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

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To calculate the area of a regular octagon cut from a square with a side length of 1 cm, the side length of the octagon can be represented as x, with the triangles cut from each corner having a side length of (1-x)/2. The area of the octagon is derived by subtracting the total area of the four triangles from the area of the square. The resulting area calculation yields approximately 0.828 cm², but it can be expressed more precisely as 1 - 2/((√2 + 2)²), which simplifies to 2(√2 - 1). This exact formulation provides a clearer understanding of the octagon's area.
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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)
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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