Another Geometry problem...

I knew that this problem can be solved in various ways.. And hence was curious to try it with my students...

                            

They stayed up to my one expectation fully that they could solve this problem quickly. However I got only 3 different approaches (as against my expectation of 5 to 6 !) 

But when I realized that there are still 3 methods from just 7 students, then I was quite happy :-))

Almost all of them had extended the two oblique lines to meet the pair of parallel lines.

After this construction, there were two (little) different routes:

Most common method

In this method, first the Alternate angle property was used to find one of the angles of the Triangle. Then, using the property of angles of triangle, the third missing angle is calculated.

Finally, this third angle of the triangle forms a linear pair with x and hence it can be easily calculated..

However, one of them - Rohit - did a little twist...

He first calculated one of the angles of the Triangle (BGC) using the Alternate angles property like others. 

Then he directly used the Exterior Angle property to calculate x. 

Using Exterior Angle Property

And finally, my most favorite one ! 

This was done by Jitu...

This is how he had done it....

Instead of extending the two oblique lines like others, he dropped a perpendicular from one of the parallel lines to other.

He then focuses his attention on the newly formed Pentagon using this perpendicular line.

This pentagon has two of its angles as 90 while other two are given as 30 and 10. So he figured out the missing fifth angle (they had discovered the formula for "sum of the angles of a polygon" few months back)....

Now he used the property of sum of angles forming a circle to find the desired angle x.

Interesting way, isn't it?  :-)

In fact all the students were delighted by this construction/ approach....

I asked them if any of them could think of more ways... They thought for a while but couldn't think of any... So I gave them a hint - I began from what Jitu had done...

"Jitu created two new points on the two lines.... I can already see two points on these lines.."

"yes sir, we can even join E and F (refer diagram above)"

So I did this construction as suggested by them...

"So what next?" I asked them.

They thought for a while.... and soon realized there is nothing that can help them forge ahead...

One of them exclaimed - Sir we can drop a perpendicular from the point E on the upper line to the lower line...

"Okay... So how will this situation be better than the present one?"

"Here, we get a Quadrilateral with three known angles, so we can find the fourth unknown i.e. X"

I looked at the class for their views... And all agreed with this.... 

Immediately the other said that we can even draw a perpendicular here similar to what Jitu did on the right hand side,... We will directly get x in one step itself... 

I could see a smile on Jitu's face :)

"How about drawing a parallel line somewhere , instead of perpendicular?" I threw this idea to them...

It didn't take much time for one of them to say--

Yes we can draw a line from the vertex of angle x, parallel to the given two parallel lines...

"Will that help?" I asked the class...

Yes... Use alternate angles property twice and we are done ! - shouted one of them.

."Good.. Any other way?"

I could see them quite surprised now :)

"Fine... So let this be homework.. If you find any other way, then share with the class tomorrow... Meanwhile, also write down your solutions systematically with reasoning at home... I will see them in our next class.. "

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1) How did you solve this problem ?

2) Your views about the approaches of students..?

3) your opinion about the way this class was taken ?

Will be happy to hear your responses.. 

Thanks and Regards

Rupesh Gesota

PS: These students are from Marathi medium Government school , studying in class 7 and 8... We play with Maths after their school hours.. To know more, check the website www.supportmentor.weebly.com

My students solved it better than me :)

This is an interesting problem that was given to about 30 Maths Teachers in one of the PD workshops that I attended recently.  I must confess that almost all of us, baring very few, struggled quite a lot and for quite a long to find its solution.. In fact, many of us could not even arrive at the desired solution :) However, I had a gut feeling that my students 'will' be able to reach the destination, and that too in a 'proper' way....

Yes, I was quite sure that my students would do this problem-solving much 'better than me' !  ..... Why?

Because I have been taking care and effort since past 2 years, to see to it that I do not pollute their minds or damage their mathematical thinking by feeding them 'my' methods or by enforcing upon them some 'standard' methods directly....

Makes you reflect?  :-)

So here is the problem:

So, I told this incident to my students while giving this problem... (that it was given to teachers and we all took lot of time, etc).. Naturally, the first thing they asked me was-

"Sir, could you solve it?"

"Yes, I could..."

I could see one of them saying to the other, "'Sir' will 'obviously' solve it....!"

"Wait ! It took me lot of time and effort.... and I was not so happy with my work.... However, I feel you will do better than me....and hence I am giving this to you.... "

Some of them smiled at me, while others had already dived into the problem.. :)

They sought clarification about some requirements mentioned in the problem... I clarified those with examples... and then there was silence for about 10 minutes...

I told them mid-way that they can even work in pairs or groups if they wish.... But none of them did.... It seems they wanted to give their own shot for some time..

Would suggest you to give a try before reading their solutions :)
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Two of them - Sahil and Rohit - told me that they are on the verge of completion... have eliminated many possibilities out of the total 120 and are working on the final six ones ....

I understood what they were saying / doing... Did you?  :-)

Soon, they declared that ---- "There is No solution."

I felt Waaaow.... but contained my excitement and rather probed them if they have a proof to support their claim...

Yes sir, we can prove it...

I asked them to pen down their solution systematically while others were still solving... 

In some time, Vaishnavi too declared her accomplishment and this was followed by the rest of the class in next 10-15 minutes....

Every student who was done, was directed to 'write down' their solution / approach in a comprehensible form...

Only when everyone claimed that they have solved it, we began with the Whole-Class Discussion...

Rohit pushed Sahil for sharing.... I had seen both of them working together in the end....

Sahil's work:

He says that there is only one way to make 14 and 13. For 14 - 2,3,4,5 should be together while for 13 - 1,3,4,5 should be together. Hence he concluded that 1 and 2 should be on the two extremes.

So this leaves 3,4,5 in the middle zone which gives rise to 6 possibilities, as listed by him.

He then ruled out each of these possibilities with reasoning. For example: 

Option 1 ----  1,3,4,5,2 does not yield 6
Option 2 ---- 1,3,5,4,2 does not yield 7

and so on.....

Since none of the options work in the favor of given conditions, so the problem does not have any solution.

I asked others if they want to ask anything to Sahil.... All agreed with his work...

Jeetu said that his approach was also very similar to this....

Tanvi came forward. This is her work:..

She too first found out that 1 and 2 will be on the two extremes.

Later, she wrote down all the possible ways of making every number from 6 to 15 (though she later said there was no need to write for 13 to 15 again :)  

She then cancelled out all the options which used both 1 and 2 to make a number.

Then she started with 6-sized bogie. She chose to test one of the options of 6 i.e. 5+1
This would mean placing 5 besides 1. This forces 7 to be made using 4,3. However the option (5,3) for making 8 fixes the position of 3 at the center and hence 5 next to 1.

Thus the train looks like this:  1,5,3,4,2

This structure cannot however make a small train of size-10. 

So this means the initial assumption of 6 to be made using 5 and 1 would be wrong, thus leaving only (4,2) for 6. 

Again, sizes 7 and 8  will fix the positions of 3 and 5 respectively, eventually and surprisingly leading to the same arrangement 1,5,3,4,2 which does not help us for giving 10.

Thus any of the options of 6 leads to an impossibility for the size 10 and hence it is not possible to make such a train....

Again, I looked at the class for their agreement/ disagreement. .. All agreed...

Vaishnavi came up with her approach:

Her initial approach is same as that of Tanvi.... Listing down and ruling out the options having 1 and 2 together. However then she chose to work from the other end (bigger numbers), unlike Tanvi who worked from 6.

Done with 15,14 and 13, she said 12 will also be possible with 3,4,5 anywhere in the center.....  1  ? ? ? 2

Analyzing the only remaining option for 11 i.e. 5,4,2, its clear that 3 will have to be next to 1. This rules out two more options (5,1) for 6 and (5,4,1) for 10.....    1 3 ? ? 2

Analyzing the only remaining option for 10 i.e. (5,3,2) its clear that its impossible to make 10 because 3 and 2 are already 2 spaces apart.

Thus its not possible to complete this order.

All agreed to this.

Kanchan said that her approach is also somewhat similar to Vaishnavi's (starting from the bigger numbers).. This is her work:

She first fixed up 1 and 2 in the two extremes like others. However, she did not list all the possibilities for 6 to 14 like her two peers.

After finishing off  the work of 12 to 15, she listed the two options for 11:
(5,4,2) and (1,2,3,5)

She ruled out the latter (since 1 and 2 are apart) and hence the former got fixed. This meant either 4 or 5 should be next to 2.  .... (note that Vaishnavi had directly concluded that 3 will be next to 1 using this case)

She then listed down the three options for making 10 and one-by-one ruled out those too with appropriate reasoning
(5,4,1) -- Not possible as 5 and 4 will be next to 2.
(1,2,3,4) -- Not possible as 1 and 2 are 3 spaces apart
(2,3,5) -- Not possible as (2,4,5) have to be together to make 11 as discussed above.

Thus it is not possible to make 10 and hence this problem has no solutions.

Remember, I told you in the beginning - about my confidence that my students would be probably solving this better than me... And I am so happy and even proud that they did so !!

You want to see my work?  This is it :))

How about you trying this problem with your students / children?

Could you arrive at the correct solution on your own? What/ How was your approach?
I will be happy if you can share your or your students' approach with me, if its different than any of the above....

What are your views about the thought processes of students above? 

Waiting to hear from you, 

Thanks and Regards

Rupesh Gesota

PS: These students are studying in class-7 and 8 in a navi-mumbai based marathi medium government school.. I am working with them on their Maths Enrichment since past couple of years after their school hours. To know more, check the link:
www.supportmentor.weebly.com

Simple Puzzle (Tin of Biscuits) - multiple approaches

I was sure they will crack this puzzle quickly, but I was more curious to know their multiple approaches.

"A tin full of biscuits weighs 5 kg 200 gm. The same tin half full of biscuits weighs 3 kg. Calculate the mass of empty tin."

Almost all of them were done in about a minute.

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Give this problem a try before you read the solutions below.

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Tanvi's approach:

I took one more tin half full of biscuits. So two half-filled tins weigh 6 kg. But a fully filled tin weighs 5 kg 200 gms. So the difference in these two weights corresponds to the weight of tin i.e. 800 gms.

Kanchan's approach:

I halved the weight of fully filled tin, thus leading to 2600 gms. Now, the other half filled tin weighs 3000 gms. So the difference 400 gms corresponds to half the weight of tin. So the tin weighs 800 gms.

Vaishnavi's approach:

She solved this algebraically. Let the weights of tin and fully filled biscuits be T and B gms resp.

T + B = 5200 ...... (1)
T + 0.5 B = 3000  ....... (2)
T + 0.5 (5200 - T) = 3000   .......... (from 1)
T + 2600 - 0.5 T = 3000
0.5 T = 400
T = 800

To this Rohit responded,

Look at your 1st and 2nd equations. We can clearly see that 0.5 B = 2200.  
Thus B = 4400 and hence T = 800

How did YOU solve this problem? Was your approach different than any of the above?

Which approach did you like the most?

How about trying this with your students/ children? Would love to know their approach :)

Thanks and Regards

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