Manipulating a general length function

Cat is tipping over bed-side lamps during the night

Why did Jodrell Bank assist the Soviet Union to collect data from their spacecraft in the mid 1960's?

Explain the objections to these measures against human trafficking

What is the etymology of the kanji 食?

What are the advantages of using `make` for small projects?

How can I deal with a significant flaw I found in my previous supervisor’s paper?

High pressure canisters of air as gun-less projectiles

Can pricing be copyrighted?

How to avoid being sexist when trying to employ someone to function in a very sexist environment?

Is there any differences between “gucken” and “schauen”?

Why zero tolerance on nudity in space?

What makes the Forgotten Realms "forgotten"?

What do you call a fact that doesn't match the settings?

Does Windows 10's telemetry include sending *.doc files if Word crashed?

How did the original light saber work?

Why is working on the same position for more than 15 years not a red flag?

Strange Sign on Lab Door

Unwarranted claim of higher degree of accuracy in zircon geochronology

Do authors have to be politically correct in article-writing?

Pendulum Rotation

1 0 1 0 1 0 1 0 1 0 1

Why did Bush enact a completely different foreign policy to that which he espoused during the 2000 Presidential election campaign?

Why can a 352GB NumPy ndarray be used on an 8GB memory macOS computer?

A flower in a hexagon

3D Geometry Contest Math ProblemFind the area of the region $ABCD$.How do you calculate the smallest cycle possible for a given tile shape?A circle is divided into equal arcs by $n$ diamtrs. Prove the feet of the $perp$s from a point $M$ in circle to them are vertices of a regular n-gonDesign a tangramDividing a polygon in desired wayTricky Geometry Problem involving rectangle rotation.What's the word for same-length-sided-ness? equilarity? equilaterity? equilateralness?What 'uniform' shapes can be used to build an approximated spherical object?Area of a regular hexagon via area of triangles

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 42 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 42 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 42 mins ago

J. W. Tanner

2,6161217

asked 5 hours ago

DAVODAVO

85

$endgroup$

share|cite|improve this question

edited 42 mins ago

J. W. Tanner

2,6161217

asked 5 hours ago

DAVODAVO

85

share|cite|improve this question

edited 42 mins ago

J. W. Tanner

2,6161217

asked 5 hours ago

DAVODAVO

85

edited 42 mins ago

J. W. Tanner

2,6161217

edited 42 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 | 

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3133044%2fa-flower-in-a-hexagon%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

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