This problem caught my attention while I was looking for something related to another geometrical problem on internet. And after studying it for a while, I was pretty sure that my students would love even this one.

To this, one of my students remarked -- "Sir, not even pencil?" :-))

So I allowed them to struggle for some time (about 15 mins). Then I asked them if they had any ideas..

"Sir, we need a square of side = square-root of (m x n)."

Okay... So how do you get that?

No answer....

"Do you remember some time back we had arrived at such term ----- sqrt(ab)"

After a while, one of them replied -- "

I saw some students flipping back their pages after hearing this. But soon, they were stumped again. I intervened again after @ 5 mins -

A right-triangle might help you here...

While few of them were still wondering, I could see some playing with their old friend - Pythagoras :)

When there was a road-block again after some time, I shot the final arrow --

Remember, we had done some exploration with the altitude of right triangle....

To this, couple of them just sprang up and started sketching again... and in no time, one of them shouted --

"Sir, I got the GM....!"

I could also see a pinch of embarrassment in his smiles... He had realized that we had worked on this in the past and he had forgotten this....

So now, he was happy that the side of the square was ready... But was stumped with the next step.... that to get this side, you need a Right Triangle... And how do you get that?

I noticed that while some students where listening to our conversation, there were few who were engrossed in their own work.... And one of them was Kanchan... She came with her piece of work after few more minutes. This is what she had done.

For a teacher having some other (simpler) solution in his mind, it took an effort to not get overwhelmed by the size of such a work :-)

1) She made the diagonal of the given rectangle to get a Right Triangle. Draw its altitude (k)

to the hypotenuse (h)

2) Area of Rectangle = Area of Square = m x n. So a Square of side = sqrt(mn) is reqd.

3) Let area of such a square = m x n = a^2 (say) , where a is some +ve number = sqrt (mn)

4) Now the Right Triangle is Half the Rectangle.

So its Area = 1/2 x k x h = 1/2 x m x n which means, k x h = m x n

5) So a^2 = k x h

6) Now she makes this square of area a^2 sit on the side of another Right Triangle a, b, c.

Thus a^2 = c^2 - b^2 (by Pythagoras Thrm)

Thus a^2 = (c+b) (c-b) which is also equal to k x h.

7) Equating (c+b) to h and (c-b) to k

8) Now, h - k = (c+b) - (c-b) = 2b

9) So now perform this h - k means with radius = k and center as bottom-left corner of rectangle, cut an arc on 'h' to divide it into two lengths: the upper portion is 2b

10) Draw the perpendicular bisector of 2b to get 'b'.

11) Now, since h = c + b, the lower part of h with b removed will be 'c'.

12) With this 'c' as radius and top-right corner of rectangle as center (i.e. b as base), cut an arc on the perpendicular bisector to get 'a' (i.e. we construct the Right Triangle a-b-c)

13) And Hurray! We can now construct a square on this side 'a'. which will have same area as that of rectangle because a^2 = m x n ...... (from (3))

Fantastic !! Isn't it ??

(the color chalks were used by me later while taking the snaps -- for easier explanation)

So now I told her to actually do this construction. And she happily got engrossed into this.

Meanwhile, I found that Rohit had teamed up with 2 other students to work out the solution together. He was still wondering as to how to get the right angled Vertex at the top so that he can form a Right Triangle.

And one of his peers - Sahil - was just trying to draw circles with the foot of altitude as center and 'm' as radius and then 'n' as radius. They soon realized that this was not working.

When I noticed that they had persevered enough, I decided to give them a small hint --

Yes, Circle will help you...

And no wonder, in next couple of minutes, they declared, with a wide and relaxed smile on their face -- "Yes Sir, we are done! "

To this, Sahil taunted to Rohit -- "Because of my mistake!" :-) :-) :-)

It was clearly evident that they knew what they have done, however the ever-doubting teacher in me couldn't resist himself & probed them for the explanation behind this...

They said it correctly...

1) Can you tell what property helped them to crack this problem eventually?

2) What are your views about the two methods, the investigation process and esp. the approach

Awaiting your comments/ views....

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.

Meanwhile if you have any other way of solving this problem, please do share it with me and I will share that with my students. They will be happy to study that.

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

**"Given a rectangle of some size m x n, construct a square of area same as that of rectangle using only compass and a straight edge (not a marked ruler)"**To this, one of my students remarked -- "Sir, not even pencil?" :-))

So I allowed them to struggle for some time (about 15 mins). Then I asked them if they had any ideas..

"Sir, we need a square of side = square-root of (m x n)."

Okay... So how do you get that?

No answer....

"Do you remember some time back we had arrived at such term ----- sqrt(ab)"

After a while, one of them replied -- "

**Yes, in GM (geometric mean)**."I saw some students flipping back their pages after hearing this. But soon, they were stumped again. I intervened again after @ 5 mins -

A right-triangle might help you here...

While few of them were still wondering, I could see some playing with their old friend - Pythagoras :)

When there was a road-block again after some time, I shot the final arrow --

Remember, we had done some exploration with the altitude of right triangle....

To this, couple of them just sprang up and started sketching again... and in no time, one of them shouted --

"Sir, I got the GM....!"

I could also see a pinch of embarrassment in his smiles... He had realized that we had worked on this in the past and he had forgotten this....

*That in a right triangle, the length of the altitude to hypotenuse equals the square root of product of lengths of two segments into which the hypotenuse is divided...**We had arrived at this result while proving Pythagoras theorem using Similarity.**Never mind, who remembers things if they are not used often?*So now, he was happy that the side of the square was ready... But was stumped with the next step.... that to get this side, you need a Right Triangle... And how do you get that?

I noticed that while some students where listening to our conversation, there were few who were engrossed in their own work.... And one of them was Kanchan... She came with her piece of work after few more minutes. This is what she had done.

Kanchan's work |

For a teacher having some other (simpler) solution in his mind, it took an effort to not get overwhelmed by the size of such a work :-)

__Let me illustrate her idea in following steps:__1) She made the diagonal of the given rectangle to get a Right Triangle. Draw its altitude (k)

to the hypotenuse (h)

2) Area of Rectangle = Area of Square = m x n. So a Square of side = sqrt(mn) is reqd.

3) Let area of such a square = m x n = a^2 (say) , where a is some +ve number = sqrt (mn)

4) Now the Right Triangle is Half the Rectangle.

So its Area = 1/2 x k x h = 1/2 x m x n which means, k x h = m x n

5) So a^2 = k x h

6) Now she makes this square of area a^2 sit on the side of another Right Triangle a, b, c.

Thus a^2 = c^2 - b^2 (by Pythagoras Thrm)

Thus a^2 = (c+b) (c-b) which is also equal to k x h.

7) Equating (c+b) to h and (c-b) to k

8) Now, h - k = (c+b) - (c-b) = 2b

9) So now perform this h - k means with radius = k and center as bottom-left corner of rectangle, cut an arc on 'h' to divide it into two lengths: the upper portion is 2b

10) Draw the perpendicular bisector of 2b to get 'b'.

11) Now, since h = c + b, the lower part of h with b removed will be 'c'.

12) With this 'c' as radius and top-right corner of rectangle as center (i.e. b as base), cut an arc on the perpendicular bisector to get 'a' (i.e. we construct the Right Triangle a-b-c)

13) And Hurray! We can now construct a square on this side 'a'. which will have same area as that of rectangle because a^2 = m x n ...... (from (3))

Fantastic !! Isn't it ??

(the color chalks were used by me later while taking the snaps -- for easier explanation)

So now I told her to actually do this construction. And she happily got engrossed into this.

Meanwhile, I found that Rohit had teamed up with 2 other students to work out the solution together. He was still wondering as to how to get the right angled Vertex at the top so that he can form a Right Triangle.

And one of his peers - Sahil - was just trying to draw circles with the foot of altitude as center and 'm' as radius and then 'n' as radius. They soon realized that this was not working.

When I noticed that they had persevered enough, I decided to give them a small hint --

Yes, Circle will help you...

And no wonder, in next couple of minutes, they declared, with a wide and relaxed smile on their face -- "Yes Sir, we are done! "

To this, Sahil taunted to Rohit -- "Because of my mistake!" :-) :-) :-)

It was clearly evident that they knew what they have done, however the ever-doubting teacher in me couldn't resist himself & probed them for the explanation behind this...

They said it correctly...

1) Can you tell what property helped them to crack this problem eventually?

2) What are your views about the two methods, the investigation process and esp. the approach

**invented**by Kanchan?Awaiting your comments/ views....

I had also posted this problem to another student over

Meanwhile if you have any other way of solving this problem, please do share it with me and I will share that with my students. They will be happy to study that.

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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