How Do Ratios of Sides Affect Areas in Regular Polygons?

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SUMMARY

The discussion focuses on the relationship between the ratios of sides and the areas of triangles and squares formed by dividing sides in specific ratios. In an equilateral triangle ABC, when a side is divided in the ratio 1:2, the area of the resulting triangle DEF can be analyzed using coordinate geometry and trigonometric rules. Participants suggest using the sine and cosine rules to derive the areas and establish conjectures regarding these ratios. The conversation also extends to squares and other regular polygons, prompting further investigation into similar relationships.

PREREQUISITES
  • Understanding of equilateral triangles and their properties
  • Familiarity with coordinate geometry
  • Knowledge of trigonometric functions, specifically sine and cosine rules
  • Basic concepts of area calculation for polygons
NEXT STEPS
  • Explore the derivation of area formulas for triangles using coordinate geometry
  • Investigate the application of the sine and cosine rules in area calculations
  • Research the properties of regular polygons and their area relationships
  • Conduct experiments with different ratios of side lengths in polygons to observe area changes
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Students studying geometry, mathematics educators, and anyone interested in the properties of polygons and their area relationships.

songoku
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Homework Statement


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

a. What is the ratio of the area of the two equilateral triangles?
b. Check the ratio of the areas for different ratio of side (1:n, pick your own value of n)
c. By analyzing the results above, conjecture a relationship between the ratios of the sides and the ratio of the areas of the triangles
d. Prove this conjecture analytically
e. Does this conjecture hold for non-equilateral triangles? Explain

2. Do a similar construction in a square where each side is divided into the ratio of 1:2.
a. Compare the area of the inner square to the area of the original square
b. How do the areas compare if each side is divided into the ratio 1:n?
c. Prove the conjecture

3. If segments were constructed in a similar manner in other regular polygons, would similar relationship exist? Investigate the relationship in another regular polygon

Homework Equations


Not sure

The Attempt at a Solution


I have already stuck at first question. I guess I use equation:
Area of triangle = 1/2. a . b . sin θ

But I can't find the side of smaller triangle DEF in terms of side of triangle ABC. Please help

Thanks
 
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Perhaps you can use coordinate geometry to get the coordinates of the points. If you let the side length be a, the two lower points could be (say) (0,0) and (a,0). Find the coordinates of the upper point, then find the points on the sides which divide them into the given ratio. You can find the coordinates of the sides of the inner triangle by calculating the equation of the lines joining the dividing with their opposite vertices.

EDIT: This seems like an IB Math assessment problem to me!
 
hi songoku! :smile:
songoku said:
… But I can't find the side of smaller triangle DEF in terms of side of triangle ABC.

my inclination would be to use the cosine and sine rules to find the angles, and then the area, of the smallest triangles, and then to find the are of the small equilateral triangle by subtraction

(btw, why do your very small diagrams take up so much space? :redface:)
 
That seems to be a much more efficient approach!
 
Pi-Bond said:
EDIT: This seems like an IB Math assessment problem to me!

Yes, it is; and now it is my homework

tiny-tim said:
hi songoku! :smile:

my inclination would be to use the cosine and sine rules to find the angles, and then the area, of the smallest triangles, and then to find the are of the small equilateral triangle by subtraction

hi tiny - tim! :smile:
let line BE meets side AC at Q. I can find the angle AQE but I still can't find the area of AEQ, if it is the smallest triangle you mean. I do not know how to find EQ or AE

(btw, why do your very small diagrams take up so much space? :redface:)

not sure :redface:
 
angle AEQ = 60° :wink:
 
tiny-tim said:
angle AEQ = 60° :wink:

I get it. Thanks a lot :smile:
 

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