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