A flower in a hexagon
Clash Royale CLAN TAG#URR8PPP
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The area of the ✽ in a ⬡
So I had a geometry problem in my maths exam test and I couldn't solve it, though it didn't seem that hard. I tried to connect the points from center to end to create an equilateral triangle but was still unable to find the solution.
Would be thankful if you could help me out.
The question is the following:
We have an equal hexagon with sides equal to $1$ and six circular arcs with radius equal to $1$ from each vertices of the regular hexagon in it creating a flower-shaped-like object. Find the area of the flower.
Here were the possible answers ↓
1. $pi$
3. $3pi/2$
4. $4sqrt2-pi$
5. $pi/2+sqrt3$
6. $2pi-3sqrt3$
P.S. for better picture check this one by @AnirbanNiloy.
geometry triangles area
$endgroup$
add a comment |
$begingroup$
The area of the ✽ in a ⬡
So I had a geometry problem in my maths exam test and I couldn't solve it, though it didn't seem that hard. I tried to connect the points from center to end to create an equilateral triangle but was still unable to find the solution.
Would be thankful if you could help me out.
The question is the following:
We have an equal hexagon with sides equal to $1$ and six circular arcs with radius equal to $1$ from each vertices of the regular hexagon in it creating a flower-shaped-like object. Find the area of the flower.
Here were the possible answers ↓
1. $pi$
3. $3pi/2$
4. $4sqrt2-pi$
5. $pi/2+sqrt3$
6. $2pi-3sqrt3$
P.S. for better picture check this one by @AnirbanNiloy.
geometry triangles area
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3
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I count six ellipses
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– Ross Millikan
Mar 3 at 2:04
3
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Are the ellipses tangent to the sides at the vertices and meeting without overlap at the center? If so, they are not ellipses. Do you mean pairs of circular arcs?
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– Oscar Lanzi
Mar 3 at 3:01
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This question has been altered to something quite different than it originally was, some hours after it was posted, and after the older answers were posted. Although the change was probably for the better as the question is concerned, this leaves a confusing state of affairs.
$endgroup$
– Marc van Leeuwen
Mar 3 at 11:01
1
$begingroup$
Actually the picture was by user @AnirbanNiloy
$endgroup$
– Deepak
Mar 3 at 13:25
add a comment |
$begingroup$
The area of the ✽ in a ⬡
So I had a geometry problem in my maths exam test and I couldn't solve it, though it didn't seem that hard. I tried to connect the points from center to end to create an equilateral triangle but was still unable to find the solution.
Would be thankful if you could help me out.
The question is the following:
We have an equal hexagon with sides equal to $1$ and six circular arcs with radius equal to $1$ from each vertices of the regular hexagon in it creating a flower-shaped-like object. Find the area of the flower.
Here were the possible answers ↓
1. $pi$
3. $3pi/2$
4. $4sqrt2-pi$
5. $pi/2+sqrt3$
6. $2pi-3sqrt3$
P.S. for better picture check this one by @AnirbanNiloy.
geometry triangles area
$endgroup$
The area of the ✽ in a ⬡
So I had a geometry problem in my maths exam test and I couldn't solve it, though it didn't seem that hard. I tried to connect the points from center to end to create an equilateral triangle but was still unable to find the solution.
Would be thankful if you could help me out.
The question is the following:
We have an equal hexagon with sides equal to $1$ and six circular arcs with radius equal to $1$ from each vertices of the regular hexagon in it creating a flower-shaped-like object. Find the area of the flower.
Here were the possible answers ↓
1. $pi$
3. $3pi/2$
4. $4sqrt2-pi$
5. $pi/2+sqrt3$
6. $2pi-3sqrt3$
P.S. for better picture check this one by @AnirbanNiloy.
geometry triangles area
geometry triangles area
edited Mar 3 at 14:45
Andrés E. Caicedo
65.8k8160251
65.8k8160251
asked Mar 3 at 1:09
DAVODAVO
567
567
3
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I count six ellipses
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– Ross Millikan
Mar 3 at 2:04
3
$begingroup$
Are the ellipses tangent to the sides at the vertices and meeting without overlap at the center? If so, they are not ellipses. Do you mean pairs of circular arcs?
$endgroup$
– Oscar Lanzi
Mar 3 at 3:01
$begingroup$
This question has been altered to something quite different than it originally was, some hours after it was posted, and after the older answers were posted. Although the change was probably for the better as the question is concerned, this leaves a confusing state of affairs.
$endgroup$
– Marc van Leeuwen
Mar 3 at 11:01
1
$begingroup$
Actually the picture was by user @AnirbanNiloy
$endgroup$
– Deepak
Mar 3 at 13:25
add a comment |
3
$begingroup$
I count six ellipses
$endgroup$
– Ross Millikan
Mar 3 at 2:04
3
$begingroup$
Are the ellipses tangent to the sides at the vertices and meeting without overlap at the center? If so, they are not ellipses. Do you mean pairs of circular arcs?
$endgroup$
– Oscar Lanzi
Mar 3 at 3:01
$begingroup$
This question has been altered to something quite different than it originally was, some hours after it was posted, and after the older answers were posted. Although the change was probably for the better as the question is concerned, this leaves a confusing state of affairs.
$endgroup$
– Marc van Leeuwen
Mar 3 at 11:01
1
$begingroup$
Actually the picture was by user @AnirbanNiloy
$endgroup$
– Deepak
Mar 3 at 13:25
3
3
$begingroup$
I count six ellipses
$endgroup$
– Ross Millikan
Mar 3 at 2:04
$begingroup$
I count six ellipses
$endgroup$
– Ross Millikan
Mar 3 at 2:04
3
3
$begingroup$
Are the ellipses tangent to the sides at the vertices and meeting without overlap at the center? If so, they are not ellipses. Do you mean pairs of circular arcs?
$endgroup$
– Oscar Lanzi
Mar 3 at 3:01
$begingroup$
Are the ellipses tangent to the sides at the vertices and meeting without overlap at the center? If so, they are not ellipses. Do you mean pairs of circular arcs?
$endgroup$
– Oscar Lanzi
Mar 3 at 3:01
$begingroup$
This question has been altered to something quite different than it originally was, some hours after it was posted, and after the older answers were posted. Although the change was probably for the better as the question is concerned, this leaves a confusing state of affairs.
$endgroup$
– Marc van Leeuwen
Mar 3 at 11:01
$begingroup$
This question has been altered to something quite different than it originally was, some hours after it was posted, and after the older answers were posted. Although the change was probably for the better as the question is concerned, this leaves a confusing state of affairs.
$endgroup$
– Marc van Leeuwen
Mar 3 at 11:01
1
1
$begingroup$
Actually the picture was by user @AnirbanNiloy
$endgroup$
– Deepak
Mar 3 at 13:25
$begingroup$
Actually the picture was by user @AnirbanNiloy
$endgroup$
– Deepak
Mar 3 at 13:25
add a comment |
5 Answers
5
active
oldest
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Assuming those are circular arcs and not "ellipses", you can indeed find the area exactly.
The "flower" has six "petals" Each of those petals has axial symmetry and can be divided into two halves. Each of those halves is the segment subtending a central angle of $fracpi3$ (radian measure) of a circle of radius $1$.
The area of one such segment is $frac 12 r^2(theta - sintheta) = frac 12(fracpi3 - fracsqrt32)$.
There are $12$ such segments, yielding the total area of the "flower" as $2pi - 3sqrt 3$.
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Since the shape of each petal is claimed to be ellipses, your answer is not valid in assuming that they consist of circular arcs. However, I do reckon that they are intented to be arcs.
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– Trebor
Mar 3 at 3:08
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@Trebor I had edited my answer (I think before your comment) to clarify that assumption. However, I think it's a valid assumption, and just a very poor question.
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– Deepak
Mar 3 at 3:18
1
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@Deepak Welcome! Anytime!!!! You deserve more than that.
$endgroup$
– Anirban Niloy
Mar 3 at 7:20
1
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@Davo I am also glad that you are satisfied with the diagram. You can install the app GeoGebra Classic from Play-Store for illustrating any graph or diagram. I can suggest only that to you.
$endgroup$
– Anirban Niloy
Mar 3 at 14:52
1
$begingroup$
@DAVO Moreover, you can check from the following link whatever you prefer and which software you want to use math.stackexchange.com/q/1985/63393]
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– Anirban Niloy
Mar 3 at 14:59
|
show 3 more comments
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Here's a solution that doesn't require trigonometry. The hexagon is made up of 6 "petals" and 6 "wedges" (the scooped out triangles between the petals). Call the area of a petal $p$, and the area of a wedge $w$.
What is the area enclosed by two adjacent sides of the hexagon and the arc drawn from the vertex between them. (For example, the region EFAGE). It's one third of a unit circle, and contains three petals and two wedges:
$3p + 2w = frac 1 3 pi r^2 = frac pi 3$
What is the area of one triangle (one sixth of the hexagon). The altitude of the triangle is $sqrt1 - frac 1 4 = fracsqrt 32$, so the area is $fracsqrt 34$. The triangle is made up of two half-petals and one wedge:
$p + w = fracsqrt 34$
Solving the two simultaneous equations:
$2p + 2w = fracsqrt 32$
$p = fracpi3 - fracsqrt 32$
And the area of the flower is $6p = 2pi - 3sqrt 3$.
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add a comment |
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[This answer was posted for the original version of the problem, which had essentially no information about the figure other than the picture itself]
We can't easily find the area exactly - we just don't have enough information for that. But then, what are the estimates we're looking at? (1) $approx 3.1$. (3) $approx 4.7$. (4) $approx 2.5$. (5) $approx 3.3$. (6) $approx 1.1$.
If the area's close to $3$, we might have difficulty deciding. If it's farther away, it'll be an easy choice.
Now, what's the area of the hexagon? That's something we can calculate exactly; an equilateral triangle of side $1$ has area $fracsqrt34$, and the hexagon is six of those stuck together for a total area of $frac32sqrt3approx 2.6$. The flower is clearly much less than that. Only one of the answers is plausible. It has to be (6), $2pi-3sqrt3$.
And, reverse-engineering from the answer... the curves that define the "petals" are the six circles centered at the vertices of the hexagon with radius $1$. That bit about "five ellipses" is just wrong.
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add a comment |
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Shaded blue area is
(Area of pie)-(Area of rhombus)=$picdot 1^2cdotfrac120360-1^2cdotsin(120)$=$ fracpi3-fracsqrt32$
And we have 6 of them, and so total area of flower is
$ 6(fracpi3-fracsqrt32)=2pi-3sqrt3$
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add a comment |
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Here's another way of looking at it. Imagine taking two circles of radius 1 and cutting them into thirds. Arrange them so that the six 120° angles you have (one from each wedge) form the corners of the hexagon.
Every point in one of the "petals" is covered by three of these wedges. Every point in the hexagon that's not in one of the petals is covered by two of these wedges. The total area covered by the six wedges is therefore twice the area of the total hexagon plus the area of the petals; or, in other words, the area of the petals is the area of two unit circles minus twice the area of the hexagon.
The area of the two circles is $2pi$; the area of the hexagon is $3sqrt3/2$. The area of the petals is therefore $2 pi - 3 sqrt3 approx 1.087...$
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add a comment |
Your Answer
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assuming those are circular arcs and not "ellipses", you can indeed find the area exactly.
The "flower" has six "petals" Each of those petals has axial symmetry and can be divided into two halves. Each of those halves is the segment subtending a central angle of $fracpi3$ (radian measure) of a circle of radius $1$.
The area of one such segment is $frac 12 r^2(theta - sintheta) = frac 12(fracpi3 - fracsqrt32)$.
There are $12$ such segments, yielding the total area of the "flower" as $2pi - 3sqrt 3$.
$endgroup$
$begingroup$
Since the shape of each petal is claimed to be ellipses, your answer is not valid in assuming that they consist of circular arcs. However, I do reckon that they are intented to be arcs.
$endgroup$
– Trebor
Mar 3 at 3:08
$begingroup$
@Trebor I had edited my answer (I think before your comment) to clarify that assumption. However, I think it's a valid assumption, and just a very poor question.
$endgroup$
– Deepak
Mar 3 at 3:18
1
$begingroup$
@Deepak Welcome! Anytime!!!! You deserve more than that.
$endgroup$
– Anirban Niloy
Mar 3 at 7:20
1
$begingroup$
@Davo I am also glad that you are satisfied with the diagram. You can install the app GeoGebra Classic from Play-Store for illustrating any graph or diagram. I can suggest only that to you.
$endgroup$
– Anirban Niloy
Mar 3 at 14:52
1
$begingroup$
@DAVO Moreover, you can check from the following link whatever you prefer and which software you want to use math.stackexchange.com/q/1985/63393]
$endgroup$
– Anirban Niloy
Mar 3 at 14:59
|
show 3 more comments
$begingroup$
Assuming those are circular arcs and not "ellipses", you can indeed find the area exactly.
The "flower" has six "petals" Each of those petals has axial symmetry and can be divided into two halves. Each of those halves is the segment subtending a central angle of $fracpi3$ (radian measure) of a circle of radius $1$.
The area of one such segment is $frac 12 r^2(theta - sintheta) = frac 12(fracpi3 - fracsqrt32)$.
There are $12$ such segments, yielding the total area of the "flower" as $2pi - 3sqrt 3$.
$endgroup$
$begingroup$
Since the shape of each petal is claimed to be ellipses, your answer is not valid in assuming that they consist of circular arcs. However, I do reckon that they are intented to be arcs.
$endgroup$
– Trebor
Mar 3 at 3:08
$begingroup$
@Trebor I had edited my answer (I think before your comment) to clarify that assumption. However, I think it's a valid assumption, and just a very poor question.
$endgroup$
– Deepak
Mar 3 at 3:18
1
$begingroup$
@Deepak Welcome! Anytime!!!! You deserve more than that.
$endgroup$
– Anirban Niloy
Mar 3 at 7:20
1
$begingroup$
@Davo I am also glad that you are satisfied with the diagram. You can install the app GeoGebra Classic from Play-Store for illustrating any graph or diagram. I can suggest only that to you.
$endgroup$
– Anirban Niloy
Mar 3 at 14:52
1
$begingroup$
@DAVO Moreover, you can check from the following link whatever you prefer and which software you want to use math.stackexchange.com/q/1985/63393]
$endgroup$
– Anirban Niloy
Mar 3 at 14:59
|
show 3 more comments
$begingroup$
Assuming those are circular arcs and not "ellipses", you can indeed find the area exactly.
The "flower" has six "petals" Each of those petals has axial symmetry and can be divided into two halves. Each of those halves is the segment subtending a central angle of $fracpi3$ (radian measure) of a circle of radius $1$.
The area of one such segment is $frac 12 r^2(theta - sintheta) = frac 12(fracpi3 - fracsqrt32)$.
There are $12$ such segments, yielding the total area of the "flower" as $2pi - 3sqrt 3$.
$endgroup$
Assuming those are circular arcs and not "ellipses", you can indeed find the area exactly.
The "flower" has six "petals" Each of those petals has axial symmetry and can be divided into two halves. Each of those halves is the segment subtending a central angle of $fracpi3$ (radian measure) of a circle of radius $1$.
The area of one such segment is $frac 12 r^2(theta - sintheta) = frac 12(fracpi3 - fracsqrt32)$.
There are $12$ such segments, yielding the total area of the "flower" as $2pi - 3sqrt 3$.
edited Mar 3 at 5:07
Anirban Niloy
8411318
8411318
answered Mar 3 at 3:02
DeepakDeepak
17.6k11539
17.6k11539
$begingroup$
Since the shape of each petal is claimed to be ellipses, your answer is not valid in assuming that they consist of circular arcs. However, I do reckon that they are intented to be arcs.
$endgroup$
– Trebor
Mar 3 at 3:08
$begingroup$
@Trebor I had edited my answer (I think before your comment) to clarify that assumption. However, I think it's a valid assumption, and just a very poor question.
$endgroup$
– Deepak
Mar 3 at 3:18
1
$begingroup$
@Deepak Welcome! Anytime!!!! You deserve more than that.
$endgroup$
– Anirban Niloy
Mar 3 at 7:20
1
$begingroup$
@Davo I am also glad that you are satisfied with the diagram. You can install the app GeoGebra Classic from Play-Store for illustrating any graph or diagram. I can suggest only that to you.
$endgroup$
– Anirban Niloy
Mar 3 at 14:52
1
$begingroup$
@DAVO Moreover, you can check from the following link whatever you prefer and which software you want to use math.stackexchange.com/q/1985/63393]
$endgroup$
– Anirban Niloy
Mar 3 at 14:59
|
show 3 more comments
$begingroup$
Since the shape of each petal is claimed to be ellipses, your answer is not valid in assuming that they consist of circular arcs. However, I do reckon that they are intented to be arcs.
$endgroup$
– Trebor
Mar 3 at 3:08
$begingroup$
@Trebor I had edited my answer (I think before your comment) to clarify that assumption. However, I think it's a valid assumption, and just a very poor question.
$endgroup$
– Deepak
Mar 3 at 3:18
1
$begingroup$
@Deepak Welcome! Anytime!!!! You deserve more than that.
$endgroup$
– Anirban Niloy
Mar 3 at 7:20
1
$begingroup$
@Davo I am also glad that you are satisfied with the diagram. You can install the app GeoGebra Classic from Play-Store for illustrating any graph or diagram. I can suggest only that to you.
$endgroup$
– Anirban Niloy
Mar 3 at 14:52
1
$begingroup$
@DAVO Moreover, you can check from the following link whatever you prefer and which software you want to use math.stackexchange.com/q/1985/63393]
$endgroup$
– Anirban Niloy
Mar 3 at 14:59
$begingroup$
Since the shape of each petal is claimed to be ellipses, your answer is not valid in assuming that they consist of circular arcs. However, I do reckon that they are intented to be arcs.
$endgroup$
– Trebor
Mar 3 at 3:08
$begingroup$
Since the shape of each petal is claimed to be ellipses, your answer is not valid in assuming that they consist of circular arcs. However, I do reckon that they are intented to be arcs.
$endgroup$
– Trebor
Mar 3 at 3:08
$begingroup$
@Trebor I had edited my answer (I think before your comment) to clarify that assumption. However, I think it's a valid assumption, and just a very poor question.
$endgroup$
– Deepak
Mar 3 at 3:18
$begingroup$
@Trebor I had edited my answer (I think before your comment) to clarify that assumption. However, I think it's a valid assumption, and just a very poor question.
$endgroup$
– Deepak
Mar 3 at 3:18
1
1
$begingroup$
@Deepak Welcome! Anytime!!!! You deserve more than that.
$endgroup$
– Anirban Niloy
Mar 3 at 7:20
$begingroup$
@Deepak Welcome! Anytime!!!! You deserve more than that.
$endgroup$
– Anirban Niloy
Mar 3 at 7:20
1
1
$begingroup$
@Davo I am also glad that you are satisfied with the diagram. You can install the app GeoGebra Classic from Play-Store for illustrating any graph or diagram. I can suggest only that to you.
$endgroup$
– Anirban Niloy
Mar 3 at 14:52
$begingroup$
@Davo I am also glad that you are satisfied with the diagram. You can install the app GeoGebra Classic from Play-Store for illustrating any graph or diagram. I can suggest only that to you.
$endgroup$
– Anirban Niloy
Mar 3 at 14:52
1
1
$begingroup$
@DAVO Moreover, you can check from the following link whatever you prefer and which software you want to use math.stackexchange.com/q/1985/63393]
$endgroup$
– Anirban Niloy
Mar 3 at 14:59
$begingroup$
@DAVO Moreover, you can check from the following link whatever you prefer and which software you want to use math.stackexchange.com/q/1985/63393]
$endgroup$
– Anirban Niloy
Mar 3 at 14:59
|
show 3 more comments
$begingroup$
Here's a solution that doesn't require trigonometry. The hexagon is made up of 6 "petals" and 6 "wedges" (the scooped out triangles between the petals). Call the area of a petal $p$, and the area of a wedge $w$.
What is the area enclosed by two adjacent sides of the hexagon and the arc drawn from the vertex between them. (For example, the region EFAGE). It's one third of a unit circle, and contains three petals and two wedges:
$3p + 2w = frac 1 3 pi r^2 = frac pi 3$
What is the area of one triangle (one sixth of the hexagon). The altitude of the triangle is $sqrt1 - frac 1 4 = fracsqrt 32$, so the area is $fracsqrt 34$. The triangle is made up of two half-petals and one wedge:
$p + w = fracsqrt 34$
Solving the two simultaneous equations:
$2p + 2w = fracsqrt 32$
$p = fracpi3 - fracsqrt 32$
And the area of the flower is $6p = 2pi - 3sqrt 3$.
$endgroup$
add a comment |
$begingroup$
Here's a solution that doesn't require trigonometry. The hexagon is made up of 6 "petals" and 6 "wedges" (the scooped out triangles between the petals). Call the area of a petal $p$, and the area of a wedge $w$.
What is the area enclosed by two adjacent sides of the hexagon and the arc drawn from the vertex between them. (For example, the region EFAGE). It's one third of a unit circle, and contains three petals and two wedges:
$3p + 2w = frac 1 3 pi r^2 = frac pi 3$
What is the area of one triangle (one sixth of the hexagon). The altitude of the triangle is $sqrt1 - frac 1 4 = fracsqrt 32$, so the area is $fracsqrt 34$. The triangle is made up of two half-petals and one wedge:
$p + w = fracsqrt 34$
Solving the two simultaneous equations:
$2p + 2w = fracsqrt 32$
$p = fracpi3 - fracsqrt 32$
And the area of the flower is $6p = 2pi - 3sqrt 3$.
$endgroup$
add a comment |
$begingroup$
Here's a solution that doesn't require trigonometry. The hexagon is made up of 6 "petals" and 6 "wedges" (the scooped out triangles between the petals). Call the area of a petal $p$, and the area of a wedge $w$.
What is the area enclosed by two adjacent sides of the hexagon and the arc drawn from the vertex between them. (For example, the region EFAGE). It's one third of a unit circle, and contains three petals and two wedges:
$3p + 2w = frac 1 3 pi r^2 = frac pi 3$
What is the area of one triangle (one sixth of the hexagon). The altitude of the triangle is $sqrt1 - frac 1 4 = fracsqrt 32$, so the area is $fracsqrt 34$. The triangle is made up of two half-petals and one wedge:
$p + w = fracsqrt 34$
Solving the two simultaneous equations:
$2p + 2w = fracsqrt 32$
$p = fracpi3 - fracsqrt 32$
And the area of the flower is $6p = 2pi - 3sqrt 3$.
$endgroup$
Here's a solution that doesn't require trigonometry. The hexagon is made up of 6 "petals" and 6 "wedges" (the scooped out triangles between the petals). Call the area of a petal $p$, and the area of a wedge $w$.
What is the area enclosed by two adjacent sides of the hexagon and the arc drawn from the vertex between them. (For example, the region EFAGE). It's one third of a unit circle, and contains three petals and two wedges:
$3p + 2w = frac 1 3 pi r^2 = frac pi 3$
What is the area of one triangle (one sixth of the hexagon). The altitude of the triangle is $sqrt1 - frac 1 4 = fracsqrt 32$, so the area is $fracsqrt 34$. The triangle is made up of two half-petals and one wedge:
$p + w = fracsqrt 34$
Solving the two simultaneous equations:
$2p + 2w = fracsqrt 32$
$p = fracpi3 - fracsqrt 32$
And the area of the flower is $6p = 2pi - 3sqrt 3$.
answered Mar 3 at 7:04
AndyBAndyB
612
612
add a comment |
add a comment |
$begingroup$
[This answer was posted for the original version of the problem, which had essentially no information about the figure other than the picture itself]
We can't easily find the area exactly - we just don't have enough information for that. But then, what are the estimates we're looking at? (1) $approx 3.1$. (3) $approx 4.7$. (4) $approx 2.5$. (5) $approx 3.3$. (6) $approx 1.1$.
If the area's close to $3$, we might have difficulty deciding. If it's farther away, it'll be an easy choice.
Now, what's the area of the hexagon? That's something we can calculate exactly; an equilateral triangle of side $1$ has area $fracsqrt34$, and the hexagon is six of those stuck together for a total area of $frac32sqrt3approx 2.6$. The flower is clearly much less than that. Only one of the answers is plausible. It has to be (6), $2pi-3sqrt3$.
And, reverse-engineering from the answer... the curves that define the "petals" are the six circles centered at the vertices of the hexagon with radius $1$. That bit about "five ellipses" is just wrong.
$endgroup$
add a comment |
$begingroup$
[This answer was posted for the original version of the problem, which had essentially no information about the figure other than the picture itself]
We can't easily find the area exactly - we just don't have enough information for that. But then, what are the estimates we're looking at? (1) $approx 3.1$. (3) $approx 4.7$. (4) $approx 2.5$. (5) $approx 3.3$. (6) $approx 1.1$.
If the area's close to $3$, we might have difficulty deciding. If it's farther away, it'll be an easy choice.
Now, what's the area of the hexagon? That's something we can calculate exactly; an equilateral triangle of side $1$ has area $fracsqrt34$, and the hexagon is six of those stuck together for a total area of $frac32sqrt3approx 2.6$. The flower is clearly much less than that. Only one of the answers is plausible. It has to be (6), $2pi-3sqrt3$.
And, reverse-engineering from the answer... the curves that define the "petals" are the six circles centered at the vertices of the hexagon with radius $1$. That bit about "five ellipses" is just wrong.
$endgroup$
add a comment |
$begingroup$
[This answer was posted for the original version of the problem, which had essentially no information about the figure other than the picture itself]
We can't easily find the area exactly - we just don't have enough information for that. But then, what are the estimates we're looking at? (1) $approx 3.1$. (3) $approx 4.7$. (4) $approx 2.5$. (5) $approx 3.3$. (6) $approx 1.1$.
If the area's close to $3$, we might have difficulty deciding. If it's farther away, it'll be an easy choice.
Now, what's the area of the hexagon? That's something we can calculate exactly; an equilateral triangle of side $1$ has area $fracsqrt34$, and the hexagon is six of those stuck together for a total area of $frac32sqrt3approx 2.6$. The flower is clearly much less than that. Only one of the answers is plausible. It has to be (6), $2pi-3sqrt3$.
And, reverse-engineering from the answer... the curves that define the "petals" are the six circles centered at the vertices of the hexagon with radius $1$. That bit about "five ellipses" is just wrong.
$endgroup$
[This answer was posted for the original version of the problem, which had essentially no information about the figure other than the picture itself]
We can't easily find the area exactly - we just don't have enough information for that. But then, what are the estimates we're looking at? (1) $approx 3.1$. (3) $approx 4.7$. (4) $approx 2.5$. (5) $approx 3.3$. (6) $approx 1.1$.
If the area's close to $3$, we might have difficulty deciding. If it's farther away, it'll be an easy choice.
Now, what's the area of the hexagon? That's something we can calculate exactly; an equilateral triangle of side $1$ has area $fracsqrt34$, and the hexagon is six of those stuck together for a total area of $frac32sqrt3approx 2.6$. The flower is clearly much less than that. Only one of the answers is plausible. It has to be (6), $2pi-3sqrt3$.
And, reverse-engineering from the answer... the curves that define the "petals" are the six circles centered at the vertices of the hexagon with radius $1$. That bit about "five ellipses" is just wrong.
edited Mar 3 at 20:11
answered Mar 3 at 1:51
jmerryjmerry
16.9k11633
16.9k11633
add a comment |
add a comment |
$begingroup$
Shaded blue area is
(Area of pie)-(Area of rhombus)=$picdot 1^2cdotfrac120360-1^2cdotsin(120)$=$ fracpi3-fracsqrt32$
And we have 6 of them, and so total area of flower is
$ 6(fracpi3-fracsqrt32)=2pi-3sqrt3$
$endgroup$
add a comment |
$begingroup$
Shaded blue area is
(Area of pie)-(Area of rhombus)=$picdot 1^2cdotfrac120360-1^2cdotsin(120)$=$ fracpi3-fracsqrt32$
And we have 6 of them, and so total area of flower is
$ 6(fracpi3-fracsqrt32)=2pi-3sqrt3$
$endgroup$
add a comment |
$begingroup$
Shaded blue area is
(Area of pie)-(Area of rhombus)=$picdot 1^2cdotfrac120360-1^2cdotsin(120)$=$ fracpi3-fracsqrt32$
And we have 6 of them, and so total area of flower is
$ 6(fracpi3-fracsqrt32)=2pi-3sqrt3$
$endgroup$
Shaded blue area is
(Area of pie)-(Area of rhombus)=$picdot 1^2cdotfrac120360-1^2cdotsin(120)$=$ fracpi3-fracsqrt32$
And we have 6 of them, and so total area of flower is
$ 6(fracpi3-fracsqrt32)=2pi-3sqrt3$
edited Mar 3 at 13:40
answered Mar 3 at 13:08
Okkes DulgerciOkkes Dulgerci
1753
1753
add a comment |
add a comment |
$begingroup$
Here's another way of looking at it. Imagine taking two circles of radius 1 and cutting them into thirds. Arrange them so that the six 120° angles you have (one from each wedge) form the corners of the hexagon.
Every point in one of the "petals" is covered by three of these wedges. Every point in the hexagon that's not in one of the petals is covered by two of these wedges. The total area covered by the six wedges is therefore twice the area of the total hexagon plus the area of the petals; or, in other words, the area of the petals is the area of two unit circles minus twice the area of the hexagon.
The area of the two circles is $2pi$; the area of the hexagon is $3sqrt3/2$. The area of the petals is therefore $2 pi - 3 sqrt3 approx 1.087...$
$endgroup$
add a comment |
$begingroup$
Here's another way of looking at it. Imagine taking two circles of radius 1 and cutting them into thirds. Arrange them so that the six 120° angles you have (one from each wedge) form the corners of the hexagon.
Every point in one of the "petals" is covered by three of these wedges. Every point in the hexagon that's not in one of the petals is covered by two of these wedges. The total area covered by the six wedges is therefore twice the area of the total hexagon plus the area of the petals; or, in other words, the area of the petals is the area of two unit circles minus twice the area of the hexagon.
The area of the two circles is $2pi$; the area of the hexagon is $3sqrt3/2$. The area of the petals is therefore $2 pi - 3 sqrt3 approx 1.087...$
$endgroup$
add a comment |
$begingroup$
Here's another way of looking at it. Imagine taking two circles of radius 1 and cutting them into thirds. Arrange them so that the six 120° angles you have (one from each wedge) form the corners of the hexagon.
Every point in one of the "petals" is covered by three of these wedges. Every point in the hexagon that's not in one of the petals is covered by two of these wedges. The total area covered by the six wedges is therefore twice the area of the total hexagon plus the area of the petals; or, in other words, the area of the petals is the area of two unit circles minus twice the area of the hexagon.
The area of the two circles is $2pi$; the area of the hexagon is $3sqrt3/2$. The area of the petals is therefore $2 pi - 3 sqrt3 approx 1.087...$
$endgroup$
Here's another way of looking at it. Imagine taking two circles of radius 1 and cutting them into thirds. Arrange them so that the six 120° angles you have (one from each wedge) form the corners of the hexagon.
Every point in one of the "petals" is covered by three of these wedges. Every point in the hexagon that's not in one of the petals is covered by two of these wedges. The total area covered by the six wedges is therefore twice the area of the total hexagon plus the area of the petals; or, in other words, the area of the petals is the area of two unit circles minus twice the area of the hexagon.
The area of the two circles is $2pi$; the area of the hexagon is $3sqrt3/2$. The area of the petals is therefore $2 pi - 3 sqrt3 approx 1.087...$
answered Mar 3 at 14:05
Michael SeifertMichael Seifert
5,182626
5,182626
add a comment |
add a comment |
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3
$begingroup$
I count six ellipses
$endgroup$
– Ross Millikan
Mar 3 at 2:04
3
$begingroup$
Are the ellipses tangent to the sides at the vertices and meeting without overlap at the center? If so, they are not ellipses. Do you mean pairs of circular arcs?
$endgroup$
– Oscar Lanzi
Mar 3 at 3:01
$begingroup$
This question has been altered to something quite different than it originally was, some hours after it was posted, and after the older answers were posted. Although the change was probably for the better as the question is concerned, this leaves a confusing state of affairs.
$endgroup$
– Marc van Leeuwen
Mar 3 at 11:01
1
$begingroup$
Actually the picture was by user @AnirbanNiloy
$endgroup$
– Deepak
Mar 3 at 13:25