Four equal circles intersect: What is the area of the small shaded portion and its height
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In terms of $R$ which is the radius of all four circles, what is the area of the intersection region of these four equal circles and the height of the marked arrow in the figure? The marked arrow is along the line CD, also the midpoint of all the circles are points A, B, C and D. Looking for a very short intuitive solution. I have checked similar questions on this site for example this and this.
geometry circle
$endgroup$
add a comment |
$begingroup$
In terms of $R$ which is the radius of all four circles, what is the area of the intersection region of these four equal circles and the height of the marked arrow in the figure? The marked arrow is along the line CD, also the midpoint of all the circles are points A, B, C and D. Looking for a very short intuitive solution. I have checked similar questions on this site for example this and this.
geometry circle
$endgroup$
1
$begingroup$
Notice anything special about $triangle ABC$?
$endgroup$
– Blue
2 hours ago
$begingroup$
Hmmn! It's equilateral!
$endgroup$
– Abdulhameed
1 hour ago
1
$begingroup$
Then just use one of the previous solutions, since you know the angle.
$endgroup$
– Andrei
1 hour ago
$begingroup$
Thanks! I think its clear now! I need to subtract the area of the sector from the triangle to get the small area and then multiply by two. What about the height any clues?
$endgroup$
– Abdulhameed
1 hour ago
$begingroup$
@Abdulhameed: For height, use the equilateral triangles again. If you know the height of a triangle, and the radius of the circle, then ...
$endgroup$
– Blue
1 hour ago
add a comment |
$begingroup$
In terms of $R$ which is the radius of all four circles, what is the area of the intersection region of these four equal circles and the height of the marked arrow in the figure? The marked arrow is along the line CD, also the midpoint of all the circles are points A, B, C and D. Looking for a very short intuitive solution. I have checked similar questions on this site for example this and this.
geometry circle
$endgroup$
In terms of $R$ which is the radius of all four circles, what is the area of the intersection region of these four equal circles and the height of the marked arrow in the figure? The marked arrow is along the line CD, also the midpoint of all the circles are points A, B, C and D. Looking for a very short intuitive solution. I have checked similar questions on this site for example this and this.
geometry circle
geometry circle
edited 2 hours ago
Abdulhameed
asked 2 hours ago
AbdulhameedAbdulhameed
122113
122113
1
$begingroup$
Notice anything special about $triangle ABC$?
$endgroup$
– Blue
2 hours ago
$begingroup$
Hmmn! It's equilateral!
$endgroup$
– Abdulhameed
1 hour ago
1
$begingroup$
Then just use one of the previous solutions, since you know the angle.
$endgroup$
– Andrei
1 hour ago
$begingroup$
Thanks! I think its clear now! I need to subtract the area of the sector from the triangle to get the small area and then multiply by two. What about the height any clues?
$endgroup$
– Abdulhameed
1 hour ago
$begingroup$
@Abdulhameed: For height, use the equilateral triangles again. If you know the height of a triangle, and the radius of the circle, then ...
$endgroup$
– Blue
1 hour ago
add a comment |
1
$begingroup$
Notice anything special about $triangle ABC$?
$endgroup$
– Blue
2 hours ago
$begingroup$
Hmmn! It's equilateral!
$endgroup$
– Abdulhameed
1 hour ago
1
$begingroup$
Then just use one of the previous solutions, since you know the angle.
$endgroup$
– Andrei
1 hour ago
$begingroup$
Thanks! I think its clear now! I need to subtract the area of the sector from the triangle to get the small area and then multiply by two. What about the height any clues?
$endgroup$
– Abdulhameed
1 hour ago
$begingroup$
@Abdulhameed: For height, use the equilateral triangles again. If you know the height of a triangle, and the radius of the circle, then ...
$endgroup$
– Blue
1 hour ago
1
1
$begingroup$
Notice anything special about $triangle ABC$?
$endgroup$
– Blue
2 hours ago
$begingroup$
Notice anything special about $triangle ABC$?
$endgroup$
– Blue
2 hours ago
$begingroup$
Hmmn! It's equilateral!
$endgroup$
– Abdulhameed
1 hour ago
$begingroup$
Hmmn! It's equilateral!
$endgroup$
– Abdulhameed
1 hour ago
1
1
$begingroup$
Then just use one of the previous solutions, since you know the angle.
$endgroup$
– Andrei
1 hour ago
$begingroup$
Then just use one of the previous solutions, since you know the angle.
$endgroup$
– Andrei
1 hour ago
$begingroup$
Thanks! I think its clear now! I need to subtract the area of the sector from the triangle to get the small area and then multiply by two. What about the height any clues?
$endgroup$
– Abdulhameed
1 hour ago
$begingroup$
Thanks! I think its clear now! I need to subtract the area of the sector from the triangle to get the small area and then multiply by two. What about the height any clues?
$endgroup$
– Abdulhameed
1 hour ago
$begingroup$
@Abdulhameed: For height, use the equilateral triangles again. If you know the height of a triangle, and the radius of the circle, then ...
$endgroup$
– Blue
1 hour ago
$begingroup$
@Abdulhameed: For height, use the equilateral triangles again. If you know the height of a triangle, and the radius of the circle, then ...
$endgroup$
– Blue
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you repeat this pattern infinitely, you'll find that each circle consists of $12$ of these football shaped areas, together with $6$ 'triangular' areas in between (see red circle in figure). So, setting the area of the footballs to $A$, and that of the triangles to $B$, we have:
$pi^2=12A +6B$ (I am setting radius to $1$ ... you can easily adjust for $R$)
OK, now consider the green rectangle formed by four of the points on a circle. We see that the height of such a rectangle is equal to the radius, so that is $1$, and the width is easily found to be $sqrt{3}$, and this area includes $4$ whole footballs, $2$ half footballs, $2$ whole 'triangles', and $4$ half triangles. So:
$sqrt{3}=5A+4B$
Now you can easily solve for $A$
$endgroup$
$begingroup$
I appreciate this effort but could you make this clearer.
$endgroup$
– Abdulhameed
1 hour ago
$begingroup$
@Abdulhameed Added a picture
$endgroup$
– Bram28
58 mins ago
$begingroup$
@Bram28 really nice picture, +1
$endgroup$
– Zubin Mukerjee
55 mins ago
$begingroup$
This is so beautiful! Its well appreciated
$endgroup$
– Abdulhameed
51 mins ago
$begingroup$
@Abdulhameed Another solution is to take $1/6$ of the area of the area of the circle of radius $1$, subtract the area of an equilateral triangle with side length $1$, then multiply by $2$: $$2left(frac{1}{6}pi^2 - frac{sqrt{3}}{4}right)$$
$endgroup$
– Zubin Mukerjee
45 mins ago
|
show 2 more comments
$begingroup$
Notice that the area of the equilateral triangle with edge $R$ plus $frac12$ the area of the shaded region is $frac16$ the area of the circle. The height of the triangle is $frac{Rsqrt3}{2}$ and the area is $frac12cdot Rcdotfrac{Rsqrt3}{2}=frac{R^2sqrt3}{4}$. The area of the shaded region is:
$$2left(frac{pi R^2}{6}-frac{R^2sqrt3}{4}right)=frac{2pi R^2}{6}-frac{3R^2sqrt3}{6}=frac{R^2(2pi-3sqrt3)}{6}$$
Also notice that the height of the triangle plus $frac12$ the height of the shaded area is equal to the radius of the circle. The height of the shaded area is:
$$2left(R-frac{Rsqrt3}{2}right)=2R-Rsqrt3=R(2-sqrt3)$$
$endgroup$
$begingroup$
The area of the triangle should be $$frac{R^2sqrt{3}}{4}$$
$endgroup$
– Zubin Mukerjee
41 mins ago
$begingroup$
@ZubinMukerjee oh, yeah... mental slip
$endgroup$
– Daniel Mathias
39 mins ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you repeat this pattern infinitely, you'll find that each circle consists of $12$ of these football shaped areas, together with $6$ 'triangular' areas in between (see red circle in figure). So, setting the area of the footballs to $A$, and that of the triangles to $B$, we have:
$pi^2=12A +6B$ (I am setting radius to $1$ ... you can easily adjust for $R$)
OK, now consider the green rectangle formed by four of the points on a circle. We see that the height of such a rectangle is equal to the radius, so that is $1$, and the width is easily found to be $sqrt{3}$, and this area includes $4$ whole footballs, $2$ half footballs, $2$ whole 'triangles', and $4$ half triangles. So:
$sqrt{3}=5A+4B$
Now you can easily solve for $A$
$endgroup$
$begingroup$
I appreciate this effort but could you make this clearer.
$endgroup$
– Abdulhameed
1 hour ago
$begingroup$
@Abdulhameed Added a picture
$endgroup$
– Bram28
58 mins ago
$begingroup$
@Bram28 really nice picture, +1
$endgroup$
– Zubin Mukerjee
55 mins ago
$begingroup$
This is so beautiful! Its well appreciated
$endgroup$
– Abdulhameed
51 mins ago
$begingroup$
@Abdulhameed Another solution is to take $1/6$ of the area of the area of the circle of radius $1$, subtract the area of an equilateral triangle with side length $1$, then multiply by $2$: $$2left(frac{1}{6}pi^2 - frac{sqrt{3}}{4}right)$$
$endgroup$
– Zubin Mukerjee
45 mins ago
|
show 2 more comments
$begingroup$
If you repeat this pattern infinitely, you'll find that each circle consists of $12$ of these football shaped areas, together with $6$ 'triangular' areas in between (see red circle in figure). So, setting the area of the footballs to $A$, and that of the triangles to $B$, we have:
$pi^2=12A +6B$ (I am setting radius to $1$ ... you can easily adjust for $R$)
OK, now consider the green rectangle formed by four of the points on a circle. We see that the height of such a rectangle is equal to the radius, so that is $1$, and the width is easily found to be $sqrt{3}$, and this area includes $4$ whole footballs, $2$ half footballs, $2$ whole 'triangles', and $4$ half triangles. So:
$sqrt{3}=5A+4B$
Now you can easily solve for $A$
$endgroup$
$begingroup$
I appreciate this effort but could you make this clearer.
$endgroup$
– Abdulhameed
1 hour ago
$begingroup$
@Abdulhameed Added a picture
$endgroup$
– Bram28
58 mins ago
$begingroup$
@Bram28 really nice picture, +1
$endgroup$
– Zubin Mukerjee
55 mins ago
$begingroup$
This is so beautiful! Its well appreciated
$endgroup$
– Abdulhameed
51 mins ago
$begingroup$
@Abdulhameed Another solution is to take $1/6$ of the area of the area of the circle of radius $1$, subtract the area of an equilateral triangle with side length $1$, then multiply by $2$: $$2left(frac{1}{6}pi^2 - frac{sqrt{3}}{4}right)$$
$endgroup$
– Zubin Mukerjee
45 mins ago
|
show 2 more comments
$begingroup$
If you repeat this pattern infinitely, you'll find that each circle consists of $12$ of these football shaped areas, together with $6$ 'triangular' areas in between (see red circle in figure). So, setting the area of the footballs to $A$, and that of the triangles to $B$, we have:
$pi^2=12A +6B$ (I am setting radius to $1$ ... you can easily adjust for $R$)
OK, now consider the green rectangle formed by four of the points on a circle. We see that the height of such a rectangle is equal to the radius, so that is $1$, and the width is easily found to be $sqrt{3}$, and this area includes $4$ whole footballs, $2$ half footballs, $2$ whole 'triangles', and $4$ half triangles. So:
$sqrt{3}=5A+4B$
Now you can easily solve for $A$
$endgroup$
If you repeat this pattern infinitely, you'll find that each circle consists of $12$ of these football shaped areas, together with $6$ 'triangular' areas in between (see red circle in figure). So, setting the area of the footballs to $A$, and that of the triangles to $B$, we have:
$pi^2=12A +6B$ (I am setting radius to $1$ ... you can easily adjust for $R$)
OK, now consider the green rectangle formed by four of the points on a circle. We see that the height of such a rectangle is equal to the radius, so that is $1$, and the width is easily found to be $sqrt{3}$, and this area includes $4$ whole footballs, $2$ half footballs, $2$ whole 'triangles', and $4$ half triangles. So:
$sqrt{3}=5A+4B$
Now you can easily solve for $A$
edited 59 mins ago
answered 1 hour ago
Bram28Bram28
61.2k44591
61.2k44591
$begingroup$
I appreciate this effort but could you make this clearer.
$endgroup$
– Abdulhameed
1 hour ago
$begingroup$
@Abdulhameed Added a picture
$endgroup$
– Bram28
58 mins ago
$begingroup$
@Bram28 really nice picture, +1
$endgroup$
– Zubin Mukerjee
55 mins ago
$begingroup$
This is so beautiful! Its well appreciated
$endgroup$
– Abdulhameed
51 mins ago
$begingroup$
@Abdulhameed Another solution is to take $1/6$ of the area of the area of the circle of radius $1$, subtract the area of an equilateral triangle with side length $1$, then multiply by $2$: $$2left(frac{1}{6}pi^2 - frac{sqrt{3}}{4}right)$$
$endgroup$
– Zubin Mukerjee
45 mins ago
|
show 2 more comments
$begingroup$
I appreciate this effort but could you make this clearer.
$endgroup$
– Abdulhameed
1 hour ago
$begingroup$
@Abdulhameed Added a picture
$endgroup$
– Bram28
58 mins ago
$begingroup$
@Bram28 really nice picture, +1
$endgroup$
– Zubin Mukerjee
55 mins ago
$begingroup$
This is so beautiful! Its well appreciated
$endgroup$
– Abdulhameed
51 mins ago
$begingroup$
@Abdulhameed Another solution is to take $1/6$ of the area of the area of the circle of radius $1$, subtract the area of an equilateral triangle with side length $1$, then multiply by $2$: $$2left(frac{1}{6}pi^2 - frac{sqrt{3}}{4}right)$$
$endgroup$
– Zubin Mukerjee
45 mins ago
$begingroup$
I appreciate this effort but could you make this clearer.
$endgroup$
– Abdulhameed
1 hour ago
$begingroup$
I appreciate this effort but could you make this clearer.
$endgroup$
– Abdulhameed
1 hour ago
$begingroup$
@Abdulhameed Added a picture
$endgroup$
– Bram28
58 mins ago
$begingroup$
@Abdulhameed Added a picture
$endgroup$
– Bram28
58 mins ago
$begingroup$
@Bram28 really nice picture, +1
$endgroup$
– Zubin Mukerjee
55 mins ago
$begingroup$
@Bram28 really nice picture, +1
$endgroup$
– Zubin Mukerjee
55 mins ago
$begingroup$
This is so beautiful! Its well appreciated
$endgroup$
– Abdulhameed
51 mins ago
$begingroup$
This is so beautiful! Its well appreciated
$endgroup$
– Abdulhameed
51 mins ago
$begingroup$
@Abdulhameed Another solution is to take $1/6$ of the area of the area of the circle of radius $1$, subtract the area of an equilateral triangle with side length $1$, then multiply by $2$: $$2left(frac{1}{6}pi^2 - frac{sqrt{3}}{4}right)$$
$endgroup$
– Zubin Mukerjee
45 mins ago
$begingroup$
@Abdulhameed Another solution is to take $1/6$ of the area of the area of the circle of radius $1$, subtract the area of an equilateral triangle with side length $1$, then multiply by $2$: $$2left(frac{1}{6}pi^2 - frac{sqrt{3}}{4}right)$$
$endgroup$
– Zubin Mukerjee
45 mins ago
|
show 2 more comments
$begingroup$
Notice that the area of the equilateral triangle with edge $R$ plus $frac12$ the area of the shaded region is $frac16$ the area of the circle. The height of the triangle is $frac{Rsqrt3}{2}$ and the area is $frac12cdot Rcdotfrac{Rsqrt3}{2}=frac{R^2sqrt3}{4}$. The area of the shaded region is:
$$2left(frac{pi R^2}{6}-frac{R^2sqrt3}{4}right)=frac{2pi R^2}{6}-frac{3R^2sqrt3}{6}=frac{R^2(2pi-3sqrt3)}{6}$$
Also notice that the height of the triangle plus $frac12$ the height of the shaded area is equal to the radius of the circle. The height of the shaded area is:
$$2left(R-frac{Rsqrt3}{2}right)=2R-Rsqrt3=R(2-sqrt3)$$
$endgroup$
$begingroup$
The area of the triangle should be $$frac{R^2sqrt{3}}{4}$$
$endgroup$
– Zubin Mukerjee
41 mins ago
$begingroup$
@ZubinMukerjee oh, yeah... mental slip
$endgroup$
– Daniel Mathias
39 mins ago
add a comment |
$begingroup$
Notice that the area of the equilateral triangle with edge $R$ plus $frac12$ the area of the shaded region is $frac16$ the area of the circle. The height of the triangle is $frac{Rsqrt3}{2}$ and the area is $frac12cdot Rcdotfrac{Rsqrt3}{2}=frac{R^2sqrt3}{4}$. The area of the shaded region is:
$$2left(frac{pi R^2}{6}-frac{R^2sqrt3}{4}right)=frac{2pi R^2}{6}-frac{3R^2sqrt3}{6}=frac{R^2(2pi-3sqrt3)}{6}$$
Also notice that the height of the triangle plus $frac12$ the height of the shaded area is equal to the radius of the circle. The height of the shaded area is:
$$2left(R-frac{Rsqrt3}{2}right)=2R-Rsqrt3=R(2-sqrt3)$$
$endgroup$
$begingroup$
The area of the triangle should be $$frac{R^2sqrt{3}}{4}$$
$endgroup$
– Zubin Mukerjee
41 mins ago
$begingroup$
@ZubinMukerjee oh, yeah... mental slip
$endgroup$
– Daniel Mathias
39 mins ago
add a comment |
$begingroup$
Notice that the area of the equilateral triangle with edge $R$ plus $frac12$ the area of the shaded region is $frac16$ the area of the circle. The height of the triangle is $frac{Rsqrt3}{2}$ and the area is $frac12cdot Rcdotfrac{Rsqrt3}{2}=frac{R^2sqrt3}{4}$. The area of the shaded region is:
$$2left(frac{pi R^2}{6}-frac{R^2sqrt3}{4}right)=frac{2pi R^2}{6}-frac{3R^2sqrt3}{6}=frac{R^2(2pi-3sqrt3)}{6}$$
Also notice that the height of the triangle plus $frac12$ the height of the shaded area is equal to the radius of the circle. The height of the shaded area is:
$$2left(R-frac{Rsqrt3}{2}right)=2R-Rsqrt3=R(2-sqrt3)$$
$endgroup$
Notice that the area of the equilateral triangle with edge $R$ plus $frac12$ the area of the shaded region is $frac16$ the area of the circle. The height of the triangle is $frac{Rsqrt3}{2}$ and the area is $frac12cdot Rcdotfrac{Rsqrt3}{2}=frac{R^2sqrt3}{4}$. The area of the shaded region is:
$$2left(frac{pi R^2}{6}-frac{R^2sqrt3}{4}right)=frac{2pi R^2}{6}-frac{3R^2sqrt3}{6}=frac{R^2(2pi-3sqrt3)}{6}$$
Also notice that the height of the triangle plus $frac12$ the height of the shaded area is equal to the radius of the circle. The height of the shaded area is:
$$2left(R-frac{Rsqrt3}{2}right)=2R-Rsqrt3=R(2-sqrt3)$$
edited 37 mins ago
answered 44 mins ago
Daniel MathiasDaniel Mathias
81917
81917
$begingroup$
The area of the triangle should be $$frac{R^2sqrt{3}}{4}$$
$endgroup$
– Zubin Mukerjee
41 mins ago
$begingroup$
@ZubinMukerjee oh, yeah... mental slip
$endgroup$
– Daniel Mathias
39 mins ago
add a comment |
$begingroup$
The area of the triangle should be $$frac{R^2sqrt{3}}{4}$$
$endgroup$
– Zubin Mukerjee
41 mins ago
$begingroup$
@ZubinMukerjee oh, yeah... mental slip
$endgroup$
– Daniel Mathias
39 mins ago
$begingroup$
The area of the triangle should be $$frac{R^2sqrt{3}}{4}$$
$endgroup$
– Zubin Mukerjee
41 mins ago
$begingroup$
The area of the triangle should be $$frac{R^2sqrt{3}}{4}$$
$endgroup$
– Zubin Mukerjee
41 mins ago
$begingroup$
@ZubinMukerjee oh, yeah... mental slip
$endgroup$
– Daniel Mathias
39 mins ago
$begingroup$
@ZubinMukerjee oh, yeah... mental slip
$endgroup$
– Daniel Mathias
39 mins ago
add a comment |
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1
$begingroup$
Notice anything special about $triangle ABC$?
$endgroup$
– Blue
2 hours ago
$begingroup$
Hmmn! It's equilateral!
$endgroup$
– Abdulhameed
1 hour ago
1
$begingroup$
Then just use one of the previous solutions, since you know the angle.
$endgroup$
– Andrei
1 hour ago
$begingroup$
Thanks! I think its clear now! I need to subtract the area of the sector from the triangle to get the small area and then multiply by two. What about the height any clues?
$endgroup$
– Abdulhameed
1 hour ago
$begingroup$
@Abdulhameed: For height, use the equilateral triangles again. If you know the height of a triangle, and the radius of the circle, then ...
$endgroup$
– Blue
1 hour ago