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A flower in a hexagon

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1

1

$begingroup$

The area of the flower

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 and create an equilateral triangle but still was 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 create a flower-shaped-like object. Find the area of the flower.

Here are the possible answers ↓

1. $pi$

3. $3pi/2$

4. $4sqrt2-pi$

5. $pi/2+sqrt3$

6. $2pi-3sqrt3$

geometry logic triangle area

share|cite|improve this question

edited 37 mins ago

J. W. Tanner

2,6161217

asked 5 hours ago

DAVODAVO

85

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I count six ellipses
  $endgroup$
  – Ross Millikan
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  3 hours ago
  

  
  
  
  

add a comment | 

1

1

$begingroup$

The area of the flower

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 and create an equilateral triangle but still was 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 create a flower-shaped-like object. Find the area of the flower.

Here are the possible answers ↓

1. $pi$

3. $3pi/2$

4. $4sqrt2-pi$

5. $pi/2+sqrt3$

6. $2pi-3sqrt3$

geometry logic triangle area

share|cite|improve this question

edited 37 mins ago

J. W. Tanner

2,6161217

asked 5 hours ago

DAVODAVO

85

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I count six ellipses
  $endgroup$
  – Ross Millikan
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  3 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

The area of the flower

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 and create an equilateral triangle but still was 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 create a flower-shaped-like object. Find the area of the flower.

Here are the possible answers ↓

1. $pi$

3. $3pi/2$

4. $4sqrt2-pi$

5. $pi/2+sqrt3$

6. $2pi-3sqrt3$

geometry logic triangle area

share|cite|improve this question

edited 37 mins ago

J. W. Tanner

2,6161217

asked 5 hours ago

DAVODAVO

85

$endgroup$

share|cite|improve this question

edited 37 mins ago

J. W. Tanner

2,6161217

asked 5 hours ago

DAVODAVO

85

share|cite|improve this question

edited 37 mins ago

J. W. Tanner

2,6161217

asked 5 hours ago

DAVODAVO

85

edited 37 mins ago

J. W. Tanner

2,6161217

edited 37 mins ago

J. W. Tanner

2,6161217

asked 5 hours ago

DAVODAVO

85

asked 5 hours ago

DAVODAVO

85

2 Answers
2

active

oldest

votes

3

$begingroup$

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

answered 4 hours ago

jmerryjmerry

11.6k1527

$endgroup$

add a comment | 

3

$begingroup$

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

edited 1 hour ago

Anirban Niloy

609218

answered 3 hours ago

DeepakDeepak

17.2k11536

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  3 hours ago
  

  
  
  
  

add a comment | 

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3

$begingroup$

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

answered 4 hours ago

jmerryjmerry

11.6k1527

$endgroup$

add a comment | 

3

$begingroup$

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

answered 4 hours ago

jmerryjmerry

11.6k1527

$endgroup$

add a comment | 

$begingroup$

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

answered 4 hours ago

jmerryjmerry

11.6k1527

$endgroup$

share|cite|improve this answer

answered 4 hours ago

jmerryjmerry

11.6k1527

answered 4 hours ago

jmerryjmerry

11.6k1527

answered 4 hours ago

jmerryjmerry

11.6k1527

3

$begingroup$

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

edited 1 hour ago

Anirban Niloy

609218

answered 3 hours ago

DeepakDeepak

17.2k11536

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  3 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

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

edited 1 hour ago

Anirban Niloy

609218

answered 3 hours ago

DeepakDeepak

17.2k11536

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  3 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 1 hour ago

Anirban Niloy

609218

answered 3 hours ago

DeepakDeepak

17.2k11536

$endgroup$

share|cite|improve this answer

edited 1 hour ago

Anirban Niloy

609218

answered 3 hours ago

DeepakDeepak

17.2k11536

edited 1 hour ago

Anirban Niloy

609218

edited 1 hour ago

Anirban Niloy

609218

answered 3 hours ago

DeepakDeepak

17.2k11536

answered 3 hours ago

DeepakDeepak

17.2k11536