Getting area of pentagon with some values missing

Clash Royale CLAN TAG#URR8PPP

up vote
2
down vote

favorite

1

I have a problem how to get the area from the picture.
Some ideas I got are not good enough to get the correct value of the whole element.

geometry area

share|cite|improve this question

edited Jul 30 at 21:49

高田航

1,120318

asked Jul 30 at 20:41

Mladen Uzelac

1112

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  This post would be much improved if you include the ideas you had, and where they failed!
  – B. Mehta
  Jul 30 at 20:44
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The ordinary word for a convex polygon of five vertices is pentagon. The word you’ve used is usually reserved for another kind of five-vertex figure.
  – Lubin
  Jul 30 at 21:26
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

1

I have a problem how to get the area from the picture.
Some ideas I got are not good enough to get the correct value of the whole element.

geometry area

share|cite|improve this question

edited Jul 30 at 21:49

高田航

1,120318

asked Jul 30 at 20:41

Mladen Uzelac

1112

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  This post would be much improved if you include the ideas you had, and where they failed!
  – B. Mehta
  Jul 30 at 20:44
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The ordinary word for a convex polygon of five vertices is pentagon. The word you’ve used is usually reserved for another kind of five-vertex figure.
  – Lubin
  Jul 30 at 21:26
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

1

up vote
2
down vote

favorite

1

1

I have a problem how to get the area from the picture.
Some ideas I got are not good enough to get the correct value of the whole element.

geometry area

share|cite|improve this question

edited Jul 30 at 21:49

高田航

1,120318

asked Jul 30 at 20:41

Mladen Uzelac

1112

I have a problem how to get the area from the picture.
Some ideas I got are not good enough to get the correct value of the whole element.

geometry area

share|cite|improve this question

edited Jul 30 at 21:49

高田航

1,120318

asked Jul 30 at 20:41

Mladen Uzelac

1112

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 21:49

高田航

1,120318

edited Jul 30 at 21:49

高田航

1,120318

edited Jul 30 at 21:49

高田航

1,120318

1,120318

asked Jul 30 at 20:41

Mladen Uzelac

1112

asked Jul 30 at 20:41

Mladen Uzelac

1112

asked Jul 30 at 20:41

Mladen Uzelac

1112

1112

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  This post would be much improved if you include the ideas you had, and where they failed!
  – B. Mehta
  Jul 30 at 20:44
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The ordinary word for a convex polygon of five vertices is pentagon. The word you’ve used is usually reserved for another kind of five-vertex figure.
  – Lubin
  Jul 30 at 21:26
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  This post would be much improved if you include the ideas you had, and where they failed!
  – B. Mehta
  Jul 30 at 20:44
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The ordinary word for a convex polygon of five vertices is pentagon. The word you’ve used is usually reserved for another kind of five-vertex figure.
  – Lubin
  Jul 30 at 21:26
  

  
  
  
  

2

2

This post would be much improved if you include the ideas you had, and where they failed!
– B. Mehta
Jul 30 at 20:44

This post would be much improved if you include the ideas you had, and where they failed!
– B. Mehta
Jul 30 at 20:44

2

2

The ordinary word for a convex polygon of five vertices is pentagon. The word you’ve used is usually reserved for another kind of five-vertex figure.
– Lubin
Jul 30 at 21:26

The ordinary word for a convex polygon of five vertices is pentagon. The word you’ve used is usually reserved for another kind of five-vertex figure.
– Lubin
Jul 30 at 21:26

add a comment | 

4 Answers
4

active

oldest

votes

up vote
2
down vote

You don’t have enough data to compute the area (if 13.62 denotes the lenght of the basis as it seems on the OP figure). The area depends on the « balance » between the left and the right sides of the house. And this seems unknown from your picture.

share|cite|improve this answer

edited Jul 31 at 8:22

answered Jul 30 at 21:10

mathcounterexamples.net

23.1k21651

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, I agree. But since those different lengths of the sides is an anomaly I decided to accept that the tip is in the middle.
  – Mladen Uzelac
  Jul 31 at 16:41
  

  
  
  
  

add a comment | 

up vote
1
down vote

Hint: Draw straight lines between the upper left corner and the midline, and upper right corner and the midline. Now you have two rectangles and two triangles. Can you compute their areas?

share|cite|improve this answer

answered Jul 30 at 20:58

Servaes

20k33484

add a comment | 

up vote
0
down vote

Hint: You know the width is $13.62$, and the maximum height it reaches is $1.65$, so it must fit within a rectangle of those dimensions.

share|cite|improve this answer

edited Jul 30 at 21:32

mathcounterexamples.net

23.1k21651

answered Jul 30 at 20:54

高田航

1,120318

add a comment | 

up vote
0
down vote

I asked someone and this answer will satisfy me

share|cite|improve this answer

answered Jul 31 at 16:39

Mladen Uzelac

1112

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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2867404%2fgetting-area-of-pentagon-with-some-values-missing%23new-answer', 'question_page');

);

Post as a guest

Name Email

4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

You don’t have enough data to compute the area (if 13.62 denotes the lenght of the basis as it seems on the OP figure). The area depends on the « balance » between the left and the right sides of the house. And this seems unknown from your picture.

share|cite|improve this answer

edited Jul 31 at 8:22

answered Jul 30 at 21:10

mathcounterexamples.net

23.1k21651

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, I agree. But since those different lengths of the sides is an anomaly I decided to accept that the tip is in the middle.
  – Mladen Uzelac
  Jul 31 at 16:41
  

  
  
  
  

add a comment | 

up vote
2
down vote

You don’t have enough data to compute the area (if 13.62 denotes the lenght of the basis as it seems on the OP figure). The area depends on the « balance » between the left and the right sides of the house. And this seems unknown from your picture.

share|cite|improve this answer

edited Jul 31 at 8:22

answered Jul 30 at 21:10

mathcounterexamples.net

23.1k21651

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, I agree. But since those different lengths of the sides is an anomaly I decided to accept that the tip is in the middle.
  – Mladen Uzelac
  Jul 31 at 16:41
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

You don’t have enough data to compute the area (if 13.62 denotes the lenght of the basis as it seems on the OP figure). The area depends on the « balance » between the left and the right sides of the house. And this seems unknown from your picture.

share|cite|improve this answer

edited Jul 31 at 8:22

answered Jul 30 at 21:10

mathcounterexamples.net

23.1k21651

You don’t have enough data to compute the area (if 13.62 denotes the lenght of the basis as it seems on the OP figure). The area depends on the « balance » between the left and the right sides of the house. And this seems unknown from your picture.

share|cite|improve this answer

edited Jul 31 at 8:22

answered Jul 30 at 21:10

mathcounterexamples.net

23.1k21651

share|cite|improve this answer

share|cite|improve this answer

edited Jul 31 at 8:22

edited Jul 31 at 8:22

edited Jul 31 at 8:22

answered Jul 30 at 21:10

mathcounterexamples.net

23.1k21651

answered Jul 30 at 21:10

mathcounterexamples.net

23.1k21651

answered Jul 30 at 21:10

mathcounterexamples.net

23.1k21651

23.1k21651

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, I agree. But since those different lengths of the sides is an anomaly I decided to accept that the tip is in the middle.
  – Mladen Uzelac
  Jul 31 at 16:41
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, I agree. But since those different lengths of the sides is an anomaly I decided to accept that the tip is in the middle.
  – Mladen Uzelac
  Jul 31 at 16:41
  

  
  
  
  

Yes, I agree. But since those different lengths of the sides is an anomaly I decided to accept that the tip is in the middle.
– Mladen Uzelac
Jul 31 at 16:41

Yes, I agree. But since those different lengths of the sides is an anomaly I decided to accept that the tip is in the middle.
– Mladen Uzelac
Jul 31 at 16:41

add a comment | 

up vote
1
down vote

Hint: Draw straight lines between the upper left corner and the midline, and upper right corner and the midline. Now you have two rectangles and two triangles. Can you compute their areas?

share|cite|improve this answer

answered Jul 30 at 20:58

Servaes

20k33484

add a comment | 

up vote
1
down vote

Hint: Draw straight lines between the upper left corner and the midline, and upper right corner and the midline. Now you have two rectangles and two triangles. Can you compute their areas?

share|cite|improve this answer

answered Jul 30 at 20:58

Servaes

20k33484

add a comment | 

up vote
1
down vote

up vote
1
down vote

Hint: Draw straight lines between the upper left corner and the midline, and upper right corner and the midline. Now you have two rectangles and two triangles. Can you compute their areas?

share|cite|improve this answer

answered Jul 30 at 20:58

Servaes

20k33484

Hint: Draw straight lines between the upper left corner and the midline, and upper right corner and the midline. Now you have two rectangles and two triangles. Can you compute their areas?

share|cite|improve this answer

answered Jul 30 at 20:58

Servaes

20k33484

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 20:58

Servaes

20k33484

answered Jul 30 at 20:58

Servaes

20k33484

answered Jul 30 at 20:58

Servaes

20k33484

20k33484

add a comment | 

add a comment | 

up vote
0
down vote

Hint: You know the width is $13.62$, and the maximum height it reaches is $1.65$, so it must fit within a rectangle of those dimensions.

share|cite|improve this answer

edited Jul 30 at 21:32

mathcounterexamples.net

23.1k21651

answered Jul 30 at 20:54

高田航

1,120318

add a comment | 

up vote
0
down vote

Hint: You know the width is $13.62$, and the maximum height it reaches is $1.65$, so it must fit within a rectangle of those dimensions.

share|cite|improve this answer

edited Jul 30 at 21:32

mathcounterexamples.net

23.1k21651

answered Jul 30 at 20:54

高田航

1,120318

add a comment | 

up vote
0
down vote

up vote
0
down vote

Hint: You know the width is $13.62$, and the maximum height it reaches is $1.65$, so it must fit within a rectangle of those dimensions.

share|cite|improve this answer

edited Jul 30 at 21:32

mathcounterexamples.net

23.1k21651

answered Jul 30 at 20:54

高田航

1,120318

Hint: You know the width is $13.62$, and the maximum height it reaches is $1.65$, so it must fit within a rectangle of those dimensions.

share|cite|improve this answer

edited Jul 30 at 21:32

mathcounterexamples.net

23.1k21651

answered Jul 30 at 20:54

高田航

1,120318

share|cite|improve this answer

share|cite|improve this answer

edited Jul 30 at 21:32

mathcounterexamples.net

23.1k21651

edited Jul 30 at 21:32

mathcounterexamples.net

23.1k21651

edited Jul 30 at 21:32

mathcounterexamples.net

23.1k21651

23.1k21651

answered Jul 30 at 20:54

高田航

1,120318

answered Jul 30 at 20:54

高田航

1,120318

answered Jul 30 at 20:54

高田航

1,120318

1,120318

add a comment | 

add a comment | 

up vote
0
down vote

I asked someone and this answer will satisfy me

share|cite|improve this answer

answered Jul 31 at 16:39

Mladen Uzelac

1112

add a comment | 

up vote
0
down vote

I asked someone and this answer will satisfy me

share|cite|improve this answer

answered Jul 31 at 16:39

Mladen Uzelac

1112

add a comment | 

up vote
0
down vote

up vote
0
down vote

I asked someone and this answer will satisfy me

share|cite|improve this answer

answered Jul 31 at 16:39

Mladen Uzelac

1112

I asked someone and this answer will satisfy me

share|cite|improve this answer

answered Jul 31 at 16:39

Mladen Uzelac

1112

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 16:39

Mladen Uzelac

1112

answered Jul 31 at 16:39

Mladen Uzelac

1112

answered Jul 31 at 16:39

Mladen Uzelac

1112

1112

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2867404%2fgetting-area-of-pentagon-with-some-values-missing%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

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

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

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email