Four equal circles intersect: What is the area of the small shaded portion and its height












3












$begingroup$


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.



enter image description here






geometry circle







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$endgroup$








  • 1




    $begingroup$
    Notice anything special about $triangle ABC$?
    $endgroup$
    – Blue
    2 hours ago










  • $begingroup$
    Hmmn! It's equilateral!
    $endgroup$
    – Abdulhameed
    2 hours ago






  • 1




    $begingroup$
    Then just use one of the previous solutions, since you know the angle.
    $endgroup$
    – Andrei
    2 hours 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
    2 hours 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
    2 hours ago
















3












$begingroup$


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.



enter image description here






geometry circle







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Notice anything special about $triangle ABC$?
    $endgroup$
    – Blue
    2 hours ago










  • $begingroup$
    Hmmn! It's equilateral!
    $endgroup$
    – Abdulhameed
    2 hours ago






  • 1




    $begingroup$
    Then just use one of the previous solutions, since you know the angle.
    $endgroup$
    – Andrei
    2 hours 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
    2 hours 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
    2 hours ago














3












3








3


1



$begingroup$


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.



enter image description here






geometry circle







share|cite|improve this question











$endgroup$




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.



enter image description here






geometry circle




geometry circle






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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
    2 hours ago






  • 1




    $begingroup$
    Then just use one of the previous solutions, since you know the angle.
    $endgroup$
    – Andrei
    2 hours 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
    2 hours 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
    2 hours ago














  • 1




    $begingroup$
    Notice anything special about $triangle ABC$?
    $endgroup$
    – Blue
    2 hours ago










  • $begingroup$
    Hmmn! It's equilateral!
    $endgroup$
    – Abdulhameed
    2 hours ago






  • 1




    $begingroup$
    Then just use one of the previous solutions, since you know the angle.
    $endgroup$
    – Andrei
    2 hours 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
    2 hours 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
    2 hours 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
2 hours ago




$begingroup$
Hmmn! It's equilateral!
$endgroup$
– Abdulhameed
2 hours ago




1




1




$begingroup$
Then just use one of the previous solutions, since you know the angle.
$endgroup$
– Andrei
2 hours ago




$begingroup$
Then just use one of the previous solutions, since you know the angle.
$endgroup$
– Andrei
2 hours 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
2 hours 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
2 hours 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
2 hours 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
2 hours ago










2 Answers
2






active

oldest

votes


















4












$begingroup$

enter image description hereIf 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 $A$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I appreciate this effort but could you make this clearer.
    $endgroup$
    – Abdulhameed
    2 hours ago










  • $begingroup$
    @Abdulhameed Added a picture
    $endgroup$
    – Bram28
    1 hour ago










  • $begingroup$
    @Bram28 really nice picture, +1
    $endgroup$
    – Zubin Mukerjee
    1 hour ago










  • $begingroup$
    This is so beautiful! Its well appreciated
    $endgroup$
    – Abdulhameed
    1 hour 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
    1 hour ago



















1












$begingroup$

enter image description here



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:
$$2left(R-frac{Rsqrt3}{2}right)=2R-Rsqrt3=R(2-sqrt3)$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The area of the triangle should be $$frac{R^2sqrt{3}}{4}$$
    $endgroup$
    – Zubin Mukerjee
    1 hour ago










  • $begingroup$
    @ZubinMukerjee oh, yeah... mental slip
    $endgroup$
    – Daniel Mathias
    1 hour ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

enter image description hereIf 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 $A$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I appreciate this effort but could you make this clearer.
    $endgroup$
    – Abdulhameed
    2 hours ago










  • $begingroup$
    @Abdulhameed Added a picture
    $endgroup$
    – Bram28
    1 hour ago










  • $begingroup$
    @Bram28 really nice picture, +1
    $endgroup$
    – Zubin Mukerjee
    1 hour ago










  • $begingroup$
    This is so beautiful! Its well appreciated
    $endgroup$
    – Abdulhameed
    1 hour 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
    1 hour ago
















4












$begingroup$

enter image description hereIf 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 $A$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I appreciate this effort but could you make this clearer.
    $endgroup$
    – Abdulhameed
    2 hours ago










  • $begingroup$
    @Abdulhameed Added a picture
    $endgroup$
    – Bram28
    1 hour ago










  • $begingroup$
    @Bram28 really nice picture, +1
    $endgroup$
    – Zubin Mukerjee
    1 hour ago










  • $begingroup$
    This is so beautiful! Its well appreciated
    $endgroup$
    – Abdulhameed
    1 hour 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
    1 hour ago














4












4








4





$begingroup$

enter image description hereIf 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 $A$






share|cite|improve this answer











$endgroup$



enter image description hereIf 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 $A$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 2 hours ago









Bram28Bram28

61.2k44591




61.2k44591












  • $begingroup$
    I appreciate this effort but could you make this clearer.
    $endgroup$
    – Abdulhameed
    2 hours ago










  • $begingroup$
    @Abdulhameed Added a picture
    $endgroup$
    – Bram28
    1 hour ago










  • $begingroup$
    @Bram28 really nice picture, +1
    $endgroup$
    – Zubin Mukerjee
    1 hour ago










  • $begingroup$
    This is so beautiful! Its well appreciated
    $endgroup$
    – Abdulhameed
    1 hour 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
    1 hour ago


















  • $begingroup$
    I appreciate this effort but could you make this clearer.
    $endgroup$
    – Abdulhameed
    2 hours ago










  • $begingroup$
    @Abdulhameed Added a picture
    $endgroup$
    – Bram28
    1 hour ago










  • $begingroup$
    @Bram28 really nice picture, +1
    $endgroup$
    – Zubin Mukerjee
    1 hour ago










  • $begingroup$
    This is so beautiful! Its well appreciated
    $endgroup$
    – Abdulhameed
    1 hour 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
    1 hour ago
















$begingroup$
I appreciate this effort but could you make this clearer.
$endgroup$
– Abdulhameed
2 hours ago




$begingroup$
I appreciate this effort but could you make this clearer.
$endgroup$
– Abdulhameed
2 hours ago












$begingroup$
@Abdulhameed Added a picture
$endgroup$
– Bram28
1 hour ago




$begingroup$
@Abdulhameed Added a picture
$endgroup$
– Bram28
1 hour ago












$begingroup$
@Bram28 really nice picture, +1
$endgroup$
– Zubin Mukerjee
1 hour ago




$begingroup$
@Bram28 really nice picture, +1
$endgroup$
– Zubin Mukerjee
1 hour ago












$begingroup$
This is so beautiful! Its well appreciated
$endgroup$
– Abdulhameed
1 hour ago




$begingroup$
This is so beautiful! Its well appreciated
$endgroup$
– Abdulhameed
1 hour 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
1 hour 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
1 hour ago











1












$begingroup$

enter image description here



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:
$$2left(R-frac{Rsqrt3}{2}right)=2R-Rsqrt3=R(2-sqrt3)$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The area of the triangle should be $$frac{R^2sqrt{3}}{4}$$
    $endgroup$
    – Zubin Mukerjee
    1 hour ago










  • $begingroup$
    @ZubinMukerjee oh, yeah... mental slip
    $endgroup$
    – Daniel Mathias
    1 hour ago
















1












$begingroup$

enter image description here



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:
$$2left(R-frac{Rsqrt3}{2}right)=2R-Rsqrt3=R(2-sqrt3)$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The area of the triangle should be $$frac{R^2sqrt{3}}{4}$$
    $endgroup$
    – Zubin Mukerjee
    1 hour ago










  • $begingroup$
    @ZubinMukerjee oh, yeah... mental slip
    $endgroup$
    – Daniel Mathias
    1 hour ago














1












1








1





$begingroup$

enter image description here



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:
$$2left(R-frac{Rsqrt3}{2}right)=2R-Rsqrt3=R(2-sqrt3)$$






share|cite|improve this answer











$endgroup$



enter image description here



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:
$$2left(R-frac{Rsqrt3}{2}right)=2R-Rsqrt3=R(2-sqrt3)$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 1 hour ago









Daniel MathiasDaniel Mathias

81917




81917












  • $begingroup$
    The area of the triangle should be $$frac{R^2sqrt{3}}{4}$$
    $endgroup$
    – Zubin Mukerjee
    1 hour ago










  • $begingroup$
    @ZubinMukerjee oh, yeah... mental slip
    $endgroup$
    – Daniel Mathias
    1 hour ago


















  • $begingroup$
    The area of the triangle should be $$frac{R^2sqrt{3}}{4}$$
    $endgroup$
    – Zubin Mukerjee
    1 hour ago










  • $begingroup$
    @ZubinMukerjee oh, yeah... mental slip
    $endgroup$
    – Daniel Mathias
    1 hour ago
















$begingroup$
The area of the triangle should be $$frac{R^2sqrt{3}}{4}$$
$endgroup$
– Zubin Mukerjee
1 hour ago




$begingroup$
The area of the triangle should be $$frac{R^2sqrt{3}}{4}$$
$endgroup$
– Zubin Mukerjee
1 hour ago












$begingroup$
@ZubinMukerjee oh, yeah... mental slip
$endgroup$
– Daniel Mathias
1 hour ago




$begingroup$
@ZubinMukerjee oh, yeah... mental slip
$endgroup$
– Daniel Mathias
1 hour ago


















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