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 :)

....
...
...
...
....
...
...

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 :)

• 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 ------------------------------------ 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