• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
  6. 6
  7. 7
  8. 8
  9. 9
  10. 10
  11. 11
  12. 12
  13. 13
  14. 14
  15. 15
  16. 16
  17. 17
  18. 18
  19. 19

Math IA- Type 1 The Segments of a Polygon

Extracts from this document...


Math IA- Type 1

The Segments of a Polygon

Abhinav Jain

IB Higher Level Assignment: Internal Assessment

Type 1: Modeling

Mr. Murgatroyd

Date: 15/03/2009

Word Count:

  1. In an equilateral triangle ABC, a line segment is drawn from each vertex to a point on the opposite side so that the segment divides the side in the ratio 1:2, creating another equilateral triangle DEF.





Measurements and drawing shown above has been made through the Geometer’s Sketchpad package.

Measure of one side of the ΔABC = 12cm

Measure of one side of the ΔDEF = 5cm


The areas have also been calculated using the Geometer’s Sketchpad package and are show in the diagram above.

For ΔABC = 62.5cm2

For ΔDEF = 8.8 cm2

In order to find the ratio, one needs to divide the area of ΔABC by the area of the ΔDEF which will give one the ratio of the area of the bigger triangle to the smaller triangle.


62.5cm2 ÷ 8.8 cm2 = 7:1

The ratio between the areas of the equilateral ΔABC to ΔDEF when the segment divides the side in the ratio 1:2 is 7:1.

b) In order to repeat the procedure above for at least two other side ratios, 1: n

The two ratios chosen are 1:3 and 1:4 for no specific reason.

Ratio of Sides = 1:3image01.png

Again, the diagram above and values were obtained and created using Geometer’s Sketchpad Package.

For ΔABC = 62.0 cm2

For ΔDEF = 19.1 cm2

In order to find the ratio, one needs to divide the area of ΔABC by the area of the ΔDEF which will give one the ratio of the area of the bigger triangle to the smaller triangle.


62.0 cm2/19.1cm2 = 13:4

...read more.



d) Before this conjecture is proved analytically, it is very important to understand the basics of similarity. Triangles are similar if their corresponding angles are congruent and the ratio of their corresponding sides is in proportion. The name for this proportion is known as the similarity ratio.

In order to prove the conjecture above, one needs to find the area of the bigger triangle and divide it by the area of the smaller triangle. If the produced equation matches the equation above, then the conjecture has been proved analytically.

The conjecture has been proved on the following page.

In to order to able to test the validity of this conjecture, another triangle was produced using the geometer’s sketch pad and the conjecture is validated if the ratio produced from GSP matches the value of the ratio provided by the conjecture.


 The ratio the segment divides the side in is 1:9

From our conjecture which is

(n2 + n + 1) ÷ (n – 1)2

Therefore if 9 is substituted in place on n, then the following answer is retrieved.

(92 + 9 + 1) ÷ (9-1)2

= 1.42

Therefore both the values from GSP and the conjecture match, proving the conjecture to be valid.

2) In order to be able to prove whether this conjecture is valid for non-equilateral triangles, another sketch is made using the GSP. However, this time a non equilateral triangle is made. If the value of the ratio of the area matches the ratio from the conjecture, then the equations holds true for non-equilateral triangles.  The ratio of sides that will be used will 1:4


The ratio between the areas of the two triangles show above is 2.34

...read more.


2. Therefore, one can also deduce that the other smaller shapes such as of a pentagon or an octagon will also follow this rule.

Limitations of this assignment:  

There was one main limitation that was related to the technological aspect of this assignment. The main problem of this assignment was using GSP itself. One can say that the values calculated by GSP for the different areas and lengths were not as accurate as they could possibly have been. This is partly due to human error as while we try to use the mouse to change the different lengths, it is nearly impossible to be able to move the hand with such precision so that every one of our lengths is consistent with each other therefore resulting in small inaccuracies caused due to the way GSP was programmed and the mouse that was in use while making the different shapes.

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Extended Essay- Math

    ;.1/4�~\<� "U� �?���'��]����%Y>����cSH���%�F,�JG��3�9��kR [�c�AE�FL�n�!�K'�*�Q�4ޯ���uF@� ggZ��[' Í8[-�21/4(r)ٷ�y`Zk��!"$��q�Ȩ�U�^�d�|��m<�MG�<���� 3/4�(tm)�7�1/4Z�3�m2�� /57�'4�&S�1/2Ë-^6m�6Ìn��"Û¦m߰��;,���(tm)-� �)�E O� *t�:���WH �qLG�>_+��4 #��1�����G( 3�'���ע�E�b��Bs���� !h�X�"�1���E9'��:�b�^�C�(c)�Oug�N �_o��LOÌ��y�Ԫ�=��-a�JY�6rÙ¦ �N{"`d�z�=������-��O��2�Ԧ��� �A��5(tm)�(��gk'��s��Y��"eT�A"9<9k�kam(r)�4���Åhd.�l���7.�"����61|z �"g�Î�A _�xAP"y�>��'�8H'(r)���;�(tm)�Ý�w��Ĵ�Ï�<BÌ(��$�í¯H~O�^�<1r4GC��� �|�u����^��^7��3/4�����T�jÌÔQ �ì¦-���%U ��P�Mp��G2��_��%�@��RuOR���PK!�ڷ �word/endnotes.xml�U�n�0}_(c)����`ZJ[+�Ru��ն�(r)1�*3/4�6���H ��Vi"� ���9�(tm)aq�*"`Ë­Z�(�cp�t.�&E��-f�(p���VZ��C�Ëo��p�+� B9�-��{C�б'K��R0"�.��i�����yx�#�}"w �}�jK:�I}-��� 3/4�6"T�c�H(r)�o����͵Ý�}(c)� �ų�"�?� 0��V'��� ��!@�le��>"�ٿz;:Ar�'iVK(r)|G/�1/4�Z�R�c�>��({J� >Û(-��sΥg-6p�}�Æ�&''�d��C[P�2z��3n�...8�C��="��k>-�c%5Z�+ ���� ����Z1/2 X�$�3�t�~*�}`4+-Kj8 $#��->W�2-��-��4�� l:n��^[&'�hu�Ln��[�13/4"�WW"�3/4����&e�+E�0�"�"�Ö�lM0�8��-�&\.B�Vӭ��:��i�...�"�|" O�7�*K(r)�"�Ib�H��c� ��PK!iI`�H-word/header1.xml�T�R�0}�L�A�g+W� �iHh3�v��~���XE�'��3/4+��N�_V�s�-����^I��� �3<���f&z-�7W1/2F>P�Si4����b���y��C��}ZY-�"�&�gW��`�x� }fTb�[�xR-'C2 �-u�q��'�=��N(tm)��e-�Y���ơ"c�1/2[� 3/4o� 2�]i{ �� n...��'i ��p�t�XU�"*� �+�n���8�6�6����&eiX(c)� �1/4�q ���...�O3/41/2 �(ZI�-�Y����:{�9��}K\���È$%bC=��ßrÄD�N�1��l�<o3/4�m�<uRe�1/2�B}v��]UV1/4m��:�8^��L�"�1/44�*��Y�)��)-(r)w�8z+A8�bG�9 &��7���u����bt���p �>����x�X�^A4...�-��0!

  2. Tide Modeling

    Difference 0.725 0.637 0.31 0 0.14 0.087 0.325 0.412 0.29 0 0.26 0.438 ??=5.675,sin-,,??-6.,??...+6.475 and ??=5.675,cos-,,??-6.,??-3...+ 6.475 Only one color is shown because graphs overlap each other since they are the same. 4. Use the regression feature of your graphing calculator to develop a best-fit function for this data.

  1. Math Studies I.A

    Malaysia 15,300 72.76 1113228 234090000 5294.0176 Mali 1,200 54.5 65400 1440000 2970.25 Marshall Islands 2,500 70.61 176525 6250000 4985.7721 Mauritius 12,100 72.88 881848 146410000 5311.4944 Mexico 14,200 76.2 1082040 201640000 5806.44 Moldova 2,500 70.2 175500 6250000 4928.04 Mongolia 3,200 66.99 214368 10240000 4487.6601 Montserrat 3,400 72.6 246840 11560000 5270.76 Mozambique

  2. Math IA - Logan's Logo

    changing the variable c will affect the horizontal shift of the sine curve, so thatare translations to the right, whileare translations to the left. The original sine curve starts (meaning it crosses the center line of its curve) at point (0,0), and using this point as a reference, I can determine how many units leftwards my curve has shifted.

  1. Math Studies - IA

    In the Masters (one of four majors) their mean score was 72. In the same way, the entire team's mean score can be determined in the majors. It is unlikely that all Ryder Cup players participate consequently, and that will be a limitation to the investigation. Concerning the Ryder Cup, Europe has won every time on this side of the millennium.

  2. Math IA type 2. In this task I will be investigating Probabilities and investigating ...

    whose probability of winning is being calculated therefore the n will always be the total points played subtracted by 1.] However, k will be the number of points that need to be won by the player whose probability of winning is being calculated.

  1. Math IA - Matrix Binomials

    Thus, we can now deduce the geometric sequence of these scalar values using the general equation listed above: In the sequence {1, 20, 400, 8000}, f=1 r=20 Here, we can express r in terms of b (b=10): r= 20 20= 10?2 r=2b We can also express in terms of b

  2. Gold Medal heights IB IA- score 15

    This function could represent majority of the data points as the correlation increases then slowly levels out. However, to specific reference to the data point (1936, 203) indicates that a line of best fit would start with a negative correlation then switch into a positive correlation with the following years.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work