# Math IA- Type 1 The Segments of a Polygon

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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:

- 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)

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 = 1:3

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

Middle

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

Conclusion

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.

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

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