Monday, June 11, 2018

When students are not directly fed the text-book methods - A.P. - Part-2

Hello folks, 

So as I had said, I am back again with the Part-2 of this story :-)

Hope you remember about the onset of an unusual activity in our class? - my (lower grade) students have started doing (& enjoying) Maths from (higher-grade) text-books. Its an unusual activity not just because of the different in the class-levels, but because we had never used any text-books till now!  :-))

In the previous post, I had shared about their exploration of problems based on AP (arithmetic progression) and I am  happy that many of you liked their different approaches to solve one of the problems. For those, who haven't read the previous post yet, this is the link:

So after solving some more AP-based problems (should we really call them problems? :-), we  moved on to GP (Geometric progression). 

In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. For example, the sequence 2, 6, 18, 54, ... is a geometric progression with common ratio 3.

As we did in AP, here too I asked them to figure out the formula for -
a) nth term of GP and then,
b) sum of first n terms of GP 

They found the former quickly but needed some help in the second case.

Then I gave them around 4-5 problems at a stretch, out of which first few were quite easy. 

This one took some time, but I am glad they could solve it.


I would again suggest you to (try to) solve this on your own first.... 
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Following are the solutions of students:

Method-1: by Vaishnavi


Method-2: by Jitu


Method-3 - by Rohit



Yes, Rohit's method is somewhat similar to that of Vaishnavi's. But the reason I have shared it because he has worked out on his own and secondly, his representation is little different than that of Vaishnavi.

1) How did you solve this problem?
2) Your views about their approaches?

Thanks and Regards
Rupesh Gesota

PS: These students are from grade-7 and 8 Marathi medium government school and are part of a maths enrichment program- MENTOR. To know more, check www.supportmentor.weebly.com

Saturday, June 9, 2018

When students are not directly fed the text-book methods - A.P. - Part-1

I have been Playing Maths with a bunch of marathi-medium municipal school students after their school-hours.

As I can now see them approach and solve quite challenging (out of the text-book) problems comfortably, I decided, for a change, to now pick up their text-book for a while and see what unfolds...Of course, I cannot make them solve the problems from the school text-books of their age (they have surpassed it long back). So I  directly picked up their SSC (grade-10) Algebra text-book. (They have just cleared their grade-7 and grade-8 exams) The first chapter was about Sequence (AP, GP). I was aware that this would be a cakewalk for them. So I just rushed through the terminologies and asked them to derive the formula for Nth term of Arithmetic Progression (AP)

(For those who are unaware or have forgotten what AP is --- let me give you couple of example, instead of its definition :)
i) 2, 5, 8, 11, 14, ......  
ii) 3, -7, -17, -27, ....
iii) 5, 10, 12, 17, 19, ....
iv) 2, 20, 200, 2000, .....

Only (i) and (ii) are APs while (iii) and (iv) are not APs.. (common difference (d) = 3 in the 1st case and d = -10 in the 2nd case)   

So what I asked them was to make a formula for nth term of AP i.e. in a given AP, if you wish to 'quickly' find the 200th term, 5369th term, etc without listing down all the middle terms, then one can do this using its formula.

They didn't take more than half a minute derive the required result. It was correct. 
Tn = T1 + (n-1)d) 

So now I immediately asked them to derive the formula for Sum of first nth terms of AP (Sn)

The book was in my hand and hence I saw its method. I was wondering if this method will click to them or will they do something else, but I was very sure they will crack it. And in almost a minute all hands were up. 

No wonder, they had used a different method.... But the main thing to wonder was they all had used the same method :-)

I forgot to take the snap of their work. So I am typing down what they did:

Sn =  T1  + T2 + T3 +T4 + ...
     =  T1  + (T1 + d) + (T1 + 2d) + (T1 + 3d) +....
     =  nT1 + d + 2d + 3d + ....
     =  nT1 + d (1 + 2 + 3 +....)  
     =  nT1 + d (n-1) n / 2 
     =  n [ T1 + d(n-1)/2 ]

If you observe they have used the formula for sum of first few (n-1) natural numbers for deriving an expression for Sn.

However the text book derivation does not use this method and formula (in fact the formula for sum of first n natural no.s is derived later in the book using the formula for Sn).. And that's also fine.

So I first appreciated them for their accuracy but then challenged them to get the expression for Sn without using the formula for sum of first few natural numbers. I asked them - what if you didn't know the formula that you have used in this derivation?

To this, one of them replied  - "Sir, we would have derived it the way we had done it long back.."

"Hmm... And how had you done that?"

"By pairing up.... First and last numbers.... second and second-last... and so on...."

"Yes, so why don't use the same idea here too?"

They looked at me puzzled...

So after a minute, I just showed them the method given in their text book.


They studied it and they responded like -- "Sir, isn't this the same pairing method?''  :-)

Then, I gave them some text-book problems just for practice (were they 'problems'? :-) Luckily, the last one among them was a little better.

problem: 4 terms are in AP. Product of 1st and 4th terms is 45. Sum of 2nd and 3rd terms is 18. Find the terms. 

I would suggest you to (try to) solve this problem before reading further...

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Yes -- because the text-book was again in my hand, I saw the method. However, let me also confess that I would have used the same method because this is what was directly taught to me till drill and kill. 

Thankfully, I recovered from being an instructor and don't do this damage anymore :)
And you know what?  This not only saves my effort but also helps me learn interesting methods from them :-))

Method-1: By Jitu
 


He argued that since 3 x d x t1 and 45 are multiples of 3, hence (t1)^2 should also be multiple of 3. Hence it can take only two values: 9 and 36 to get the required sum 45. Substituting (t1)^2 as 36 does not satisfy the equation and substituting as 9 does satisfy. He solved further to arrive at the required AP.

He was the first to complete the problem. To engage him, I gave him another similar problem, but of 3 consecutive numbers. (sum of 3 no.s. is 27 & their product is 504) and he was quick to solve this one too:


 Method-2 - By Sahil (original 4 numbers problem)
 
He too argued similarly about the addition equation. Since '2a' and '3d' sum up to an Even number, and '2a' is already Even, so '3d' too should be Even. Hence 'd' should be Even. (did you understand this? :-)

So then he found the possible values for d. He also said that 'd' cannot be 0 since that would make all terms equal, which is in turn not possible 45 (given as product of two terms) is not a perfect square !  He also said that 'd' cannot be '6' because that would make a=0 leading to the product of 1st and 4th terms as 0, thus contradicting the given information.

He then found the corresponding 2 values for a, substituted each pair in the equation. 
(a,d) = (6,2) didn't satisfy but (3,4) did satisfy. He then got the required sequence.


Method-3 - by Kanchan

 
She first told me that Sum of 2nd & 3th terms would be same as the Sum of 1st and 4th terms. 


When I asked her why, she said - its obvious. :)  1st term is less than 2nd term by the same amount as the 4th term is more than 3rd term.

Did you get this?   :-) :-)

I was happy she could visualize this relation, however I challenged her to prove it while she was writing on the board, and if you notice, she has done it partially.

What drove me crazy was her next move.She went into Quadratic Equation. 
"we know the sum of two numbers, we know the product of same two numbers.. So we can make an equation, and find its roots.."  

Hope you will take some time to study her solution.
 
(let me mention that quadratic equation is formally taught in 10th, that too in a dry way... she has not only used quadratics while in 8th but has also applied that knowledge in another topic altogether!)

And her method, gave us both the possible sequences - 3,7,11,15   and  15,11,7,3 which was not the case with other methods.

Finally, I asked them if they wanted to know how its solved in the text-book. And they all were curious to know :-)

So I simply copy pasted this method on the board, leaving for them to analyze and understand.... And they could easily do so.... 



I waited for them to study this approach. And then -
 
"So if I give you another problem -- but this time of 3 consecutive numbers in AP, then how will you solve it?"

"We will then take those terms as  a-d, a, a+d"

"Oh ! you will use this text-book method?" 

"Yes sir, this is an interesting and even an efficient one!" 

I could see them smiling.

"So I should have directly showed this method to you, isn't it?"

"No.... then how would we think & discover our own methods?", Vaishnavi reacted. :-)

                                                                                                                 ..... (Part-2 soon)
Thanks & Regards
Rupesh Gesota
www.supportmentor.weebly.com

Sunday, April 8, 2018

Adding Fractions through Visuals/ Understanding

I really liked the visuals drawn by students while solving this problem.

The problem at hand was 4/3 + 5/2

Students had reached a point where they had understood the need / reason for fractions to be of same size i.e to have same denominators to add easily. However we had not arrived at any particular method yet to achieve this.

One of them said that each of the unit fractions above i.e. 1/2 and 1/3 to be split into 1/4. Most probably, the reason for this could be the pictorial representations of both the quantities (that were drawn on the board) looked bigger than Quarter... Quarter is generally a common / familiar fraction to students after Half..... All of them agreed with this suggestion... So I simply went ahead without showing any hesitation. This is how the picture looked like:



So now they said that we have 14 quarters in all plus 4 smaller pieces. When I asked them how to add the smaller (green) pieces to these 14 quarters, one of them argued -

Green piece is half of Quarter. So, 2 green pieces would make up 1 Quarter.

This is not the first time I have witnessed any student giving this specific argument. I have heard it several times i.e. students misinterpreting this left over piece as half of 1/4th (why do you feel would so many students be saying/ seeing it this way?)

I chose to ask the class about this viewpoint. And to no surprise, the whole class completely agreed with this, except for one student.

She said - If 2 smaller (green) pieces sum up to 1 Quarter, then 2 pieces of one-third should sum up to 3 quarters! That isn't true. So green piece is not half of Quarter.

Isn't this a beautiful argument?

I looked at the class. Not everyone understood this. So a picture was drawn where a whole was 1st divided into three thirds, and then one 1/3rd was erased. This visual made the game easier for them as they were very familiar with the visual of 3/4th. 



So, now the problem was - Whats the size/ name of this smaller piece??

It did not take much time for one of them to shout this - 

So then, THREE green pieces would make one Quarter! 

I must confess that when I heard this claim at first, I thought it was just a random guess and hence would get eliminated through another line of argument. I did not pay attention to this or did not evaluate this new claim, probably because of the tone in which it was broadcasted and also probably because of its nature ( since TWO didn't work, so it must THREE) !

However, I am glad that couple of them took it seriously and they not just agreed with this claim but even proved it correct with the help of this diagram.
 


 Now, isn't this too beautiful? :-)

Finally, when I probed them, they could also find out the name of this green piece.

"Because 3 pieces make one quarter, so 12 such pieces would make one whole, hence its 1/12th "

So now, we knew that 3 thirds is same as 4 quarters and the remaining one more third also had one quarter. But then we are left with one green piece alone. 

To this, one of them proposed - So lets represent each quarter in terms of this green piece now, because we know that 3 greens make one quarter.

I looked at the class again for their approval. Some required one more round of explanation but soon everyone was on the boat.

So now finally, they transformed the original problem 4/3 + 5/2  i.e.

4 thirds + 5 halves  --->  16 twelfths + 30 twelfths  =  46 twelfths.

You might have noted that they did not multiply the Numerator and Denominator by the same number to get a common denominator.... Neither they took any LCM, nor they did any cross multiplication....

So what is your view about this approach?

Thanks and Regards
Rupesh Gesota


PS: Students study in marathi medium municipal schools and hail from disadvantaged backgrounds. To know more about and support this maths enrichment program, check the website www.supportmentor.weebly.com

Tuesday, April 3, 2018

An interesting date today :)

"Sir, tomorrow is an interesting day!", my students drew my attention to an old post-it note stuck on our notice board.

"Oh, really? And what's that?"

"Its 4th day of the 4th month (April), and its also falling on the 4th day of the week i.e. Wednesday."

Flash back - we had figured out this special day many months back, but I could not recollect the specific instance that had triggered us for this exploration.So I asked them.

"What had made us landed up on this?"

It did not take much time for them to recall - 

"Some one had given you the problem last year that if 1st January 2017 falls on Sunday, then 1st of which month will be Sunday again? While working on its solution, we had extended that problem to this present one..."

"Oh yes...!"

Then I thought if the year 2018 too could give us a 4 in some way. I tried for a while, but in vain.I shared my disappointment with them.

Probably, one of my students, Kanchan, took this seriously and almost within couple of minutes, came up with this marvel:

 
I hope you will spare at least half a minute to appreciate the beauty of this expression. No doubt I could notice several interesting properties in this, however, the villian in me pushed the student to explain everything :)

And here she goes -
1) The digits of the year 2018 are used in order.
2) Each digit has an exponent
3) Exponents too are in order (descending as well as consecutive).

And this was her final stroke, which I had failed to notice -

4) The expression evaluates to 4.

Yes, it was a 4 again... She used the year 2018 beautifully to yield a 4, which her teacher had failed to do. Now, isn't that a proud moment for a teacher?  :-)


---  Did you figure out all these interesting properties?

---  And if this is not enough to delight you, let me also tell you that my students have also figured out the next immediate date which has similar interesting property like 4th April 2018 (falling on the 4th day of the week). Can you find out this date? And probably even next date after that?  i.e. Xth day of Xth month which falls on the Xth day of the week. 

--- How about doing this activity with your students?

Do let me know your views and experiences about this.. Will be eager to know...

Regards
Rupesh Gesota

PS: Students belong to marathi medium government school (class-7 and 8) based at Navi-Mumbai.  They hail from disadvantaged backgrounds. To know more about their Maths Enrichment program, check this link: www.supportmentor.weebly.com

Thursday, January 4, 2018

The 1729 Hangover :)

So I was really surprised when this familiar number showed up; unexpectedly, while I was computing for something else.....



Incidentally, I was with my students when this 'accident' happened...  And I could not contain my excitement, but had to call them to celebrate this...

I shouted -- "Hey guys... Did you all know that 1729 can be made using the first five natural numbers, and that too in the same order?? "

With this announcement, they simply left what they were doing and just dived into my calculator....No wonder they too were awestruck with this revelation :)

(though the machine just showed 54 x 32 = 1728, but they could infer that adding 1 one would yield that special number -- 1729)

For those, who are unaware about the story behind 1729 (the taxicab number or ramanujan-hardy number), plz check this video: https://www.youtube.com/watch?v=1g2E3Yf9BF8

To this, one of the students immediately exclaimed - "Sir, it means this expression 54 x 32 should be equivalent to 12 cubed..."

And I was delighted with this remark of this 7th grade chap....   
(do you wonder, what would have made him think so?)

"Yes, shouldn't it be?", I countered him.

"Yes, it should be!", affirmed the other student with confidence.

And in next moment, he picked up the chalk to prove this statement on the board.... :)
And others too surrounded him to - as if to witness the climax !


I know at first, this incident might seem quite normal or 'what's so great about this?' types...

However, if one observes closely - esp school maths teachers like me - there are definitely some surprising and interesting moments for us, in this short incident...

Will be glad if you can find at least couple of those :)

Thanks and Regards
Rupesh Gesota

PS: Students belong to marathi medium government school (class-7 and 8) based at Navi-Mumbai. To know more about their Maths Enrichment program, check this link:
www.supportmentor.weebly.com

Monday, December 25, 2017

Easy, yet interesting problem...

I think this is one of the rare problems where all the students solved it in the same way (whoever could solve it :)

While they were still working on this problem, I asked them about their opinion. Most of them guessed that the two areas will be equal. Couple of them thought the greener one might be bigger :)

Vaishnavi was the one who cracked it first...... She proved it beautifully that both the regions are of same size..... Though she could not explain and represent the solution clearly to me in the beginning, however, I was quite impressed with the way she pulled off this stint on the board to the class ....probably the after-effect of long interrogation session she had to go through :)


She said -

1) Radius of the smaller circles is half that of the bigger one. Hence the total area of four small circles is equal to the area of big circle... Note how she has explained this above!

2) Now, this bigger circle is made up of many smaller regions as shown above, labelled as a,b,c......j,k,l.

3) However, the 4 smaller circles include only 8 of these regions (overlapping parts considered twice) ...... ...Is she right? 

4) So, if we equate these two expressions, we get

a + b + c + d + e + f + g + h + i + j + k + l = (a + b + c + d) +  2( e + f + g + h)

This leads to   ( i + j + k + l )  =  ( + f + g + h )

thus proving that the yellow and green regions have same areas.

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Jeetu explained this in similar yet little different way. He said -

1) Lets imagine that we have cut the four smaller circles out of paper.
2) Cut these circles further into pieces of shape and size as given in the diagram(12 pieces)
3) Place these pieces on their respective sections on the diagram. 
4) We will be able to place only 8 pieces on the diagram. The four petal shaped pieces will be left over. 

5) Now, because the total area of 4 circles = area of big circle, it implies that -

these spare petal-shaped pieces should be of the same size as that of the 4 sections between smaller & outer circles (i,j,k,l)

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How did you solve this problem?
What was your guess?
What's your view about the solutions of students?
How about you trying this with your children/ students?
If so, we (I and my students) will be happy to know your experience and their solutions :)

Thanks and Regards

Rupesh Gesota

PS: Students belong to marathi medium government school (class-7 and 8) based at Navi-Mumbai. To know more about their Maths Enrichment program, check this link:
www.supportmentor.weebly.com

Sunday, December 24, 2017

Area of Flower


It seems I am getting addicted to listening to (& learning from) the different beautiful approaches of my students...  :-) 

And I am also delighted to see the growing interest of my students to solve more of the Geometry problems these days... 

So the above problem I saw on facebook and I was quite sure, they would like it - Area of Flower :)  The bounding shape is Square of side 2 units and the petals are formed using semi-circles. I had given this problem to them as home-work...

Three of them had solved it in the following way: (Method-1)


Consider one petal of the flower. Divide it into two halves as shown. Now, this half-petal is the circular segment whose area is the difference between the areas of circular sector with radius as 1 and right isosceles triangle with sides as 1 i.e. (pi / 4) - 1/2.  

Now the flower comprises of 8 such segments, thus leading to its area as [2(pi) - 4] 

Method-2

One of the students, Vaishnavi visualized this in a different way. She sliced the upper semicircular section and just placed it below the lower section. It will look like this:


Now, the Area of flower (i.e. 4 petals) = Area of Square - Area of regions around the petals

Look at the figure above. She said that the Area of regions A and B are same. There are eight such identical regions, each with Area = (4 - pi) / 4. 

So, the total area of such 8 regions = (4 - pi)/4  x  8  = 2(4 - pi)


And while she did this, she also came up with one more method  - almost immediately -

Method-3

If we remove the area of inscribed square from the area of circle, we get the area of 4 outer half-petals...... Doubling that will give the area of 4 full-petals i.e flower......  Side of the inner square can be found using Pythagoras theorem which is Sq.root of (2). Thus its area is 2.

So, the area of flower = (pi - 2) x 2 = same as above :)

Method-4


Kanchan said that if we remove the two left and right semicircles from the square, we are left with two regions 'a' and 'd'

So the Total area of regions 'a' and 'd' = (4 - pi)

Now comes the beautiful part -- She saw these two regions 'a' and 'd' as part of the two upper and lower semicircles. So,if these two regions are removed from these 2 semicircles, then what will be left over is the 4 petals !

Hence Area of flower = ( pi - (4 - pi) )

Method-5

Before simplifying the above expression, she further continued that there is one more method..... Regions 'c' and 'd' are same as 'a' and 'b'.... Hence if twice the total area of regions 'a' and 'b' is removed from the area of square, this too will yield the area of flower.

This is what she has shown adjacent to her first approach above... Now if you observe the two expressions obtained don't appear equal directly. So I teased her that the two answers are different. I was glad that she countered me saying that they are equivalent. She simplified her first solution to (2 pi - 4) orally :)

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An Observation - Interpreting the answer.....

I was quite intrigued by the answer we got as the area of flower.... 2 (pi) - 4..... 

This meant if we remove a Square from 2 Circles (diameter = side), then we will get a Flower ! Really? ---  Oh !!

I shared this interpretation with wonder with my students and asked them if we could actually 'see' this...?  They stared at the diagrams for a while... I further thought aloud -

It implies that:-  1 Circle - Half Square = 2 petals !   (isn't it?)

Now, can this be 'seen' ?

And.... in less than half a minute while I was still pondering over this, I heard two YESes :) 

Would suggest you to think about this before reading further...

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He said that Half square is the upper Rectangle. This upper rectangle comprises of upper semicircle and upper regions 'c' and 'b' as shown above.

So, 

Circle - Half Square = Circle - (upper semicircle + upper regions 'c' and 'b')
                                = lower semicircle - upper region 'c' - upper region 'b'
                                
Now, we have two regions identical to upper regions 'b' and 'c' even in the lower semicircle if you observe the above diagram.... So, the above expression simplifies to:

lower semicircle - lower regions 'b' and 'c' = just two lower petals of the flower.


Now, isn't that just Waaaowww ??  :)

And yes, definitely this becomes super-exciting when this happens happens along with your students - all driven by them :)

  • Do let me know your thoughts about this lesson.....
  • How did you solve this problem?
  • Which approach did you like the most?
  • How about you trying this out with your students/ children?
  • If so, please do share your experiences and their approaches... I and even my students would love to know....  :)
Thanks and Regards
Rupesh Gesota
PS: Students belong to marathi medium government school (class-7 and 8) based at Navi-Mumbai. To know more about their Maths Enrichment program, check this link:
www.supportmentor.weebly.com

Saturday, December 2, 2017

Another Geometry Problem (Extension) : Part-2

I was pleasantly surprised to know about the amount of interest / attention drawn by my previous post on the geometric problem on various facebook groups... So many people had not just read and liked it, but had even left their comments with their approach of solving this problem. I would first like to thank all these people for sharing their methods. 

I did share some of the different methods with my students (construction of auxiliary lines in different ways) and they were quite intrigued by this fact that there can be so many ways of seeing and solving a geometric problem. 

For those who haven't read the previous post or are not aware of the original problem, you can find it here: http://rupeshgesota.blogspot.in/2017/11/another-geometry-problem.html

However, Special Thanks to following remark made by a professor/ researcher Michael de Villiers on this post:

"Good, but did you perhaps encourage your students to generalize further by exploring more 'kinks' as in this online, dynamic activity?        http://dynamicmathematicslearning.com/kinkylines.html"


This motivated me to try the extension of this problem with my students...

So then the next day, 


I asked them -- 

"The other day, in our original problem (fig (1)), when there were just two inclined lines between the two parallel lines, we saw that the measure of angle b was sum of the measures of angles a and c......, What will happen if there are three such lines between these parallel lines as in fig (2)...... or may be if there are four or more such lines between them as in fig 3 and fig 4? ........ I mean, will there still be any relation between these angles a, b, c and d in fig (2), the way we had  a + c = b in fig (1) ?"

They thought for a while and started working on their notebooks.. I stopped them...

"Wait. Can you first guess what could be the possible relation?"

These were their different guesses about the four angles in fig (2)

i) a + b = c + d

ii) a + d = b + c 

iii) a + b + c + d = 180

Then I asked them to continue with their work...

While some were engrossed in arriving at the solution, I could see that few were unable to find a break though... So I asked them to "actually construct this diagram using a scale and then measure the angles a,b, c and d to really find out if there is any relation..."

This idea was well received by this bunch... However I could see that not all were ready to do this construction / verification exercise....

So I could see the class divided into two groups - One who were working on the geometric solution without any construction, and the Other group who was doing the construction to 'actually see' the relation first....

Almost both the groups arrived at their respective results simultaneously i.e.

sum of the measures of angles a and c equals the sum of the measures of angles b & d.
a + c = b + d

And then each of these groups was asked to try with the other method - for verification of their result arrived using the first method. It was a beautiful exercise...

Finally they presented their approach to the class :

Kanchan's way

You can see that she has extended the two inclined lines to meet the parallel lines to form two triangles. Then she claimed that the angles 180-(180-c+d) and 180-(180-b+a) will be equal as they form a pair of alternate angles and hence, d-c = a-b  means 

a + c = b + d

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While Vaishnavi had construction the auxiliary lines in a different way:

Vaishnavi's way
She has extended the upper inclined line beyond the upper horizontal line and then dropped a perpendicular from there to the lower horizontal line.

She now saw a pentagon and used the property of interior angles of pentagon to prove that a + c = b + d

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Jitu instead drew two lines which were parallel to the two given parallel lines as follows: 

Jeetu's way
He then defined a new angle 'x' as interior alternate angles between these two newly constructed parallel lines.

Now if you focus on the upper two parallel lines, then x = b - a  while if you look at the two lower parallel lines, x = c - d , thus leading to 

b - a = c - d   which means a + c = b + d

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It didn't take much time for them to arrive at the relation between angles in fig (3) and fig (4) i.e. more kinks between the two parallel lines. This is what one of them did immediately: 


She proved that 

a + c + e = b + d

and even others could infer the relation for more number of angles in fig (4),  

a + c + e = b + d + f

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I would also like to share that before I gave these new set of (extended) problems to them, I also showed them couple of different approaches of solving the original problem that they had already solved the other day:



And if you observe closely, this will reveal to you the secret behind Vaishnavi's idea/ approach to solve the extended problems.

In fact, Its also interesting to note that the three methods of drawing the auxiliary lines used by the students in solving this extended problem are completely different than those used by them in solving the original problem (fig (1) few days back.

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While we all could happily derive the relation between the angles that will be formed due to many kinks , I wanted my students to 'wonder' at this result..... I asked them if this result /relation was obvious / natural or does it 'surprise' them or 'makes them wonder' as to why this must happen so?

And this nasty question of mine made them stare at this problem again :-) :-) :-)

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Meanwhile, what are your thoughts/ reactions about this result? Does that make you 'wonder'?   :-)


Regards

Rupesh Gesota