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Confusing problem determining the exact formula for area of 'complex' shape
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Draw a square with length n on each side, and then draw arcs of 1/4 circle along the DIAGONAL LINES inside this square. So we have $4$ arcs of 1/4 circle inside this square. My question is, What is the exact formula for area of the region at the center of this square?. I don't know any methods to find this, and I gave up. I want to draw it here using latex but I don't know how to do it, I'm new here. I will be amazed if someone can solve this problem quickly.I wish the formula only contain n, constant pi, rational numbers,sin/cos/tan/log/in, roots of any degree.
geometry
migrated from mathematica.stackexchange.com Aug 1 at 8:51
This question came from our site for users of Wolfram Mathematica.
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up vote
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down vote
favorite
Draw a square with length n on each side, and then draw arcs of 1/4 circle along the DIAGONAL LINES inside this square. So we have $4$ arcs of 1/4 circle inside this square. My question is, What is the exact formula for area of the region at the center of this square?. I don't know any methods to find this, and I gave up. I want to draw it here using latex but I don't know how to do it, I'm new here. I will be amazed if someone can solve this problem quickly.I wish the formula only contain n, constant pi, rational numbers,sin/cos/tan/log/in, roots of any degree.
geometry
migrated from mathematica.stackexchange.com Aug 1 at 8:51
This question came from our site for users of Wolfram Mathematica.
Seems to be Math homework and not Mathematica related...
– mgamer
Aug 1 at 8:33
Can you draw a rough sketch at least, even if not in LaTeX? I find it very hard to understand what shape you have in mind.
– Rahul
Aug 1 at 9:19
Is this the shape you want?
– John Glenn
Aug 1 at 10:33
There is no such thing as a circular arc along some line.
– Christian Blatter
Aug 1 at 11:07
It's unclear from the description, but if @JohnGlenn's image represents what OP wants, then this question is a duplicate of "Area of Bulging Square".
– Blue
Aug 1 at 23:36
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Draw a square with length n on each side, and then draw arcs of 1/4 circle along the DIAGONAL LINES inside this square. So we have $4$ arcs of 1/4 circle inside this square. My question is, What is the exact formula for area of the region at the center of this square?. I don't know any methods to find this, and I gave up. I want to draw it here using latex but I don't know how to do it, I'm new here. I will be amazed if someone can solve this problem quickly.I wish the formula only contain n, constant pi, rational numbers,sin/cos/tan/log/in, roots of any degree.
geometry
Draw a square with length n on each side, and then draw arcs of 1/4 circle along the DIAGONAL LINES inside this square. So we have $4$ arcs of 1/4 circle inside this square. My question is, What is the exact formula for area of the region at the center of this square?. I don't know any methods to find this, and I gave up. I want to draw it here using latex but I don't know how to do it, I'm new here. I will be amazed if someone can solve this problem quickly.I wish the formula only contain n, constant pi, rational numbers,sin/cos/tan/log/in, roots of any degree.
geometry
asked Aug 1 at 8:12
Bobby F
migrated from mathematica.stackexchange.com Aug 1 at 8:51
This question came from our site for users of Wolfram Mathematica.
migrated from mathematica.stackexchange.com Aug 1 at 8:51
This question came from our site for users of Wolfram Mathematica.
Seems to be Math homework and not Mathematica related...
– mgamer
Aug 1 at 8:33
Can you draw a rough sketch at least, even if not in LaTeX? I find it very hard to understand what shape you have in mind.
– Rahul
Aug 1 at 9:19
Is this the shape you want?
– John Glenn
Aug 1 at 10:33
There is no such thing as a circular arc along some line.
– Christian Blatter
Aug 1 at 11:07
It's unclear from the description, but if @JohnGlenn's image represents what OP wants, then this question is a duplicate of "Area of Bulging Square".
– Blue
Aug 1 at 23:36
add a comment |Â
Seems to be Math homework and not Mathematica related...
– mgamer
Aug 1 at 8:33
Can you draw a rough sketch at least, even if not in LaTeX? I find it very hard to understand what shape you have in mind.
– Rahul
Aug 1 at 9:19
Is this the shape you want?
– John Glenn
Aug 1 at 10:33
There is no such thing as a circular arc along some line.
– Christian Blatter
Aug 1 at 11:07
It's unclear from the description, but if @JohnGlenn's image represents what OP wants, then this question is a duplicate of "Area of Bulging Square".
– Blue
Aug 1 at 23:36
Seems to be Math homework and not Mathematica related...
– mgamer
Aug 1 at 8:33
Seems to be Math homework and not Mathematica related...
– mgamer
Aug 1 at 8:33
Can you draw a rough sketch at least, even if not in LaTeX? I find it very hard to understand what shape you have in mind.
– Rahul
Aug 1 at 9:19
Can you draw a rough sketch at least, even if not in LaTeX? I find it very hard to understand what shape you have in mind.
– Rahul
Aug 1 at 9:19
Is this the shape you want?
– John Glenn
Aug 1 at 10:33
Is this the shape you want?
– John Glenn
Aug 1 at 10:33
There is no such thing as a circular arc along some line.
– Christian Blatter
Aug 1 at 11:07
There is no such thing as a circular arc along some line.
– Christian Blatter
Aug 1 at 11:07
It's unclear from the description, but if @JohnGlenn's image represents what OP wants, then this question is a duplicate of "Area of Bulging Square".
– Blue
Aug 1 at 23:36
It's unclear from the description, but if @JohnGlenn's image represents what OP wants, then this question is a duplicate of "Area of Bulging Square".
– Blue
Aug 1 at 23:36
add a comment |Â
1 Answer
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In case this is your diagram:
Think of the area in the middle as:
You have the twice the total area of the areas of two sectors, $FEC$ and $BEG$, less areas of $triangle PGB$ and $triangle FQC$, less the area of $triangle PEQ$. Since the polygons are symmetric, you just have to solve the area of one each:
$$frac12 A=2 A_FEC-2A_triangle FQC-A_triangle PEQtag1$$
WLOG, consider your square to lie on the $xy$ -plane, centered at $(0,0)$. Thus, your vertices will be a combination of the points $left(pmfrac n2,pmfrac n2right)$.
Consider the intersection between the two circles centered at $B,C$. Using the formula for circles, with $r=n$, we know they intersect at:
$$E=left(0,frac12left(sqrt3n-nright)right)$$
The circles intersect the $x$-axis at the following points:
$$F=left(-frac12left(sqrt3n-nright),0right)\
G=left(frac12left(sqrt3n-nright),0right)$$
For $O=(0,0)$, it's very clear that $FO=EO$, therefore $FE=FOsqrt2$, and using the cosine law, we get that:
$$left(fracn-sqrt3 nsqrt2right)^2=-2 n^2 cos (angle FCE)+n^2+n^2\
angle FCE=fracpi6=30^circ$$
Therefore, $$A_FEC=pi n^2 cdot frac30360=fracpi n^212$$
To find $A_FQC$, consider the line $EC:=frac12left(sqrt3-1right)n-sqrt3x$, which intersects the $x$-axis at $Q=left(fracleft(sqrt3-1right)n2sqrt3,0right)$. Since we already know $F$, then $FQ=fracnsqrt3$. Using the distance formula tells us that $QC=fracnsqrt3$ thus $A_ FQC $ is :
$$A_FQC=frac12cdot ncdot fracnsqrt3cdot sin(30^circ)=fracn^24 sqrt3$$
To find $A_triangle PEQ$, notice that $PQ=2QO=fracleft(sqrt3-1right) nsqrt3$. Solving for the length of $EQ$, tells us that $triangle PEQ$ is equilateral, therefore:
$$A_triangle PEQ=frac12 left(fracleft(sqrt3-1right) nsqrt3right)^2cdot sin frac pi3=frac16 left(2 sqrt3-3right) n^2$$
Going back to $(1)$, we now get that the area of the figure in the middle is:
$$A=frac13 left(-3 sqrt3+pi +3right) n^2 tag2$$
Consider the circle centered at $B:=frac12left(sqrt3n^2-4nx-4x^2-nright)$, for $n=2$, the area in question, for $G=frac12 left(2 sqrt3-2right)$, should be:
$$A=4int_0^G frac12 left(sqrt-4 x^2-8 x+12-2right) , dx$$
Which evaluates to:
$$A=4cdotleft(frac13 left(-3 sqrt3+pi +3right)right)$$
which is exactly what we get from $(2)$
answered Aug 1 at 12:06
John Glenn
1,617123
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
In case this is your diagram:
Think of the area in the middle as:
You have the twice the total area of the areas of two sectors, $FEC$ and $BEG$, less areas of $triangle PGB$ and $triangle FQC$, less the area of $triangle PEQ$. Since the polygons are symmetric, you just have to solve the area of one each:
$$frac12 A=2 A_FEC-2A_triangle FQC-A_triangle PEQtag1$$
WLOG, consider your square to lie on the $xy$ -plane, centered at $(0,0)$. Thus, your vertices will be a combination of the points $left(pmfrac n2,pmfrac n2right)$.
Consider the intersection between the two circles centered at $B,C$. Using the formula for circles, with $r=n$, we know they intersect at:
$$E=left(0,frac12left(sqrt3n-nright)right)$$
The circles intersect the $x$-axis at the following points:
$$F=left(-frac12left(sqrt3n-nright),0right)\
G=left(frac12left(sqrt3n-nright),0right)$$
For $O=(0,0)$, it's very clear that $FO=EO$, therefore $FE=FOsqrt2$, and using the cosine law, we get that:
$$left(fracn-sqrt3 nsqrt2right)^2=-2 n^2 cos (angle FCE)+n^2+n^2\
angle FCE=fracpi6=30^circ$$
Therefore, $$A_FEC=pi n^2 cdot frac30360=fracpi n^212$$
To find $A_FQC$, consider the line $EC:=frac12left(sqrt3-1right)n-sqrt3x$, which intersects the $x$-axis at $Q=left(fracleft(sqrt3-1right)n2sqrt3,0right)$. Since we already know $F$, then $FQ=fracnsqrt3$. Using the distance formula tells us that $QC=fracnsqrt3$ thus $A_ FQC $ is :
$$A_FQC=frac12cdot ncdot fracnsqrt3cdot sin(30^circ)=fracn^24 sqrt3$$
To find $A_triangle PEQ$, notice that $PQ=2QO=fracleft(sqrt3-1right) nsqrt3$. Solving for the length of $EQ$, tells us that $triangle PEQ$ is equilateral, therefore:
$$A_triangle PEQ=frac12 left(fracleft(sqrt3-1right) nsqrt3right)^2cdot sin frac pi3=frac16 left(2 sqrt3-3right) n^2$$
Going back to $(1)$, we now get that the area of the figure in the middle is:
$$A=frac13 left(-3 sqrt3+pi +3right) n^2 tag2$$
Consider the circle centered at $B:=frac12left(sqrt3n^2-4nx-4x^2-nright)$, for $n=2$, the area in question, for $G=frac12 left(2 sqrt3-2right)$, should be:
$$A=4int_0^G frac12 left(sqrt-4 x^2-8 x+12-2right) , dx$$
Which evaluates to:
$$A=4cdotleft(frac13 left(-3 sqrt3+pi +3right)right)$$
which is exactly what we get from $(2)$
answered Aug 1 at 12:06
John Glenn
1,617123
add a comment |Â
up vote
3
down vote
In case this is your diagram:
Think of the area in the middle as:
You have the twice the total area of the areas of two sectors, $FEC$ and $BEG$, less areas of $triangle PGB$ and $triangle FQC$, less the area of $triangle PEQ$. Since the polygons are symmetric, you just have to solve the area of one each:
$$frac12 A=2 A_FEC-2A_triangle FQC-A_triangle PEQtag1$$
WLOG, consider your square to lie on the $xy$ -plane, centered at $(0,0)$. Thus, your vertices will be a combination of the points $left(pmfrac n2,pmfrac n2right)$.
Consider the intersection between the two circles centered at $B,C$. Using the formula for circles, with $r=n$, we know they intersect at:
$$E=left(0,frac12left(sqrt3n-nright)right)$$
The circles intersect the $x$-axis at the following points:
$$F=left(-frac12left(sqrt3n-nright),0right)\
G=left(frac12left(sqrt3n-nright),0right)$$
For $O=(0,0)$, it's very clear that $FO=EO$, therefore $FE=FOsqrt2$, and using the cosine law, we get that:
$$left(fracn-sqrt3 nsqrt2right)^2=-2 n^2 cos (angle FCE)+n^2+n^2\
angle FCE=fracpi6=30^circ$$
Therefore, $$A_FEC=pi n^2 cdot frac30360=fracpi n^212$$
To find $A_FQC$, consider the line $EC:=frac12left(sqrt3-1right)n-sqrt3x$, which intersects the $x$-axis at $Q=left(fracleft(sqrt3-1right)n2sqrt3,0right)$. Since we already know $F$, then $FQ=fracnsqrt3$. Using the distance formula tells us that $QC=fracnsqrt3$ thus $A_ FQC $ is :
$$A_FQC=frac12cdot ncdot fracnsqrt3cdot sin(30^circ)=fracn^24 sqrt3$$
To find $A_triangle PEQ$, notice that $PQ=2QO=fracleft(sqrt3-1right) nsqrt3$. Solving for the length of $EQ$, tells us that $triangle PEQ$ is equilateral, therefore:
$$A_triangle PEQ=frac12 left(fracleft(sqrt3-1right) nsqrt3right)^2cdot sin frac pi3=frac16 left(2 sqrt3-3right) n^2$$
Going back to $(1)$, we now get that the area of the figure in the middle is:
$$A=frac13 left(-3 sqrt3+pi +3right) n^2 tag2$$
Consider the circle centered at $B:=frac12left(sqrt3n^2-4nx-4x^2-nright)$, for $n=2$, the area in question, for $G=frac12 left(2 sqrt3-2right)$, should be:
$$A=4int_0^G frac12 left(sqrt-4 x^2-8 x+12-2right) , dx$$
Which evaluates to:
$$A=4cdotleft(frac13 left(-3 sqrt3+pi +3right)right)$$
which is exactly what we get from $(2)$
answered Aug 1 at 12:06
John Glenn
1,617123
add a comment |Â
up vote
3
down vote
up vote
3
down vote
In case this is your diagram:
Think of the area in the middle as:
You have the twice the total area of the areas of two sectors, $FEC$ and $BEG$, less areas of $triangle PGB$ and $triangle FQC$, less the area of $triangle PEQ$. Since the polygons are symmetric, you just have to solve the area of one each:
$$frac12 A=2 A_FEC-2A_triangle FQC-A_triangle PEQtag1$$
WLOG, consider your square to lie on the $xy$ -plane, centered at $(0,0)$. Thus, your vertices will be a combination of the points $left(pmfrac n2,pmfrac n2right)$.
Consider the intersection between the two circles centered at $B,C$. Using the formula for circles, with $r=n$, we know they intersect at:
$$E=left(0,frac12left(sqrt3n-nright)right)$$
The circles intersect the $x$-axis at the following points:
$$F=left(-frac12left(sqrt3n-nright),0right)\
G=left(frac12left(sqrt3n-nright),0right)$$
For $O=(0,0)$, it's very clear that $FO=EO$, therefore $FE=FOsqrt2$, and using the cosine law, we get that:
$$left(fracn-sqrt3 nsqrt2right)^2=-2 n^2 cos (angle FCE)+n^2+n^2\
angle FCE=fracpi6=30^circ$$
Therefore, $$A_FEC=pi n^2 cdot frac30360=fracpi n^212$$
To find $A_FQC$, consider the line $EC:=frac12left(sqrt3-1right)n-sqrt3x$, which intersects the $x$-axis at $Q=left(fracleft(sqrt3-1right)n2sqrt3,0right)$. Since we already know $F$, then $FQ=fracnsqrt3$. Using the distance formula tells us that $QC=fracnsqrt3$ thus $A_ FQC $ is :
$$A_FQC=frac12cdot ncdot fracnsqrt3cdot sin(30^circ)=fracn^24 sqrt3$$
To find $A_triangle PEQ$, notice that $PQ=2QO=fracleft(sqrt3-1right) nsqrt3$. Solving for the length of $EQ$, tells us that $triangle PEQ$ is equilateral, therefore:
$$A_triangle PEQ=frac12 left(fracleft(sqrt3-1right) nsqrt3right)^2cdot sin frac pi3=frac16 left(2 sqrt3-3right) n^2$$
Going back to $(1)$, we now get that the area of the figure in the middle is:
$$A=frac13 left(-3 sqrt3+pi +3right) n^2 tag2$$
Consider the circle centered at $B:=frac12left(sqrt3n^2-4nx-4x^2-nright)$, for $n=2$, the area in question, for $G=frac12 left(2 sqrt3-2right)$, should be:
$$A=4int_0^G frac12 left(sqrt-4 x^2-8 x+12-2right) , dx$$
Which evaluates to:
$$A=4cdotleft(frac13 left(-3 sqrt3+pi +3right)right)$$
which is exactly what we get from $(2)$
answered Aug 1 at 12:06
John Glenn
1,617123
In case this is your diagram:
Think of the area in the middle as:
You have the twice the total area of the areas of two sectors, $FEC$ and $BEG$, less areas of $triangle PGB$ and $triangle FQC$, less the area of $triangle PEQ$. Since the polygons are symmetric, you just have to solve the area of one each:
$$frac12 A=2 A_FEC-2A_triangle FQC-A_triangle PEQtag1$$
WLOG, consider your square to lie on the $xy$ -plane, centered at $(0,0)$. Thus, your vertices will be a combination of the points $left(pmfrac n2,pmfrac n2right)$.
Consider the intersection between the two circles centered at $B,C$. Using the formula for circles, with $r=n$, we know they intersect at:
$$E=left(0,frac12left(sqrt3n-nright)right)$$
The circles intersect the $x$-axis at the following points:
$$F=left(-frac12left(sqrt3n-nright),0right)\
G=left(frac12left(sqrt3n-nright),0right)$$
For $O=(0,0)$, it's very clear that $FO=EO$, therefore $FE=FOsqrt2$, and using the cosine law, we get that:
$$left(fracn-sqrt3 nsqrt2right)^2=-2 n^2 cos (angle FCE)+n^2+n^2\
angle FCE=fracpi6=30^circ$$
Therefore, $$A_FEC=pi n^2 cdot frac30360=fracpi n^212$$
To find $A_FQC$, consider the line $EC:=frac12left(sqrt3-1right)n-sqrt3x$, which intersects the $x$-axis at $Q=left(fracleft(sqrt3-1right)n2sqrt3,0right)$. Since we already know $F$, then $FQ=fracnsqrt3$. Using the distance formula tells us that $QC=fracnsqrt3$ thus $A_ FQC $ is :
$$A_FQC=frac12cdot ncdot fracnsqrt3cdot sin(30^circ)=fracn^24 sqrt3$$
To find $A_triangle PEQ$, notice that $PQ=2QO=fracleft(sqrt3-1right) nsqrt3$. Solving for the length of $EQ$, tells us that $triangle PEQ$ is equilateral, therefore:
$$A_triangle PEQ=frac12 left(fracleft(sqrt3-1right) nsqrt3right)^2cdot sin frac pi3=frac16 left(2 sqrt3-3right) n^2$$
Going back to $(1)$, we now get that the area of the figure in the middle is:
$$A=frac13 left(-3 sqrt3+pi +3right) n^2 tag2$$
Consider the circle centered at $B:=frac12left(sqrt3n^2-4nx-4x^2-nright)$, for $n=2$, the area in question, for $G=frac12 left(2 sqrt3-2right)$, should be:
$$A=4int_0^G frac12 left(sqrt-4 x^2-8 x+12-2right) , dx$$
Which evaluates to:
$$A=4cdotleft(frac13 left(-3 sqrt3+pi +3right)right)$$
which is exactly what we get from $(2)$
answered Aug 1 at 12:06
John Glenn
1,617123
edited Aug 1 at 23:27
answered Aug 1 at 12:06
John Glenn
1,617123
answered Aug 1 at 12:06
John Glenn
1,617123
answered Aug 1 at 12:06
John Glenn
1,617123
1,617123
add a comment |Â
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Seems to be Math homework and not Mathematica related...
– mgamer
Aug 1 at 8:33
Can you draw a rough sketch at least, even if not in LaTeX? I find it very hard to understand what shape you have in mind.
– Rahul
Aug 1 at 9:19
Is this the shape you want?
– John Glenn
Aug 1 at 10:33
There is no such thing as a circular arc along some line.
– Christian Blatter
Aug 1 at 11:07
It's unclear from the description, but if @JohnGlenn's image represents what OP wants, then this question is a duplicate of "Area of Bulging Square".
– Blue
Aug 1 at 23:36