knowledge and thus become a multibillionare

You sound like a software developer … is it possible with your model Software development is very much possible in SA however I feel I need to be groumed in UK or USA so as to be sure what to release . I feel I have to maintain eye contact with powerful scientist and engineers outside of the world .

Can you go down on yourelf and make avision of your model live and be lived ?

I am not a true believer .. I am a scientist and not a witch … however I trully believe that a world united can share a vision . I have did my best and now is the time for people in GIS and IT, Sciene and Engineering field to study careful from what has been discovered to carry this visions ahead .

How do you think engineering will change from your edvancement ?

Especially when you look at how engieering suffer from twist and turns of lack of integration between maths and science .. you can see that Sir Maphumul’s theory is overloaded with solutions to science vision . For example he overcome tangent and radius but also speak of aiport planning and traffic congenstion . At the same time he speak of culculus and even apply it corectly . This opens the universal chhannel never known to human life . It encourage youth to study further and explore science software beyond their thinking capability . It might take time but with injection of the new theory .. their analytic skills can never be compares to past generations What software can you think of ?

I think of marine system to monitor biodiversity and predict what underline their presence and co-exstence . We can reveal some pattern of creation of some entity such as motive for ants for digging soil . There We discover that soil can also be turned acidic . That has potential to poison the entire sorrounding environment and eventually crack the cliff walls . When that cliff fall apart new

rock are exposed and it becomes a habit of snakes , hyracks and even people . .

So ants a projected to carry such beautiful task even before the cliff come to real . The question is what is the need to create such big holes behind the rocks .

Each species is important in the environment as it help universe to endure over a long period of time .

Sir Maphumulo then watch at the lizard and see it habits to like solar energy .

He argued that we can anlyse the amount of heat and wave required by lizard to live . The lizard could happen to be a living animal specimen that sustain solar .

There is nothing to undermine . Then we look other mountain elevations and their architextural shape forms . Some are hyperbolic and some have right angles cit deep . These shapes interact to models and science design to produce response to antenna and even cloud formations . But all that is work of air temperature against the ground . Well we all know properties of air . And we also aware of what brings precipitation forces .

Can you share anything you wish in physics or engineering ?

Beutifuy I wish to invent somesort of trees and animals which could adapt to ecosystem . I could also like idea of creating new forms of computer with ability to behave as human .. to manipulate what is going on outsiede , in the space or ocean .

What is your model of a new universe ?

I feel it will begin with human brain evolution and them science ideas . It is certainly plausible with huma being but there has to be some lelev of observation and participation .

Do you like lizards and ants and why did you left fishes and or human being in your model ?

I like lizards and ants because I have discovered their major roles in expanding the biodiversity creation . As I told you about the ants function to create cliffs and deep cuts . I also like lizard when I watch it in the graph.. to look at it temperature and organise the function of its presence in the universe or habitat to which it leaves .

What else do you think we may expect out of yourtheory in the neafuture ?

I may expect crazy system of engineering , crazy road and air transport plan , crazy animal creation and crazy engineering and astronomy plans ?

Your words of closure ?

Sky is unlimited keep on searching and look up to theorist such as Isaac Newton , Pythagoras theorem , Piagent and Skemp tyep ideas etc .

What do you think has changed mathematics thinking ?

It is not autonomous to deal with that quastion . It subject to the history of how you individually has been shaped by environment . As you can see there has been contrasting view of how maths is dicoverd as some think it is internal and some think it is external . That is subject to interpretation ?

What on your point about maths changes ?

I feel it level of dedication and power to observe . I feel the shift to adopt constructivist ways of teaching expose risks to maths such as Hillviw Kid question concering the use of tangent theory without proper proof .

But also I feel it ability to say to look to ancient theorist such as Isaac Newstons and Pythagoras theorem . You cannot change world without ability to look to others failures and successes . We all cheat one anothers strengths and weaknesses . We do that coincidentally and soemtimes intentionally .

The end

https://www.mathpapa.com/algebracalculator.html? q=2%5E3%20*%202%5E2 However this is not a tangent to a curve graph but it a secant .

However this is tangent to two identical circles .

The end

Some quotes of Sir Maphumulo

If tangent line was parallel to y- axis … they would have equal gradients However I lines are perpendicular then their product of their gradient = -1 Those two circles are still a true reflection of Sir Maphumulo’s Tangent theory because we can still reflect them now using the co-ordinate geometry , trigonometry and Theorem of pythagoras to remain true . Infact you should join the line sothat you produce a line to form into a straight line .

For millions of years ago , scientist and mathematician failed to prove this theory . Today the chains are broken at University of Natal –Edgewood campus. This theory was proven by Sir MN Maphumulo in year 2018. We congratulate all university students who joint South African Department of Education with love and compassion.

Schools may now begin to apply this theory when teaching tangent-chord theorem , the tangent –radius theorem and much more. NB Sir MN Maphumulo is the very same guy who also break chains about a theory of universal evolution in which he has accomplished the following theories of physics and science.

His most famous theories are

13) Dark and light carrying forces cannot occupy same space , meaning where dark carrying forces are dominant then the light carrying forces are removed or extinguished .

14) The brain evolution and Congruent Status of Human Brain 15) Chesterville Alternatve Wavesection 16) Megatheory of New Universal Invention 17) Banjura Era , Historical Theory of extinct group of people who ever lived 18) The theory which point out that new universe similar to earth will come to exist in near future where animal , people and ecosystem willl co-exist .

19) The Multipurpose Hyperbola Theory 20) The theory of universal evolution : the first and seond time 21) The human brain evolution theory NB all these theories have been proven correctly , this is such as senior theorist to ever exist on earth.

What holded previous gnerations from being able to prove tangent theories ?

Solutions to the question above : Teching style : Most teachers refused to adopt different approach from teaching . Teachers need to focus on different approach and just rely on one method of teaching . In the past traditinal classroom approch dominated.

However it limit student right to participate because it is techer â&#x20AC;&#x201C;centred .

Learners are not icluded to be part of solution. Constructivist approach is part of solutions as it make teachers to remain facilitators . Learners play their participative role where they engage not only with procedures and methods but also relational knowledge for doing things.

During the development of this theory lesson plans and lesson plans were drafted to teach learners . Each lesson plan has topic and content including objective and skills outlined for learners to cover. Physical instruments such s ruler , protractor , pencil and curdboard material were used. In professional constructivist classroom approach language we call that material of teaching.

Methodologies and methodologies were set to suit activities being carried .

Group works were encouraged .

Record of Works to Which thought of theory were imagined … you may leave a record or look if you are looking to take your research paper about logic and cognitive dvelopment .

COPIES OF WORK DONE: To Influence Sir Maphumulo Name of School:

ROUTINE INFORMATION Hillview Secondary School

Student Surname and Name:

Maphumulo Mthokozisi

Student Number:

207503878

Grade:

10, 11, 12

Subject:

Mathematics

Topic:

Area and Volume Calculations

Content /Concept Area:

Cylinder, properties of circle and perimeter.

CAPS page number:

199

Duration of Lesson:

45 minutes

Date

August 2018

Specific Aims: To develop learner’s skills for being able to name, describe and determine areas and values of geometric figures and shapes.

To develop learner’s ability to learn to question, interpret and convert statement of geometric figures.

To train leaners to be able to device ways of converting shapes forms for calculations beginning with 2D to 3D shaped cylinder and then the rest of other shapes.

Lesson Objectives: Knowledge

Learners should gain

Skills Learners should be able to do

Value /Attitude Learners should acquire

knowledge and

following :

values and attitudes

understanding of the

conducive to :

following : Define, determine and analyse

Draw, interpret and identify

Acknowledge ability to be able

area and volume using various

prism shape forms and trace

to

prism forms for illustration.

their

geometric shape forms and

area

and

volume

recall

properties

of

calculations.

their relevant application from

Describe the area, volume and

Analyse how cylinder area and

area and volume calculations.

Acknowledge ability to

perimeter of cylinder as case

volume develops.

determine volume and area of

study

to

understanding

of

cylinder

adopted

to

illustrations and demonstration

approaches cylinder.

relying

from

lying down their conjectures through observations.

Approaches / Teaching Strategies:

Role plays.

Case studies Class discussions.

Questioning.

Whole class-discussion Group Work Resources:Maps , orthophoto and topographic map) Calculators Chalkboard Newspaper Worksheet Geogebra and HP Introduction:To introduce this lesson I use a cardboard where it must be opened so that learners use it to develop their own understanding of area and then a volume. In this lesson I use rectangular prism box.

Learners will also receive open ended questions aimed at making them discuss properties and areas and perimeters of prisms and cylinder. In doing this learners develop ideas and share opinion required for volume and area formula calculations.

Prism Cylinder

Name Cylinder

Volume 2

V =Ď&#x20AC; r h

Surface Tsa Âż Ď&#x20AC; r 2 +

2 Ď&#x20AC; rh

Cube

Lxlxl i.e. area of

Sum of area of all

a base x height.

six squares 2l2 + 2l2 + 2l2

Triangular prism

Area of a base x

Sum of areas of

h

two triangles at top and bottom plus

the

three

rectangles Cuboid

or

rectangular prism

Hexagonal Prism

Area of a base x

Sum of the areas

h

of six rectangle

lxbxh

= 2lb x 2lb x2lb

Area of a base x

Sum of the area

height

of the base and three triangles.

Triangular pyramids

1 ( A x H) 3 A = area and

Sum of the area of the base and three triangles.

H = height

Rectangular Pyramid

1 3

(A x H)

Where A = area of base H = height

Sum of the area of the base and four triangles.

Hexagonal pyramid

1 3

(A x H)

A = area of a

Sum of the area of hexagon and six triangles.

base H = height Sphere Ie not prism nor a

4 2 πr h 3

4πr

1 2 πr h 3

π r h +πr2

2

polyhedron

Cone

Development: A case study of Cylinder A cylinder is a three (3) dimensional shaped structure with two circular lids.

Given any cylinder with both vertical and horizontal plane on which 2D = horizontal plane and 3D = Vertical plane as shown above.

Suppose Radius (r) = 7 cm and Height (h) = 20 cm.

1) Use a grid pattern to indicate a cylinder any way you like

2) 3) 4) 5)

Calculate the circumference of a circle Identify line EF and G H What is volume of a cylinder What is total surface area

Conclusion and Consolidation The volume of any object is given by the area of its base multiplied by its height.

C = circumference of a circle, a distance around a circle point C ¿ 2 π r , in this case c = perimeter of 3D cylinder.

C = 2x3.14x 7 = 43, 96 cm Volume = Area of base x height Where a base = circle There A ¿ π r 2

= 3.14 x (7)2 = 153, 86

Therefore Volume = Area of base x height =

πr

2

xh

= 3.14 x (7)2 x 20 =153, 86 cm2 x 20 cm = 3077, 20cm3 The total surface = area of top circle + area of bottom circle + area of 2D of cylinder = 2(

π r 2 ) + 2( π r ) x h

= 2(3.14) (7)2+ 2(3.14) (7) x 20 = 2(3.14x7x7) + 20 x 43.94cm = 2(153.86) + 878.8 = 307.72 + 878.8 = 1186.52cm2 Assessment: Homework page 204: learner’s book.

The end

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:

CAPS page number:

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 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 learnerâ&#x20AC;&#x2122;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 Hieleâ&#x20AC;&#x2122;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

and concepts in line with the

produce a line from at least two vertices of a circle where the

investigative lesson.  Therefore foster

against the circumference so that

applications.  Did not intervene with learners

they share common points.  Where possible learners indicate

choice

key

questions,

participate through discussion and

of

their

knowledge

generation, their solutions, proofs

those vertices.  Ask questions for demonstration, from

of

properties of geometry and their

vertices of a quadrilateral also lie

respond

revision

and discoveries were absolutely independent.  Eliminate errors

write down solutions.

Task Division:Group 1- G:11 Task

Group 2- G:12

Cyclic

Cyclic

Quadrilateral

and

Outcome

Group

3

G:10 Quad Properties of a Tan circle :

: parallelogram use

are

vectors

supplementary

scalars.

Theorem

Euclidian

Opposite angles drawn through Geometry of cyclic quad the

– Group 4 : G 12

of Algebra and Outcome :-

vs

Chord

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

was this group was about very

Cosine and Sine listing rule

concerned

to numerous

much

of struck from not knowing

key

determine

the properties

value

the cyclic quad and vertices, secant,

of

angle at centre. naming That angle was triangles

of deflations

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

the elaborating

circumference.

what difference

made when I on brief them on is what conditions 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

of See rubrics

See rubrics

See rubrics

of

the

lesson.

See rubrics

marking Section C â&#x20AC;&#x201C; observations.

Report based on what were Learnerâ&#x20AC;&#x2122;s feedback and discoveries during the lesson commencement. The groups are selected randomly.

The 1st group. â&#x20AC;&#x201C; 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 â&#x20AC;&#x201C; angle at centre is twice the angle at circumference. <O1 = 2< ABD - angle at centre is twice the angle at circumference. <O1 + <O2 = 3600 â&#x20AC;&#x201C; 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(Âż AED +Âż 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.

Challenges Associated With Sir Maphumuloâ&#x20AC;&#x2122; Theory .

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Step 1: Define Your Questions. ...

Step 2: Set Clear Measurement Priorities. ...

Step 3: Collect Data. ...

Step 4: Analyze Data. ...

Step 5: Interpret Results.