In my previous post, I had shared two methods figured out by students for Squaring the given Rectangle (area unchanged) with only compass & straight edge.
If you haven't read the previous post, you can click here:
http://rupeshgesota.blogspot.com/2018/08/square-of-size-of-rectangle.html
And If you remember I had also mentioned in the end of the post that -
I had also posted this problem to another student over whatsapp and after almost a day-long struggle, she too sent me an interesting solution. I will share her solution in the Part-2 of this post along with another interesting variation of this problem, thanks to this student.
So here we go.... This was her solution:
As you can see, she has used the knowledge of generation of Pythagorean Triplets to find the square of required length i.e. sqrt(l x b) to get the area same as that of rectangle with sides l and b.
Perfect !!
I challenged her further -
What if you didn't know how to generate Triplets??
I knew that she had already persevered to get this result, but still it was important for her to know that there can be more methods...
Meanwhile, before Squaring the rectangle, she messaged me saying --
"I have converted the Rectangle into Rhombus... But still working on Squaring it.. "
This was her method:
Oh !! And this stimulated me to pose this (new) problem to my students over here now, who had solved the Squaring problem.
So when I did this the next day, it didn't take much time for them to crack it --
Vaishnavi had solved it the same way as above:
And this is how a group of three students had solved it, working together:
One of them came fwd to explain -
1) Extend side AB of Rectangle ABCD to get line l.
2) With center as C & radius = CD, get a point E on this line
3) Get F such that EF = EC
4) Join ED
5) So DCEF is Rhombus because all sides are equal.
How would you respond if your students give this explanation to you?
So I asked them to prove why DE = other 3 sides?
It was interesting to note that almost the whole class roared at me when I asked this question.
"Sir,.... It will be equal.... It's so obvious..."
" Not obvious to me.... Please prove it !"
So, you too may pause now -- and think for a while, as to how will you prove it?
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So I, in my mind, had done this way --
"because one pair of opposite sides of quadr are equal and parallel, so its a parallelogram and since its one pair of adj sides are equal, so the parallelogram is a rhombus"
But, as I had suspected and as usual, they did this in a different way:
They proved the two triangles AFD and BFC are congruent by SSS test and hence by c.s.c.t. FD and EC are equal.
Well done !!
So I asked them if anyone has tried differently?
Kanchan came forward. This was her stroke --
She showed her work in the notebook to me first, which was perfect. But while explaining it to others on the board, she forgot a part of procedure and goofed up in between as you can see in the Lower diagram. Later when questioned for veracity, she could recollect and constructed the Upper diagram. This is how she explained --
1) Draw the diagonal and altitude of rectangle ABCD.
2) Extend BD such that you get AF = DB
3) From points A and E, make equal arcs of length DB to intersect at P.
4) AFEP is Rhombus.
What do you feel? :)
So when I asked her how EF equals other 3 sides, she realized the flaw in her explanation, and she corrected it in the figure above it.
1) & 2) same as above
3) She obtained side of length BD on the extended diagonal. But its interesting to see how she saw and achieved this. An easier way would be to make an arc of radius BD with F as center on FB.... But instead, she subtracted FD from FB to get the point P so that FP = BD
Interesting, isn't it? :-)
Further steps were same as above...
So then I asked them, what's the similarity and differences in these 3 methods?
And they could answer this very well...
1) Can you answer this question precisely?
2) How would you / your students solve this problem?
Will be happy to know your views/ responses on this --
----
Thanks and Regards
Rupesh Gesota
PS: These students are from grade-8 and 9 Marathi medium government school based at Navi-Mumbai and are part of a maths enrichment program- MENTOR. To know more, check www.supportmentor.weebly.com
If you haven't read the previous post, you can click here:
http://rupeshgesota.blogspot.com/2018/08/square-of-size-of-rectangle.html
And If you remember I had also mentioned in the end of the post that -
I had also posted this problem to another student over whatsapp and after almost a day-long struggle, she too sent me an interesting solution. I will share her solution in the Part-2 of this post along with another interesting variation of this problem, thanks to this student.
So here we go.... This was her solution:
As you can see, she has used the knowledge of generation of Pythagorean Triplets to find the square of required length i.e. sqrt(l x b) to get the area same as that of rectangle with sides l and b.
Perfect !!
I challenged her further -
What if you didn't know how to generate Triplets??
I knew that she had already persevered to get this result, but still it was important for her to know that there can be more methods...
Meanwhile, before Squaring the rectangle, she messaged me saying --
"I have converted the Rectangle into Rhombus... But still working on Squaring it.. "
This was her method:
Oh !! And this stimulated me to pose this (new) problem to my students over here now, who had solved the Squaring problem.
So when I did this the next day, it didn't take much time for them to crack it --
Vaishnavi had solved it the same way as above:
 |
| Method-1 |
And this is how a group of three students had solved it, working together:
 |
| Method-2 |
One of them came fwd to explain -
1) Extend side AB of Rectangle ABCD to get line l.
2) With center as C & radius = CD, get a point E on this line
3) Get F such that EF = EC
4) Join ED
5) So DCEF is Rhombus because all sides are equal.
How would you respond if your students give this explanation to you?
So I asked them to prove why DE = other 3 sides?
It was interesting to note that almost the whole class roared at me when I asked this question.
"Sir,.... It will be equal.... It's so obvious..."
" Not obvious to me.... Please prove it !"
So, you too may pause now -- and think for a while, as to how will you prove it?
-
-
-
-
-
-
-
-
-
-
-
So I, in my mind, had done this way --
"because one pair of opposite sides of quadr are equal and parallel, so its a parallelogram and since its one pair of adj sides are equal, so the parallelogram is a rhombus"
But, as I had suspected and as usual, they did this in a different way:
They proved the two triangles AFD and BFC are congruent by SSS test and hence by c.s.c.t. FD and EC are equal.
Well done !!
So I asked them if anyone has tried differently?
Kanchan came forward. This was her stroke --
 |
| Method-3 |
She showed her work in the notebook to me first, which was perfect. But while explaining it to others on the board, she forgot a part of procedure and goofed up in between as you can see in the Lower diagram. Later when questioned for veracity, she could recollect and constructed the Upper diagram. This is how she explained --
1) Draw the diagonal and altitude of rectangle ABCD.
2) Extend BD such that you get AF = DB
3) From points A and E, make equal arcs of length DB to intersect at P.
4) AFEP is Rhombus.
What do you feel? :)
So when I asked her how EF equals other 3 sides, she realized the flaw in her explanation, and she corrected it in the figure above it.
1) & 2) same as above
3) She obtained side of length BD on the extended diagonal. But its interesting to see how she saw and achieved this. An easier way would be to make an arc of radius BD with F as center on FB.... But instead, she subtracted FD from FB to get the point P so that FP = BD
Interesting, isn't it? :-)
Further steps were same as above...
So then I asked them, what's the similarity and differences in these 3 methods?
And they could answer this very well...
1) Can you answer this question precisely?
2) How would you / your students solve this problem?
Will be happy to know your views/ responses on this --
----
Thanks and Regards
Rupesh Gesota
PS: These students are from grade-8 and 9 Marathi medium government school based at Navi-Mumbai and are part of a maths enrichment program- MENTOR. To know more, check www.supportmentor.weebly.com
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