- Level: International Baccalaureate
- Subject: Maths
- Word count: 3050
The segment of a polygon
Extracts from this document...
Introduction
THE SEGMENT OF A POLYGON
By: Risa Santoso 11DP-2
The purpose of this investigation is to find the relationship of the ratios of sides and the ratios of area in a polygon. By doing this investigation, we will be able to acquire more skills and knowledge about shapes and the relationship between the lines of sides and area of a shape with the help of trigonometry.
To find the relationship of the ratio of sides and area in a polygon, we use the “Geometer’s Sketchpad” program.
Draw an equilateral triangle using the straightedge tool in the “Geometer’s Sketchpad”. | Draw a line from the vertex to the opposite side with the ratio of 1:2 | Do the same with all of the other vertex. |
SIDES RATIO 1 : 2 | The results of the drawings are shown bellow. By letting the sides be 6 cm, the length of the AR (that made the ratio of 1:2 to RB) and the other two sides of PB and QC is 2 cm. The side of the smaller (inner) triangle is 2.27. Which have a ratio of 2.64:1 or 1:0.378 to the side length of the outer triangle (outer:inner) . The outer equilateral triangle’s area obtained from the “Geometer’s Sketchpad” is 15.59 cm² while the inner equilateral triangle’s area is 2.227cm². The sketchpad ratio of the two area are 1:0.143 or 7:1 (outer : inner). | |
SIDES RATIO 1 : 3 | This is the results of the drawings of the ratio 1:3 (length of sides). Like the graph on 1:2, let the sides be 6 cm, the length of the AR (that made the ratio of 1:3 to RB) and the other two sides of PB and QC is 1.5 cm (obtained from ). The side of the smaller (inner) triangle is 3.32 cm. It has the ratio of 1.81:1 or 1:0.55 to the side length of the outer triangle (outer:inner) . The outer equilateral triangle’s area obtained from the “Geometer’s Sketchpad” is 15.59 cm² while the inner equilateral triangle’s area is 4.77cm². The sketchpad ratio of the two area is 3.268:1 or 13:4 (outer : inner). | |
SIDES RATIO 1 : 4 | This is the results of the drawings of the ratio 1:4 (length of sides). Like the graph on 1:2 and 1:3, let the sides be 6 cm, the length of the AR (that made the ratio of 1:4 to RB) and the other two sides of PB and QC is 1.2 cm (counted from ). The side of the smaller (inner) triangle is 3.93cm. It has the ratio of 1.527:1 or 1:0.655 to the side length of the outer triangle (outer:inner) . The outer equilateral triangle’s area obtained from the “Geometer’s Sketchpad” is 15.59 cm² while the inner equilateral triangle’s area is 6.688cm². The sketchpad ratio of the two area is 1.53:1 or 7:3 (outer : inner). |
Table of result
Ratio of sides | Length of the outer (big) sides of triangle | Length of inner (smaller) sides of triangle | Ratio between the length of sides of the outer and the inner triangle | Area of outer (big) triangle | Area of the inner (small) triangle | Ratio between the area of the outer and inner triangle. |
1:2 | 6 cm | 2.27cm | 1:0.378 | 15.59cm² | 2.227cm² | 7:1 |
1:3 | 6 cm | 3.32 cm | 1:0.55 | 15.59cm² | 4.77cm² | 13:4 |
1:4 | 6 cm | 3.93 cm | 1:0.655 | 15.59cm² | 6.688cm² | 21:9 or 7:3 |
Middle
From the Pythagoras theory (tan= ) I got
=
from this point, we know that sinΘ = (sinΘ=)
while cosΘ = (cosΘ=)
To find AT we use the sine rule.
The ratio of the bigger equilateral triangle to the smaller equilateral triangle is acquired from
So, the ratio (after crossing the is 4:28 or simplified to be 1:7. The calculation of the “Geometer’s sketchpad” and the formula is true.
Of course, checking the other ratios (of the triangle shaped polygons) are a bit different from the one calculated above. Every one of the triangle have only 3 parts overlaying another part. When it is 1:2, the portion of the section is , which makes the whole part of the triangle covered if not minus the overlaying parts. In other ratios though, an additional calculation is needed.
An example is for the 1:3
In this drawing of triangle 1:3, let the sides be ‘S’.
Exactly the same as the 1:2, the area of ∆CAB can be determined as below
AR=PB=QC
∆PBU = ∆ATR = ∆QCV let the smallest triangular be ‘t’
=
=
=
There will be an additional to get the area of the smaller triangle.
For 1:4 it is basically
= +3t
Since
=
=
=
Therefore, to find the area of the smaller triangle, we are able to construct a formula of t. is the formula of the smaller triangle (S for the length of the sides)
Conclusion
I believe that the pattern of 1: n for other polygons will be similar. So if the sides are 5 (pentagon), the difference should be 8 and 10 so after the first side segment ratio of 1:2 is 14:9, the next ratio would be 22:16 and for the side segment ratio of 1:3 is 32:25. While for hexagon, the difference will be 9 and 11. So for the segment ratio of 1:2 is 19:16, 1:3 is 28:25 and for 1:4 side segment ratios is 39:36.
So from the predictions of relationship above, I made this table below to summarize my findings and my predictions.
Side segment ratio | Triangle area ratio (outer : inner) | Square area ratio (outer : inner) | Pentagon area ratio (outer : inner) | Hexagon area ratio (outer : inner) |
1:2 | 7:1 | 10:4 | 14:9 | 19:16 |
1:3 | 13:4 | 17:9 | 22:16 | 28:25 |
1:4 | 21:9 | 26:16 | 32:25 | 39:36 |
The ratio of the outer shape addition should increase since the inner shape also increases (the square of the next number). If we look at the outer ratio of the triangle and square, the pattern value increases every time, from 4 (17-13) for the 1:2, 5 (22-17) for 1:3. That is why I predict the pattern of the other polygons would be similar to that. Even though I cannot verify whether the patterns of the other kinds of polygon are exactly like my prediction, I know that the segments were constructed in a similar manner for other regular polygons since to find the triangles inside the shapes; we basically use the same formula of trigonometry (sine rule, etc).
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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