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

Math IA- Type 1 The Segments of a Polygon

Extracts from this document...

Introduction

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.

a)

 i)

image00.png

ii)

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

iii)

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.

Therefore

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.

Therefore

62.0 cm2/19.1cm2 = 13:4

...read more.

Middle

2

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.

image04.png

 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

image05.png

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

...read more.

Conclusion

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. Math Studies I.A

    1211040000 6229.9449 Gibraltar 38,200 80.9 3090380 1459240000 6544.81 Greenland 20,000 70.23 1404600 400000000 4932.2529 Guam 15,000 75.5 1132500 225000000 5700.25 Guernsey 44,600 80.42 3586732 1989160000 6467.3764 Guinea-Bissan 600 47.18 28308 360000 2225.9524 Haiti 1,300 57.03 74139 1690000 3252.4209 Hong Kong 43,800 81.77 3581526 1918440000 6686.3329 Iceland 39,900 80.43 3209157 1592010000

  1. Math Studies - IA

    As mentioned the Majors have been chosen as the basis of determining whether the new Ryder Cup trend is coherent with reality. The different sample sizes are at this point already a potential limitation to the investigation, since it can change the data.

  2. Math IA - Logan's Logo

    From this equation we see that as the value of b is increased (the horizontal dilation of the curve is greater), the period becomes smaller. We know this because b is the denominator, meaning that increasing its value would decrease the overall fraction's value.

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

    Data Table 3 Total Points played in the game. (Y) The available points* Binomial Distribution for Adam The number of possibilities for the game being played when Adam loses all games. Binomial distribution for Ben The number of possibilities for the game being played when Ben loses all games.

  2. Mathematics IA - Particles

    At this point I decide that I want to find out how long the patient will live for if he goes untreated. I presume that once the immune system responds the particles will not double every four hours, but instead they will increase by 160%.

  1. Gold Medal heights IB IA- score 15

    to its origin The curvature of a square root function is fairly simple, in Figure 4 the root curve begins with a slope greater than one. It eventually hits the slope of one and lastly it continues to decrease in slope meaning it will infinitely approach closer to a value but never reaching it.

  2. MATH Lacsap's Fractions IA

    To find the sixth row, the row number will be inputted for n and the element will also be inputted for r. The working is shown in Table 6. Table 6: Calculating the values on the sixth row E6 (1)

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