A flower in a hexagon












9












$begingroup$


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.
a flower in a hexagon




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.










share|cite|improve this question











$endgroup$








  • 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
















9












$begingroup$


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.
a flower in a hexagon




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.










share|cite|improve this question











$endgroup$








  • 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














9












9








9


2



$begingroup$


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.
a flower in a hexagon




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.










share|cite|improve this question











$endgroup$




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.
a flower in a hexagon




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 triangle area






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 3 at 14:45









Andrés E. Caicedo

65.7k8160250




65.7k8160250










asked Mar 3 at 1:09









DAVODAVO

526




526








  • 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




    $begingroup$
    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










5 Answers
5






active

oldest

votes


















13












$begingroup$

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 $frac{pi}{3}$ (radian measure) of a circle of radius $1$.



The area of one such segment is $frac 12 r^2(theta - sintheta) = frac 12(frac{pi}{3} - frac{sqrt{3}}{2})$.



There are $12$ such segments, yielding the total area of the "flower" as $2pi - 3sqrt 3$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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





















6












$begingroup$

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 $sqrt{1 - {frac 1 4}} = frac{sqrt 3}{2}$, so the area is $frac{sqrt 3}{4}$. The triangle is made up of two half-petals and one wedge:



$p + w = frac{sqrt 3}{4}$



Solving the two simultaneous equations:



$2p + 2w = frac{sqrt 3}{2}$



$p = frac{pi}{3} - frac{sqrt 3}{2}$



And the area of the flower is $6p = 2pi - 3sqrt 3$.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    [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 $frac{sqrt{3}}{4}$, and the hexagon is six of those stuck together for a total area of $frac32sqrt{3}approx 2.6$. The flower is clearly much less than that. Only one of the answers is plausible. It has to be (6), $2pi-3sqrt{3}$.



    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.






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      enter image description here



      Shaded blue area is



      (Area of pie)-(Area of rhombus)=$picdot 1^2cdotfrac{120}{360}-1^2cdotsin(120)$=$ frac{pi}{3}-frac{sqrt{3}}{2}$



      And we have 6 of them, and so total area of flower is



      $ 6(frac{pi}{3}-frac{sqrt{3}}{2})=2pi-3sqrt{3}$






      share|cite|improve this answer











      $endgroup$





















        1












        $begingroup$

        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 $3sqrt{3}/2$. The area of the petals is therefore $2 pi - 3 sqrt{3} approx 1.087...$






        share|cite|improve this answer









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          5 Answers
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          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          13












          $begingroup$

          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 $frac{pi}{3}$ (radian measure) of a circle of radius $1$.



          The area of one such segment is $frac 12 r^2(theta - sintheta) = frac 12(frac{pi}{3} - frac{sqrt{3}}{2})$.



          There are $12$ such segments, yielding the total area of the "flower" as $2pi - 3sqrt 3$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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


















          13












          $begingroup$

          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 $frac{pi}{3}$ (radian measure) of a circle of radius $1$.



          The area of one such segment is $frac 12 r^2(theta - sintheta) = frac 12(frac{pi}{3} - frac{sqrt{3}}{2})$.



          There are $12$ such segments, yielding the total area of the "flower" as $2pi - 3sqrt 3$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          13












          13








          13





          $begingroup$

          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 $frac{pi}{3}$ (radian measure) of a circle of radius $1$.



          The area of one such segment is $frac 12 r^2(theta - sintheta) = frac 12(frac{pi}{3} - frac{sqrt{3}}{2})$.



          There are $12$ such segments, yielding the total area of the "flower" as $2pi - 3sqrt 3$.






          share|cite|improve this answer











          $endgroup$



          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 $frac{pi}{3}$ (radian measure) of a circle of radius $1$.



          The area of one such segment is $frac 12 r^2(theta - sintheta) = frac 12(frac{pi}{3} - frac{sqrt{3}}{2})$.



          There are $12$ such segments, yielding the total area of the "flower" as $2pi - 3sqrt 3$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 3 at 5:07









          Anirban Niloy

          8271218




          8271218










          answered Mar 3 at 3:02









          DeepakDeepak

          17.4k11539




          17.4k11539












          • $begingroup$
            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










          • $begingroup$
            @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













          6












          $begingroup$

          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 $sqrt{1 - {frac 1 4}} = frac{sqrt 3}{2}$, so the area is $frac{sqrt 3}{4}$. The triangle is made up of two half-petals and one wedge:



          $p + w = frac{sqrt 3}{4}$



          Solving the two simultaneous equations:



          $2p + 2w = frac{sqrt 3}{2}$



          $p = frac{pi}{3} - frac{sqrt 3}{2}$



          And the area of the flower is $6p = 2pi - 3sqrt 3$.






          share|cite|improve this answer









          $endgroup$


















            6












            $begingroup$

            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 $sqrt{1 - {frac 1 4}} = frac{sqrt 3}{2}$, so the area is $frac{sqrt 3}{4}$. The triangle is made up of two half-petals and one wedge:



            $p + w = frac{sqrt 3}{4}$



            Solving the two simultaneous equations:



            $2p + 2w = frac{sqrt 3}{2}$



            $p = frac{pi}{3} - frac{sqrt 3}{2}$



            And the area of the flower is $6p = 2pi - 3sqrt 3$.






            share|cite|improve this answer









            $endgroup$
















              6












              6








              6





              $begingroup$

              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 $sqrt{1 - {frac 1 4}} = frac{sqrt 3}{2}$, so the area is $frac{sqrt 3}{4}$. The triangle is made up of two half-petals and one wedge:



              $p + w = frac{sqrt 3}{4}$



              Solving the two simultaneous equations:



              $2p + 2w = frac{sqrt 3}{2}$



              $p = frac{pi}{3} - frac{sqrt 3}{2}$



              And the area of the flower is $6p = 2pi - 3sqrt 3$.






              share|cite|improve this answer









              $endgroup$



              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 $sqrt{1 - {frac 1 4}} = frac{sqrt 3}{2}$, so the area is $frac{sqrt 3}{4}$. The triangle is made up of two half-petals and one wedge:



              $p + w = frac{sqrt 3}{4}$



              Solving the two simultaneous equations:



              $2p + 2w = frac{sqrt 3}{2}$



              $p = frac{pi}{3} - frac{sqrt 3}{2}$



              And the area of the flower is $6p = 2pi - 3sqrt 3$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Mar 3 at 7:04









              AndyBAndyB

              612




              612























                  3












                  $begingroup$

                  [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 $frac{sqrt{3}}{4}$, and the hexagon is six of those stuck together for a total area of $frac32sqrt{3}approx 2.6$. The flower is clearly much less than that. Only one of the answers is plausible. It has to be (6), $2pi-3sqrt{3}$.



                  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.






                  share|cite|improve this answer











                  $endgroup$


















                    3












                    $begingroup$

                    [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 $frac{sqrt{3}}{4}$, and the hexagon is six of those stuck together for a total area of $frac32sqrt{3}approx 2.6$. The flower is clearly much less than that. Only one of the answers is plausible. It has to be (6), $2pi-3sqrt{3}$.



                    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.






                    share|cite|improve this answer











                    $endgroup$
















                      3












                      3








                      3





                      $begingroup$

                      [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 $frac{sqrt{3}}{4}$, and the hexagon is six of those stuck together for a total area of $frac32sqrt{3}approx 2.6$. The flower is clearly much less than that. Only one of the answers is plausible. It has to be (6), $2pi-3sqrt{3}$.



                      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.






                      share|cite|improve this answer











                      $endgroup$



                      [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 $frac{sqrt{3}}{4}$, and the hexagon is six of those stuck together for a total area of $frac32sqrt{3}approx 2.6$. The flower is clearly much less than that. Only one of the answers is plausible. It has to be (6), $2pi-3sqrt{3}$.



                      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.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Mar 3 at 20:11

























                      answered Mar 3 at 1:51









                      jmerryjmerry

                      13.2k1628




                      13.2k1628























                          1












                          $begingroup$

                          enter image description here



                          Shaded blue area is



                          (Area of pie)-(Area of rhombus)=$picdot 1^2cdotfrac{120}{360}-1^2cdotsin(120)$=$ frac{pi}{3}-frac{sqrt{3}}{2}$



                          And we have 6 of them, and so total area of flower is



                          $ 6(frac{pi}{3}-frac{sqrt{3}}{2})=2pi-3sqrt{3}$






                          share|cite|improve this answer











                          $endgroup$


















                            1












                            $begingroup$

                            enter image description here



                            Shaded blue area is



                            (Area of pie)-(Area of rhombus)=$picdot 1^2cdotfrac{120}{360}-1^2cdotsin(120)$=$ frac{pi}{3}-frac{sqrt{3}}{2}$



                            And we have 6 of them, and so total area of flower is



                            $ 6(frac{pi}{3}-frac{sqrt{3}}{2})=2pi-3sqrt{3}$






                            share|cite|improve this answer











                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              enter image description here



                              Shaded blue area is



                              (Area of pie)-(Area of rhombus)=$picdot 1^2cdotfrac{120}{360}-1^2cdotsin(120)$=$ frac{pi}{3}-frac{sqrt{3}}{2}$



                              And we have 6 of them, and so total area of flower is



                              $ 6(frac{pi}{3}-frac{sqrt{3}}{2})=2pi-3sqrt{3}$






                              share|cite|improve this answer











                              $endgroup$



                              enter image description here



                              Shaded blue area is



                              (Area of pie)-(Area of rhombus)=$picdot 1^2cdotfrac{120}{360}-1^2cdotsin(120)$=$ frac{pi}{3}-frac{sqrt{3}}{2}$



                              And we have 6 of them, and so total area of flower is



                              $ 6(frac{pi}{3}-frac{sqrt{3}}{2})=2pi-3sqrt{3}$







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Mar 3 at 13:40

























                              answered Mar 3 at 13:08









                              Okkes DulgerciOkkes Dulgerci

                              1653




                              1653























                                  1












                                  $begingroup$

                                  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 $3sqrt{3}/2$. The area of the petals is therefore $2 pi - 3 sqrt{3} approx 1.087...$






                                  share|cite|improve this answer









                                  $endgroup$


















                                    1












                                    $begingroup$

                                    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 $3sqrt{3}/2$. The area of the petals is therefore $2 pi - 3 sqrt{3} approx 1.087...$






                                    share|cite|improve this answer









                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      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 $3sqrt{3}/2$. The area of the petals is therefore $2 pi - 3 sqrt{3} approx 1.087...$






                                      share|cite|improve this answer









                                      $endgroup$



                                      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 $3sqrt{3}/2$. The area of the petals is therefore $2 pi - 3 sqrt{3} approx 1.087...$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Mar 3 at 14:05









                                      Michael SeifertMichael Seifert

                                      5,052625




                                      5,052625






























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