Babushka Doll


Babushka doll

Ms. Rania Ali taught in a school located in a slum.  The kids had very little interest in learning.  Ms. Ali was aware of her challenge but the great teacher found new ways to get attention of her students.  Today she was at a loss. She did not know how to approach the next geometry lesson until she saw something on the principal’s desk.  She asked the principal to borrow the doll on her desk.  The principal agreed but asked Ms. Ali to make sure to return it.

Ms. Ali came to the class and put the doll on the teacher’s table.  Of course, all the  boys and girls were curious why Ms. Ali would bring a doll to the class.

Ms.  Ali: Class, I have this special doll to show you.  Some of you may have already seen it in the principal’s office. It is a babushka doll. This doll can be opened.

She opened the Babushka doll.   There was a second doll inside. She took it out and closed the first doll. Then she opened the second one and took out a third doll. She repeated the process two more times. She laid all the five dolls on the table.
fig.g6.1           Many students had never seen a babushka doll before and were mesmerized.  Everyone in the class liked the display until Barak spoke up.

Similar or identical?

Barak: Ms. Ali, all the five dolls are the same.

Mehak: No they are not. They are all different in size.

This followed a class discussion, and the agreement was that all the five dolls were similar because they had the same shape and design.  They would be identical if they were all of the same size too but then they could not go one inside the other.

Ms.  Ali:  Very good class.  If one of the dolls was lying down and the others were upright. Would they still be similar dolls?

The students started to whisper but then Taheen stood up and said,” I think they would still be similar because the doll is only lying down.  It could easily stand up.”   The class agreed that the doll in a different orientation would still be similar to the other dolls.

Now, shapes and cookies

Ms. Ali:  Here is a picture of different cookies.  Which of them are similar to each other.          fig.g6.2  Arisha:  They are all different in shape.  So no cookie is similar to any other.

Ms. Ali:  Very good Arisha. Now here is another picture of different cookies.
fig.g6.3.jpg           Arisha:  I could flip around any of the cookies and they would come out to be similar.  Cookies 1 to 6 are of the same size but 7 and 8 are smaller but still similar in shape.

Ms. Ali:  Rehan, you have been quiet.  Tell me if cookies 1 to 6 are similar and also the same size, would you say they are identical.

Rehan:  Yes, I could flip them and stack them one on top of the other. Cookies 1 to 6 would all come out to be exactly the same.

Ms. Ali:  In geometry, we would say that triangles 1 to 8 are similar but triangles 1 to 6 are also congruent. I want to add that the concepts of similarity and congruence apply to all the shapes – not just to the triangles. Here is an example with quadrilaterals.  Mehak, tell me which of these figures are similar and which are congruent?

fig.g6.4            Mehak: Ms. Ali, the shapes  of 1 to 4 are all squares and therefore similar.  The shape 2 is congruent  to shape 1 and the shape 4 is congruent to shape 3.  The shape  7 is similar and congruent to  the shape 5.   I would have to measure it make sure that shape 6 is also similar to 5 and 7.

Taheen: Ms. Ali, would it be right if I said that all equilateral triangles are similar to each other, all squares are similar, all pentagons are similar and so on to the extent that all the circles are also similar to each other.

Ms. Ali: Similar but not congruent.  I am impressed with your thinking.

Advantage in understanding the similarity

Taheen:  Thank you Ms. Ali but I don’t understand what is the big advantage is in understanding the similarity idea.

Ms. Ali: I drew two similar triangles – triangle 1 and triangle 2.  They are similar because angle BAC of triangle 1 equals B’A’C’ of triangle 2, angle ACB equals angle A’C’B’ and, therefore, the third angle CBA = C’B’A’.  Now if ratio of the sides AB of triangle 1 and A’B’ was two, who can tell me the ratio of BC to B’C’, and AC to A’C’?

fig.g6.5            Almost the whole class raised hands but Ms. Ali asked Feran.

Feran: 2 because if the ratio of any of sides was different in one triangle, it would change its angle and the triangles will no longer be similar.

Ms. Ali:  Very good.  From the number of students who had raised hands, I think everyone got this idea.  Taheen, so the advantage is that you do not have to measure all three sides, their ratios will be the same.  Although I used triangles to show the constant ratios of sides, the idea applies to any shapes.

The bell rang and the students disappeared without Ms. Ali dismissing the class.



ABCD is a square.  Show that AD is perpendicular to BC.

Solution: Given ABCD is a square or sides AB=BD=CD=AC, and CD is parallel to AB and AC is parallel to BD.

The triangle ABD is an isosceles triangle because AB = BD.  Therefore, angles DAB and ADB are equal.   Triangles AEB and DEB are congruent because AB = BD (given), side EB is common to both and angles DAB and ADB are equal. Therefore, angle AEB = angle BED,  But angle AEB = angle AEC and angle CED and angle AEB (opposite angles).  Therefore the angles AEB =  angle BED = angle DEC = angle AEC. Together, these angles form a complete revolution = 360°.  Hence each of the angles is 90°.  Hence line AD is perpendicular to BC.


  1. ABC is a right angle triangle with ∠ACB being a right angle. Draw a rectangle with side NL parallel to AC and ML parallel to AC such that L touches AB. Let the line segments AN be 7 units long and BM be 12 units.  Determine the area of the rectangle.

Draw the triangle ABC and the rectangle LMCN as shown.   Let the sides of the rectangle  be CM = ML = a, and CN = ML = b.


Note that angles at each corner of the rectangle are 90⁰.

Since angles on a straight line are 180⁰, ∠ANL = ∠LMB = 90⁰.

AC and LM are parallel being opposite sides of a rectangle.  Therefore, the corresponding angle ∠NAL = ∠MLB.

Similarly, ∠ALN = ∠LBM.

Since the corresponding angles of ANL and LMB are all equal, triangles ANL and LMB are similar and their sides have the same ratios.

Therefore,  the ratio a/7 = 12/b ,  Ceoss multiplying the two sides gives ab = 84.

Therefore, the area of the rectangle = ab = 84 units.

  1. Determine the radius of a semicircle inscribed from the longest side of a triangle whose sides are 3, 4 and 5 units.

Since the lenghts of the sides is 3, 4 and 5, Pythagorus states that it is a right angle triangle with the hypotenuse  BC  because 52 = 42 + 32 i.e.  25 = 16 +9.


The figure shows the triangle ABC and the semicircle on   BC inscribed in it.  Draw a perpendicular DE from the center of the circle on  AC and EF on AB.

Let the radius be r.  Then  DE = AF and EF = AD = r.  Also FB = 4-r and CD = 3-r.

Since DE is parallel to AB, ∠DEC =∠EBF. 

Since CDE and EFB are right angle triangles, ∠BEF  also equals ∠DCE. Therefore, these are similar triangles  and their corresponding sides have the same ratios:              

DC/DE = (3-r)/r =  EF/FR = r/(4-r).

(3-r)/r  = r/(4-r).             

Cross multipiication of the two sides  of the equations by  gives:

(3-r)(4-r) =  r 2 or 12 + r 2  -7r = r 2 or r = 12/7.  That is the radius of the semicircle.

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