I came across this interesting problem and thought to share this with my students...

To understand Sahil's method, let me re-draw the figure again...

1) Did you understand her method?

2) How did you solve this problem?

3) How about trying this with your students? Plz do share your classroom experience...

4) Which approach did you like the most? Why?

5) Plz share your views and comments on this article......

Rupesh Gesota

Students started working on this and after about 5 minutes, one of them was ready with her solution...

*I would suggest you to try solving this problem on your own first, and when you are ready with your solution, you can read further to see how these students have 'seen, approached and solved' it....*

__Method-1__Vaishnavi's 1st attempt |

Her approach was to count all the sides of 6 rectangles (222 x 6) and remove the ones that are not to be counted. So according to her, the answer was 999. I think, you would have figured out the mistake she did.

Yes, she forgot that the overlapping sides need to be removed twice and not just once..

But, I am glad she could quickly think and work out her error after overhearing the final answers of two of her peers, while they were explaining me :)

So this is how she finally presented her solution to the class. Note her work below. See her erased work in the yellowed part. Here, she explained that how the red segments are actually two segments when the rectangles are separated and hence need to be removed twice.

Vaishnavi's 2nd attempt |

If you are wondering what her 222 and 111 represents, then this is what she had done..

She has combined the two red breadths and two red lengths (see the 4 sides with vertical strokes) to form one rectangle (222)......what remains is one length (made up of two pieces) and one breadth which is same as half the perimeter i.e. 111

__Method-2__Tanvi's explanation to me |

Tanvi's explanation to the class |

To explain her approach, let me just make another diagram:

Diagram - to understand Tanvi's approach |

Her two 444s represents perimeters of four rectangles.. Now she understands that this sum 'includes' the segments that were not to be included --- the overlapping segments.

From here, she used the same approach as that of Vaishnavi to get 222 and 111. Thus she finally gets 999 - 333 = 666

__Method-3__Siddharth's method |

He "traveled along the boundary" of the shape and counted the lengths and breadths of rectangles it included... He calculated 6 lengths and 6 breadths, thus amounting to thrice the perimeter which leads to 666.

__Method-4__

Sahil's method |

To understand Sahil's method, let me re-draw the figure again...

Diagram - to understand Sahil's method |

I think the above diagram would be self explanatory...

Yes,.... this is how he beautifully saw and made three rectangles out of all the pieces on boundary to get three equivalent rectangles thus amounting to 666.

I thought at least one of the two remaining two students would solve the problem using my method.. However, Rohit's method was same as that of Sahil and Laxmi too had used the same approach as that of Siddharth....

So I asked them if they wanted to know how I solved this problem...

And there was a loud Yes ! :-)

So the perimeter = 6 breadths + 6 lengths = 222 x 3 = 666

They understood the "sliding" that I had done.... But I thought to interrogate them further....

"How can I be sure that when I slide upper block to the left such that pt. D coincides with C, then F too will exactly coincide with E?

Rohit jumped in...

"Sir its obvious..... that x = a...."

"There is nothing obvious in Maths, Rohit... Can you prove this statement ?"

He began...

k = y + a

k = x + y

k = a + b

So, y + a = x + y = a + b

So, a = x and y = b

---------

Two students were absent when we did this problem.. So they solved it the next day... One of them, Jitu, used the same approach as that of Siddharth......

While the other student, Kanchan, gave me a surprise !!

I thought at least one of the two remaining two students would solve the problem using my method.. However, Rohit's method was same as that of Sahil and Laxmi too had used the same approach as that of Siddharth....

So I asked them if they wanted to know how I solved this problem...

And there was a loud Yes ! :-)

__Method-5__So the perimeter = 6 breadths + 6 lengths = 222 x 3 = 666

They understood the "sliding" that I had done.... But I thought to interrogate them further....

"How can I be sure that when I slide upper block to the left such that pt. D coincides with C, then F too will exactly coincide with E?

Rohit jumped in...

"Sir its obvious..... that x = a...."

"There is nothing obvious in Maths, Rohit... Can you prove this statement ?"

He began...

k = y + a

k = x + y

k = a + b

So, y + a = x + y = a + b

So, a = x and y = b

---------

Two students were absent when we did this problem.. So they solved it the next day... One of them, Jitu, used the same approach as that of Siddharth......

While the other student, Kanchan, gave me a surprise !!

__Method-6__

Kanchan just removed the three bricks on the top layer and bottom layer. So according to her. the given shape gets converted into this equivalent shape for finding the perimeter:

So, perimeter of the given shape = 222 x 3 = 666

--------------------

Diagram to explain Kanchan's method |

So, perimeter of the given shape = 222 x 3 = 666

--------------------

1) Did you understand her method?

2) How did you solve this problem?

3) How about trying this with your students? Plz do share your classroom experience...

4) Which approach did you like the most? Why?

5) Plz share your views and comments on this article......

**Thanks and Regards**

Rupesh Gesota

Another method:

ReplyDeleteSurround the shape with a large, single rectangle. Focus on one corner of this rectangle, where you could cut out a smaller rectangle to match the original shape.

Convince yourself that the original "bumped-in" distance around this smaller rectangle is the same as the "bumped-out" distance along the surrounding rectangle (because both are one width and one length of the smaller rectangle).

This must be true at all four corners.

Then find the perimeter of the bigger rectangle. Each side is three times as long as one brick, so 3x222.