Sunday, December 24, 2017

Area of Flower

It seems I am getting addicted to listening to (& learning from) the different beautiful approaches of my students...  :-) 

And I am also delighted to see the growing interest of my students to solve more of the Geometry problems these days... 

So the above problem I saw on facebook and I was quite sure, they would like it - Area of Flower :)  The bounding shape is Square of side 2 units and the petals are formed using semi-circles. I had given this problem to them as home-work...

Three of them had solved it in the following way: (Method-1)

Consider one petal of the flower. Divide it into two halves as shown. Now, this half-petal is the circular segment whose area is the difference between the areas of circular sector with radius as 1 and right isosceles triangle with sides as 1 i.e. (pi / 4) - 1/2.  

Now the flower comprises of 8 such segments, thus leading to its area as [2(pi) - 4] 


One of the students, Vaishnavi visualized this in a different way. She sliced the upper semicircular section and just placed it below the lower section. It will look like this:

Now, the Area of flower (i.e. 4 petals) = Area of Square - Area of regions around the petals

Look at the figure above. She said that the Area of regions A and B are same. There are eight such identical regions, each with Area = (4 - pi) / 4. 

So, the total area of such 8 regions = (4 - pi)/4  x  8  = 2(4 - pi)

And while she did this, she also came up with one more method  - almost immediately -


If we remove the area of inscribed square from the area of circle, we get the area of 4 outer half-petals...... Doubling that will give the area of 4 full-petals i.e flower......  Side of the inner square can be found using Pythagoras theorem which is Sq.root of (2). Thus its area is 2.

So, the area of flower = (pi - 2) x 2 = same as above :)


Kanchan said that if we remove the two left and right semicircles from the square, we are left with two regions 'a' and 'd'

So the Total area of regions 'a' and 'd' = (4 - pi)

Now comes the beautiful part -- She saw these two regions 'a' and 'd' as part of the two upper and lower semicircles. So,if these two regions are removed from these 2 semicircles, then what will be left over is the 4 petals !

Hence Area of flower = ( pi - (4 - pi) )


Before simplifying the above expression, she further continued that there is one more method..... Regions 'c' and 'd' are same as 'a' and 'b'.... Hence if twice the total area of regions 'a' and 'b' is removed from the area of square, this too will yield the area of flower.

This is what she has shown adjacent to her first approach above... Now if you observe the two expressions obtained don't appear equal directly. So I teased her that the two answers are different. I was glad that she countered me saying that they are equivalent. She simplified her first solution to (2 pi - 4) orally :)


An Observation - Interpreting the answer.....

I was quite intrigued by the answer we got as the area of flower.... 2 (pi) - 4..... 

This meant if we remove a Square from 2 Circles (diameter = side), then we will get a Flower ! Really? ---  Oh !!

I shared this interpretation with wonder with my students and asked them if we could actually 'see' this...?  They stared at the diagrams for a while... I further thought aloud -

It implies that:-  1 Circle - Half Square = 2 petals !   (isn't it?)

Now, can this be 'seen' ?

And.... in less than half a minute while I was still pondering over this, I heard two YESes :) 

Would suggest you to think about this before reading further...


He said that Half square is the upper Rectangle. This upper rectangle comprises of upper semicircle and upper regions 'c' and 'b' as shown above.


Circle - Half Square = Circle - (upper semicircle + upper regions 'c' and 'b')
                                = lower semicircle - upper region 'c' - upper region 'b'
Now, we have two regions identical to upper regions 'b' and 'c' even in the lower semicircle if you observe the above diagram.... So, the above expression simplifies to:

lower semicircle - lower regions 'b' and 'c' = just two lower petals of the flower.

Now, isn't that just Waaaowww ??  :)

And yes, definitely this becomes super-exciting when this happens happens along with your students - all driven by them :)

  • Do let me know your thoughts about this lesson.....
  • How did you solve this problem?
  • Which approach did you like the most?
  • How about you trying this out with your students/ children?
  • If so, please do share your experiences and their approaches... I and even my students would love to know....  :)
Thanks and Regards
Rupesh Gesota
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:

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