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Construction of $sqrtab$ user ruler and compass [duplicate]

0

1

$begingroup$

This question already has an answer here:

• 
  Compass-and-straightedge construction of the square root of a given line?
  
  3 answers
  
  

Q. Given two line segments of length a and b.
Draw a line segment of length $sqrtab$ using a ruler and compass.

I didn't get any idea how to approach to the solution.

geometry geometric-construction

share|cite|improve this question

edited Apr 3 at 15:43

Jean Marie

31.6k42355

asked Apr 2 at 10:53

user579689user579689

113

$endgroup$

marked as duplicate by Jaap Scherphuis, Lord Shark the Unknown, Trần Thúc Minh Trí, Javi, Théophile Apr 3 at 16:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Lines have infinite length. You mean segments (or line segments).
  $endgroup$
  – steven gregory
  Apr 2 at 10:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Take a look at math.stackexchange.com/q/708
  $endgroup$
  – Jean Marie
  Apr 2 at 13:56
  

  
  
  
  

add a comment | 

0

1

$begingroup$

This question already has an answer here:

• 
  Compass-and-straightedge construction of the square root of a given line?
  
  3 answers
  
  

Q. Given two line segments of length a and b.
Draw a line segment of length $sqrtab$ using a ruler and compass.

I didn't get any idea how to approach to the solution.

geometry geometric-construction

share|cite|improve this question

edited Apr 3 at 15:43

Jean Marie

31.6k42355

asked Apr 2 at 10:53

user579689user579689

113

$endgroup$

marked as duplicate by Jaap Scherphuis, Lord Shark the Unknown, Trần Thúc Minh Trí, Javi, Théophile Apr 3 at 16:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Lines have infinite length. You mean segments (or line segments).
  $endgroup$
  – steven gregory
  Apr 2 at 10:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Take a look at math.stackexchange.com/q/708
  $endgroup$
  – Jean Marie
  Apr 2 at 13:56
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Compass-and-straightedge construction of the square root of a given line?
  
  3 answers
  
  

Q. Given two line segments of length a and b.
Draw a line segment of length $sqrtab$ using a ruler and compass.

I didn't get any idea how to approach to the solution.

geometry geometric-construction

share|cite|improve this question

edited Apr 3 at 15:43

Jean Marie

31.6k42355

asked Apr 2 at 10:53

user579689user579689

113

$endgroup$

share|cite|improve this question

edited Apr 3 at 15:43

Jean Marie

31.6k42355

asked Apr 2 at 10:53

user579689user579689

113

share|cite|improve this question

edited Apr 3 at 15:43

Jean Marie

31.6k42355

asked Apr 2 at 10:53

user579689user579689

113

edited Apr 3 at 15:43

Jean Marie

31.6k42355

edited Apr 3 at 15:43

Jean Marie

31.6k42355

asked Apr 2 at 10:53

user579689user579689

113

asked Apr 2 at 10:53

user579689user579689

113

3 Answers
3

active

oldest

votes

1

$begingroup$

Let the base of the large triangle be $a+b$, and the height $h$. By similarity of the small triangles,

$$frac ha=frac bh$$ so that $$h=sqrtab.$$

share|cite|improve this answer

answered Apr 3 at 15:51

Yves DaoustYves Daoust

133k676232

$endgroup$

add a comment | 

0

$begingroup$

Draw a line of length $a+b$. Construct the perpendicular line in the point they joint. The semicircle over $a+b$ cuts that perpendicular. Now the distance between that point and the joining point is $sqrtab$ due to Euclid.

share|cite|improve this answer

answered Apr 2 at 11:02

Michael HoppeMichael Hoppe

11.3k31837

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you don't mind , Can you show it analytically please??
  $endgroup$
  – user579689
  Apr 2 at 11:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See en.wikipedia.org/wiki/Geometric_mean_theorem, please.
  $endgroup$
  – Michael Hoppe
  Apr 2 at 11:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The word "diameter" is maybe missing in "The semi-circle over"
  $endgroup$
  – Jean Marie
  Apr 2 at 13:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Being not a native speaker of the English language: What is ambiguous in the phrase "semicircle over a line segment"?
  $endgroup$
  – Michael Hoppe
  Apr 2 at 13:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nothing ambiguous... Don't bother with my remark.
  $endgroup$
  – Jean Marie
  Apr 2 at 13:58
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

Join segments of length a and length b together on the same line. Call where they join Point J. Call their Midpoint M. Construct a circle centered at M through either end point of a+b. Construct a line perpendicular to a+b through point J. Call where it intersects the circle point Q. The length of QJ is $sqrtab$ as proven elsewhere. These above constructions follow from Euclid's postulates I, III, SAS, and ASA. So the constructions should be valid in neutral geometry.

A related approach. Construct a rectangle having one sidelength a and one side length b. Find a square having the same area as the starting rectangle. The side length of thes square will be $sqrtab$.

share|cite|improve this answer

answered Apr 3 at 16:26

TurlocTheRedTurlocTheRed

1,034311

$endgroup$

add a comment | 

1

$begingroup$

Let the base of the large triangle be $a+b$, and the height $h$. By similarity of the small triangles,

$$frac ha=frac bh$$ so that $$h=sqrtab.$$

share|cite|improve this answer

answered Apr 3 at 15:51

Yves DaoustYves Daoust

133k676232

$endgroup$

add a comment | 

1

$begingroup$

Let the base of the large triangle be $a+b$, and the height $h$. By similarity of the small triangles,

$$frac ha=frac bh$$ so that $$h=sqrtab.$$

share|cite|improve this answer

answered Apr 3 at 15:51

Yves DaoustYves Daoust

133k676232

$endgroup$

add a comment | 

$begingroup$

Let the base of the large triangle be $a+b$, and the height $h$. By similarity of the small triangles,

$$frac ha=frac bh$$ so that $$h=sqrtab.$$

share|cite|improve this answer

answered Apr 3 at 15:51

Yves DaoustYves Daoust

133k676232

$endgroup$

share|cite|improve this answer

answered Apr 3 at 15:51

Yves DaoustYves Daoust

133k676232

answered Apr 3 at 15:51

Yves DaoustYves Daoust

133k676232

answered Apr 3 at 15:51

Yves DaoustYves Daoust

133k676232

0

$begingroup$

Draw a line of length $a+b$. Construct the perpendicular line in the point they joint. The semicircle over $a+b$ cuts that perpendicular. Now the distance between that point and the joining point is $sqrtab$ due to Euclid.

share|cite|improve this answer

answered Apr 2 at 11:02

Michael HoppeMichael Hoppe

11.3k31837

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you don't mind , Can you show it analytically please??
  $endgroup$
  – user579689
  Apr 2 at 11:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See en.wikipedia.org/wiki/Geometric_mean_theorem, please.
  $endgroup$
  – Michael Hoppe
  Apr 2 at 11:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The word "diameter" is maybe missing in "The semi-circle over"
  $endgroup$
  – Jean Marie
  Apr 2 at 13:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Being not a native speaker of the English language: What is ambiguous in the phrase "semicircle over a line segment"?
  $endgroup$
  – Michael Hoppe
  Apr 2 at 13:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nothing ambiguous... Don't bother with my remark.
  $endgroup$
  – Jean Marie
  Apr 2 at 13:58
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

Draw a line of length $a+b$. Construct the perpendicular line in the point they joint. The semicircle over $a+b$ cuts that perpendicular. Now the distance between that point and the joining point is $sqrtab$ due to Euclid.

share|cite|improve this answer

answered Apr 2 at 11:02

Michael HoppeMichael Hoppe

11.3k31837

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you don't mind , Can you show it analytically please??
  $endgroup$
  – user579689
  Apr 2 at 11:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See en.wikipedia.org/wiki/Geometric_mean_theorem, please.
  $endgroup$
  – Michael Hoppe
  Apr 2 at 11:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The word "diameter" is maybe missing in "The semi-circle over"
  $endgroup$
  – Jean Marie
  Apr 2 at 13:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Being not a native speaker of the English language: What is ambiguous in the phrase "semicircle over a line segment"?
  $endgroup$
  – Michael Hoppe
  Apr 2 at 13:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nothing ambiguous... Don't bother with my remark.
  $endgroup$
  – Jean Marie
  Apr 2 at 13:58
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Draw a line of length $a+b$. Construct the perpendicular line in the point they joint. The semicircle over $a+b$ cuts that perpendicular. Now the distance between that point and the joining point is $sqrtab$ due to Euclid.

share|cite|improve this answer

answered Apr 2 at 11:02

Michael HoppeMichael Hoppe

11.3k31837

$endgroup$

share|cite|improve this answer

answered Apr 2 at 11:02

Michael HoppeMichael Hoppe

11.3k31837

answered Apr 2 at 11:02

Michael HoppeMichael Hoppe

11.3k31837

answered Apr 2 at 11:02

Michael HoppeMichael Hoppe

11.3k31837

0

$begingroup$

Join segments of length a and length b together on the same line. Call where they join Point J. Call their Midpoint M. Construct a circle centered at M through either end point of a+b. Construct a line perpendicular to a+b through point J. Call where it intersects the circle point Q. The length of QJ is $sqrtab$ as proven elsewhere. These above constructions follow from Euclid's postulates I, III, SAS, and ASA. So the constructions should be valid in neutral geometry.

A related approach. Construct a rectangle having one sidelength a and one side length b. Find a square having the same area as the starting rectangle. The side length of thes square will be $sqrtab$.

share|cite|improve this answer

answered Apr 3 at 16:26

TurlocTheRedTurlocTheRed

1,034311

$endgroup$

add a comment | 

0

$begingroup$

Join segments of length a and length b together on the same line. Call where they join Point J. Call their Midpoint M. Construct a circle centered at M through either end point of a+b. Construct a line perpendicular to a+b through point J. Call where it intersects the circle point Q. The length of QJ is $sqrtab$ as proven elsewhere. These above constructions follow from Euclid's postulates I, III, SAS, and ASA. So the constructions should be valid in neutral geometry.

A related approach. Construct a rectangle having one sidelength a and one side length b. Find a square having the same area as the starting rectangle. The side length of thes square will be $sqrtab$.

share|cite|improve this answer

answered Apr 3 at 16:26

TurlocTheRedTurlocTheRed

1,034311

$endgroup$

add a comment | 

$begingroup$

Join segments of length a and length b together on the same line. Call where they join Point J. Call their Midpoint M. Construct a circle centered at M through either end point of a+b. Construct a line perpendicular to a+b through point J. Call where it intersects the circle point Q. The length of QJ is $sqrtab$ as proven elsewhere. These above constructions follow from Euclid's postulates I, III, SAS, and ASA. So the constructions should be valid in neutral geometry.

A related approach. Construct a rectangle having one sidelength a and one side length b. Find a square having the same area as the starting rectangle. The side length of thes square will be $sqrtab$.

share|cite|improve this answer

answered Apr 3 at 16:26

TurlocTheRedTurlocTheRed

1,034311

$endgroup$

share|cite|improve this answer

answered Apr 3 at 16:26

TurlocTheRedTurlocTheRed

1,034311

answered Apr 3 at 16:26

TurlocTheRedTurlocTheRed

1,034311

answered Apr 3 at 16:26

TurlocTheRedTurlocTheRed

1,034311