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
asked Jan 27 at 1:50
AbdulhameedAbdulhameed
153115
$endgroup$
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
asked Jan 27 at 1:50
AbdulhameedAbdulhameed
153115
$endgroup$
1
$begingroup$
Notice anything special about $triangle ABC$?
$endgroup$
– Blue
Jan 27 at 1:58
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Hmmn! It's equilateral!
$endgroup$
– Abdulhameed
Jan 27 at 2:02
1
$begingroup$
Then just use one of the previous solutions, since you know the angle.
$endgroup$
– Andrei
Jan 27 at 2:03
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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
Jan 27 at 2:06
$begingroup$
@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
Jan 27 at 2:08
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
asked Jan 27 at 1:50
AbdulhameedAbdulhameed
153115
$endgroup$
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
asked Jan 27 at 1:50
AbdulhameedAbdulhameed
153115
asked Jan 27 at 1:50
AbdulhameedAbdulhameed
153115
edited Jan 27 at 1:56
Abdulhameed
asked Jan 27 at 1:50
AbdulhameedAbdulhameed
153115
asked Jan 27 at 1:50
AbdulhameedAbdulhameed
153115
asked Jan 27 at 1:50
AbdulhameedAbdulhameed
153115
153115
1
$begingroup$
Notice anything special about $triangle ABC$?
$endgroup$
– Blue
Jan 27 at 1:58
$begingroup$
Hmmn! It's equilateral!
$endgroup$
– Abdulhameed
Jan 27 at 2:02
1
$begingroup$
Then just use one of the previous solutions, since you know the angle.
$endgroup$
– Andrei
Jan 27 at 2:03
$begingroup$
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
Jan 27 at 2:06
$begingroup$
@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
Jan 27 at 2:08
add a comment |
1
$begingroup$
Notice anything special about $triangle ABC$?
$endgroup$
– Blue
Jan 27 at 1:58
$begingroup$
Hmmn! It's equilateral!
$endgroup$
– Abdulhameed
Jan 27 at 2:02
1
$begingroup$
Then just use one of the previous solutions, since you know the angle.
$endgroup$
– Andrei
Jan 27 at 2:03
$begingroup$
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
Jan 27 at 2:06
$begingroup$
@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
Jan 27 at 2:08
1
1
$begingroup$
Notice anything special about $triangle ABC$?
$endgroup$
– Blue
Jan 27 at 1:58
$begingroup$
Notice anything special about $triangle ABC$?
$endgroup$
– Blue
Jan 27 at 1:58
$begingroup$
Hmmn! It's equilateral!
$endgroup$
– Abdulhameed
Jan 27 at 2:02
$begingroup$
Hmmn! It's equilateral!
$endgroup$
– Abdulhameed
Jan 27 at 2:02
1
1
$begingroup$
Then just use one of the previous solutions, since you know the angle.
$endgroup$
– Andrei
Jan 27 at 2:03
$begingroup$
Then just use one of the previous solutions, since you know the angle.
$endgroup$
– Andrei
Jan 27 at 2:03
$begingroup$
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
Jan 27 at 2:06
$begingroup$
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
Jan 27 at 2:06
$begingroup$
@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
Jan 27 at 2:08
$begingroup$
@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
Jan 27 at 2:08
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If you repeat this pattern infinitely, you'll find that each circle consists of $12$ of these (American!) 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:
$pi=12A +6B$ (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, $4$ half footballs, $2$ whole 'triangles', and $4$ half triangles. So:
$sqrt{3}=6A+4B$
Now you can easily solve for $A$
answered Jan 27 at 2:18
Bram28Bram28
63.2k44793
$endgroup$
$begingroup$
@Bram28 really nice picture, +1
$endgroup$
– Zubin Mukerjee
Jan 27 at 3:06
$begingroup$
This is so beautiful! Its well appreciated
$endgroup$
– Abdulhameed
Jan 27 at 3:10
$begingroup$
@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
Jan 27 at 3:16
$begingroup$
@Abdulhameed You're welcome! :)
$endgroup$
– Bram28
Jan 27 at 3:33
2
$begingroup$
"football shaped"?? You mean the lens-shaped part highlighted in the question? Note that footballs actually are spherical (or nearly so) in many if not most parts of the world.
$endgroup$
– ilkkachu
Jan 27 at 16:03
|
show 3 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:
$$2left(R-frac{Rsqrt3}{2}right)=2R-Rsqrt3=R(2-sqrt3)$$
answered Jan 27 at 3:17
Daniel MathiasDaniel Mathias
1,33018
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add a comment |
$begingroup$
$AC=BC=AB=AD=BD=R$.Now $AB$ and $CD$ are perpendicular bisectors of each other, say at $O$.Then $DO^2=AD^2- AO^2= R^2- ({R over 2})^2$ i.e. $DO={ sqrt{3}R over 2}$.Then $height=2(R- { sqrt{3}R over 2})=(2- sqrt{3})R$.
Now if the shaded area be $∆$, as $angle ADB=60°$ , ${∆ over 2}={πR^2 over 6}- { sqrt{3}R^2 over 4}$(it's area of $∆ABD$)so, required shaded region area $∆={2π-3√3 over 6}R$.
answered Jan 27 at 16:38
Supriyo HalderSupriyo Halder
663113
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
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 (American!) 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:
$pi=12A +6B$ (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, $4$ half footballs, $2$ whole 'triangles', and $4$ half triangles. So:
$sqrt{3}=6A+4B$
Now you can easily solve for $A$
answered Jan 27 at 2:18
Bram28Bram28
63.2k44793
$endgroup$
$begingroup$
@Bram28 really nice picture, +1
$endgroup$
– Zubin Mukerjee
Jan 27 at 3:06
$begingroup$
This is so beautiful! Its well appreciated
$endgroup$
– Abdulhameed
Jan 27 at 3:10
$begingroup$
@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
Jan 27 at 3:16
$begingroup$
@Abdulhameed You're welcome! :)
$endgroup$
– Bram28
Jan 27 at 3:33
2
$begingroup$
"football shaped"?? You mean the lens-shaped part highlighted in the question? Note that footballs actually are spherical (or nearly so) in many if not most parts of the world.
$endgroup$
– ilkkachu
Jan 27 at 16:03
|
show 3 more comments
$begingroup$
If you repeat this pattern infinitely, you'll find that each circle consists of $12$ of these (American!) 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:
$pi=12A +6B$ (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, $4$ half footballs, $2$ whole 'triangles', and $4$ half triangles. So:
$sqrt{3}=6A+4B$
Now you can easily solve for $A$
answered Jan 27 at 2:18
Bram28Bram28
63.2k44793
$endgroup$
$begingroup$
@Bram28 really nice picture, +1
$endgroup$
– Zubin Mukerjee
Jan 27 at 3:06
$begingroup$
This is so beautiful! Its well appreciated
$endgroup$
– Abdulhameed
Jan 27 at 3:10
$begingroup$
@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
Jan 27 at 3:16
$begingroup$
@Abdulhameed You're welcome! :)
$endgroup$
– Bram28
Jan 27 at 3:33
2
$begingroup$
"football shaped"?? You mean the lens-shaped part highlighted in the question? Note that footballs actually are spherical (or nearly so) in many if not most parts of the world.
$endgroup$
– ilkkachu
Jan 27 at 16:03
|
show 3 more comments
$begingroup$
If you repeat this pattern infinitely, you'll find that each circle consists of $12$ of these (American!) 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:
$pi=12A +6B$ (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, $4$ half footballs, $2$ whole 'triangles', and $4$ half triangles. So:
$sqrt{3}=6A+4B$
Now you can easily solve for $A$
answered Jan 27 at 2:18
Bram28Bram28
63.2k44793
$endgroup$
If you repeat this pattern infinitely, you'll find that each circle consists of $12$ of these (American!) 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:
$pi=12A +6B$ (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, $4$ half footballs, $2$ whole 'triangles', and $4$ half triangles. So:
$sqrt{3}=6A+4B$
Now you can easily solve for $A$
answered Jan 27 at 2:18
Bram28Bram28
63.2k44793
edited Jan 27 at 18:58
answered Jan 27 at 2:18
Bram28Bram28
63.2k44793
answered Jan 27 at 2:18
Bram28Bram28
63.2k44793
answered Jan 27 at 2:18
Bram28Bram28
63.2k44793
63.2k44793
$begingroup$
@Bram28 really nice picture, +1
$endgroup$
– Zubin Mukerjee
Jan 27 at 3:06
$begingroup$
This is so beautiful! Its well appreciated
$endgroup$
– Abdulhameed
Jan 27 at 3:10
$begingroup$
@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
Jan 27 at 3:16
$begingroup$
@Abdulhameed You're welcome! :)
$endgroup$
– Bram28
Jan 27 at 3:33
2
$begingroup$
"football shaped"?? You mean the lens-shaped part highlighted in the question? Note that footballs actually are spherical (or nearly so) in many if not most parts of the world.
$endgroup$
– ilkkachu
Jan 27 at 16:03
|
show 3 more comments
$begingroup$
@Bram28 really nice picture, +1
$endgroup$
– Zubin Mukerjee
Jan 27 at 3:06
$begingroup$
This is so beautiful! Its well appreciated
$endgroup$
– Abdulhameed
Jan 27 at 3:10
$begingroup$
@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
Jan 27 at 3:16
$begingroup$
@Abdulhameed You're welcome! :)
$endgroup$
– Bram28
Jan 27 at 3:33
2
$begingroup$
"football shaped"?? You mean the lens-shaped part highlighted in the question? Note that footballs actually are spherical (or nearly so) in many if not most parts of the world.
$endgroup$
– ilkkachu
Jan 27 at 16:03
$begingroup$
@Bram28 really nice picture, +1
$endgroup$
– Zubin Mukerjee
Jan 27 at 3:06
$begingroup$
@Bram28 really nice picture, +1
$endgroup$
– Zubin Mukerjee
Jan 27 at 3:06
$begingroup$
This is so beautiful! Its well appreciated
$endgroup$
– Abdulhameed
Jan 27 at 3:10
$begingroup$
This is so beautiful! Its well appreciated
$endgroup$
– Abdulhameed
Jan 27 at 3:10
$begingroup$
@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
Jan 27 at 3:16
$begingroup$
@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
Jan 27 at 3:16
$begingroup$
@Abdulhameed You're welcome! :)
$endgroup$
– Bram28
Jan 27 at 3:33
$begingroup$
@Abdulhameed You're welcome! :)
$endgroup$
– Bram28
Jan 27 at 3:33
2
2
$begingroup$
"football shaped"?? You mean the lens-shaped part highlighted in the question? Note that footballs actually are spherical (or nearly so) in many if not most parts of the world.
$endgroup$
– ilkkachu
Jan 27 at 16:03
$begingroup$
"football shaped"?? You mean the lens-shaped part highlighted in the question? Note that footballs actually are spherical (or nearly so) in many if not most parts of the world.
$endgroup$
– ilkkachu
Jan 27 at 16:03
|
show 3 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:
$$2left(R-frac{Rsqrt3}{2}right)=2R-Rsqrt3=R(2-sqrt3)$$
answered Jan 27 at 3:17
Daniel MathiasDaniel Mathias
1,33018
$endgroup$
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:
$$2left(R-frac{Rsqrt3}{2}right)=2R-Rsqrt3=R(2-sqrt3)$$
answered Jan 27 at 3:17
Daniel MathiasDaniel Mathias
1,33018
$endgroup$
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:
$$2left(R-frac{Rsqrt3}{2}right)=2R-Rsqrt3=R(2-sqrt3)$$
answered Jan 27 at 3:17
Daniel MathiasDaniel Mathias
1,33018
$endgroup$
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:
$$2left(R-frac{Rsqrt3}{2}right)=2R-Rsqrt3=R(2-sqrt3)$$
answered Jan 27 at 3:17
Daniel MathiasDaniel Mathias
1,33018
edited Jan 27 at 3:25
answered Jan 27 at 3:17
Daniel MathiasDaniel Mathias
1,33018
answered Jan 27 at 3:17
Daniel MathiasDaniel Mathias
1,33018
answered Jan 27 at 3:17
Daniel MathiasDaniel Mathias
1,33018
1,33018
add a comment |
add a comment |
$begingroup$
$AC=BC=AB=AD=BD=R$.Now $AB$ and $CD$ are perpendicular bisectors of each other, say at $O$.Then $DO^2=AD^2- AO^2= R^2- ({R over 2})^2$ i.e. $DO={ sqrt{3}R over 2}$.Then $height=2(R- { sqrt{3}R over 2})=(2- sqrt{3})R$.
Now if the shaded area be $∆$, as $angle ADB=60°$ , ${∆ over 2}={πR^2 over 6}- { sqrt{3}R^2 over 4}$(it's area of $∆ABD$)so, required shaded region area $∆={2π-3√3 over 6}R$.
answered Jan 27 at 16:38
Supriyo HalderSupriyo Halder
663113
$endgroup$
add a comment |
$begingroup$
$AC=BC=AB=AD=BD=R$.Now $AB$ and $CD$ are perpendicular bisectors of each other, say at $O$.Then $DO^2=AD^2- AO^2= R^2- ({R over 2})^2$ i.e. $DO={ sqrt{3}R over 2}$.Then $height=2(R- { sqrt{3}R over 2})=(2- sqrt{3})R$.
Now if the shaded area be $∆$, as $angle ADB=60°$ , ${∆ over 2}={πR^2 over 6}- { sqrt{3}R^2 over 4}$(it's area of $∆ABD$)so, required shaded region area $∆={2π-3√3 over 6}R$.
answered Jan 27 at 16:38
Supriyo HalderSupriyo Halder
663113
$endgroup$
add a comment |
$begingroup$
$AC=BC=AB=AD=BD=R$.Now $AB$ and $CD$ are perpendicular bisectors of each other, say at $O$.Then $DO^2=AD^2- AO^2= R^2- ({R over 2})^2$ i.e. $DO={ sqrt{3}R over 2}$.Then $height=2(R- { sqrt{3}R over 2})=(2- sqrt{3})R$.
Now if the shaded area be $∆$, as $angle ADB=60°$ , ${∆ over 2}={πR^2 over 6}- { sqrt{3}R^2 over 4}$(it's area of $∆ABD$)so, required shaded region area $∆={2π-3√3 over 6}R$.
answered Jan 27 at 16:38
Supriyo HalderSupriyo Halder
663113
$endgroup$
$AC=BC=AB=AD=BD=R$.Now $AB$ and $CD$ are perpendicular bisectors of each other, say at $O$.Then $DO^2=AD^2- AO^2= R^2- ({R over 2})^2$ i.e. $DO={ sqrt{3}R over 2}$.Then $height=2(R- { sqrt{3}R over 2})=(2- sqrt{3})R$.
Now if the shaded area be $∆$, as $angle ADB=60°$ , ${∆ over 2}={πR^2 over 6}- { sqrt{3}R^2 over 4}$(it's area of $∆ABD$)so, required shaded region area $∆={2π-3√3 over 6}R$.
answered Jan 27 at 16:38
Supriyo HalderSupriyo Halder
663113
edited Jan 27 at 16:46
answered Jan 27 at 16:38
Supriyo HalderSupriyo Halder
663113
answered Jan 27 at 16:38
Supriyo HalderSupriyo Halder
663113
answered Jan 27 at 16:38
Supriyo HalderSupriyo Halder
663113
663113
add a comment |
add a comment |
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1
$begingroup$
Notice anything special about $triangle ABC$?
$endgroup$
– Blue
Jan 27 at 1:58
$begingroup$
Hmmn! It's equilateral!
$endgroup$
– Abdulhameed
Jan 27 at 2:02
1
$begingroup$
Then just use one of the previous solutions, since you know the angle.
$endgroup$
– Andrei
Jan 27 at 2:03
$begingroup$
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
Jan 27 at 2:06
$begingroup$
@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
Jan 27 at 2:08