How Do You Calculate the Area of a Regular Octagon Cut from a Square?

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Homework Help Overview

The problem involves calculating the area of a regular octagon formed by cutting off four equal triangles from the corners of a square with a side length of 1 cm. Participants are exploring the geometric relationships and equations that arise from this setup.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the side length of the octagon and the dimensions of the triangles being cut from the square. There are attempts to derive equations based on these relationships, including the use of quadratic equations to find the side length of the octagon.

Discussion Status

Multiple participants have shared their calculations and reasoning, with some expressing uncertainty about the correctness of their results. There is an ongoing exploration of different methods to arrive at the area, and some participants have suggested alternative expressions for the area that may provide more exact results.

Contextual Notes

Participants are working under the constraints of the problem as stated, with a focus on the geometric properties of the shapes involved and the implications of their calculations. There is a noted concern about the accuracy of decimal approximations versus exact values.

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Homework Statement


The question says that there is a square that has a side of length 1cm. It then says 4 equal triangles are cut, one off each corner. The resulting shape is a regular octagon. What is its area?


Homework Equations


I believe the sides of the triangles to both be (1-x)/2 assuming we let each side of the octagon be length x. This means that length x is represented by (triangle side length)^2 + (triangle side length)^2 = x^2.



The Attempt at a Solution


I essentially let the above equations equal and I get...
(0.5 - x^2/2)^2 + (0.5 - x^2/2)^2 = x^2
Using quadratic after simplifying the above...
X= rt2 - 1
From there I let that equal x again. I find the total area of all 4 triangles and get 1- area of four triangles which leaves me with 0.828cm^2

Is this correct? I'm not convinced it is actually correct. Any guidance?
 
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Idiotinabox said:

Homework Statement


The question says that there is a square that has a side of length 1cm. It then says 4 equal triangles are cut, one off each corner. The resulting shape is a regular octagon. What is its area?


Homework Equations


I believe the sides of the triangles to both be (1-x)/2 assuming we let each side of the octagon be length x. This means that length x is represented by (triangle side length)^2 + (triangle side length)^2 = x^2.



The Attempt at a Solution


I essentially let the above equations equal and I get...
(0.5 - x^2/2)^2 + (0.5 - x^2/2)^2 = x^2
Using quadratic after simplifying the above...
X= rt2 - 1
From there I let that equal x again. I find the total area of all 4 triangles and get 1- area of four triangles which leaves me with 0.828cm^2

Is this correct? I'm not convinced it is actually correct. Any guidance?

I got that answer calculating a different way, so it is probably correct.
 
Idiotinabox said:

Homework Statement


The question says that there is a square that has a side of length 1cm. It then says 4 equal triangles are cut, one off each corner. The resulting shape is a regular octagon. What is its area?


Homework Equations


I believe the sides of the triangles to both be (1-x)/2 assuming we let each side of the octagon be length x. This means that length x is represented by (triangle side length)^2 + (triangle side length)^2 = x^2.



The Attempt at a Solution


I essentially let the above equations equal and I get...
(0.5 - x^2/2)^2 + (0.5 - x^2/2)^2 = x^2
Using quadratic after simplifying the above...
X= rt2 - 1
From there I let that equal x again. I find the total area of all 4 triangles and get 1- area of four triangles which leaves me with 0.828cm^2

That's correct, but why give a decimal approximation when you have an exact answer:$$
1-\frac 2 {(\sqrt 2 + 2)^2}$$
 
A better simplification would be 2(√2 - 1)
 

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