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Inscribed circle in right-angled triangle

Calculate incircle radius.Two circles inside a right angled triangle!How to find the radius of small inscribed circle?Inradius in Right angled triangles.Find length of triangle as well as radius of inscribed circleRadii of inscribed and circumscribed circles in right-angled triangleFind the length of the sides of an equilateral triangle inscribed in a circle of radius $6$cm.Four circles tangent to each other and an equilateral triangleA circle of radius $r$ is inscribed into a triangle.Find the minimal value the hypotenuse of a right triangle whose radius of inscribed circle is $r$

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5

1

$begingroup$

In right-angled $triangle ABC$ with catheti $a = 11,textcm, b=7,textcm$ a circle has been inscribed. Find the radius and altitude from $C$ to the hypotenuse.

I found that the hypotenuse is $c = sqrt170$ and the radius is $r = fraca+b-c2=frac 18 - sqrt1702$. I think that the altitude $CH=$ the sum of radii of circles inscribed in $triangle ABC, triangle AHC, triangle HBC$ but I don't understand how I should calculate them. Thank you in advance!

geometry euclidean-geometry triangles circles

share|cite|improve this question

edited Apr 21 at 16:13

TheSimpliFire

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asked Apr 21 at 13:07

Nikol DimitrovaNikol Dimitrova

31099 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By the height do you mean the altitude of the triangle through the right angle?
  $endgroup$
  – TheSimpliFire
  Apr 21 at 13:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @TheSimpleFire, yes.
  $endgroup$
  – Nikol Dimitrova
  Apr 21 at 13:29
  

  
  
  
  

add a comment
 | 

5

1

$begingroup$

In right-angled $triangle ABC$ with catheti $a = 11,textcm, b=7,textcm$ a circle has been inscribed. Find the radius and altitude from $C$ to the hypotenuse.

I found that the hypotenuse is $c = sqrt170$ and the radius is $r = fraca+b-c2=frac 18 - sqrt1702$. I think that the altitude $CH=$ the sum of radii of circles inscribed in $triangle ABC, triangle AHC, triangle HBC$ but I don't understand how I should calculate them. Thank you in advance!

geometry euclidean-geometry triangles circles

share|cite|improve this question

edited Apr 21 at 16:13

TheSimpliFire

15k77 gold badges3333 silver badges7474 bronze badges

asked Apr 21 at 13:07

Nikol DimitrovaNikol Dimitrova

31099 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By the height do you mean the altitude of the triangle through the right angle?
  $endgroup$
  – TheSimpliFire
  Apr 21 at 13:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @TheSimpleFire, yes.
  $endgroup$
  – Nikol Dimitrova
  Apr 21 at 13:29
  

  
  
  
  

add a comment
 | 

$begingroup$

In right-angled $triangle ABC$ with catheti $a = 11,textcm, b=7,textcm$ a circle has been inscribed. Find the radius and altitude from $C$ to the hypotenuse.

I found that the hypotenuse is $c = sqrt170$ and the radius is $r = fraca+b-c2=frac 18 - sqrt1702$. I think that the altitude $CH=$ the sum of radii of circles inscribed in $triangle ABC, triangle AHC, triangle HBC$ but I don't understand how I should calculate them. Thank you in advance!

geometry euclidean-geometry triangles circles

share|cite|improve this question

edited Apr 21 at 16:13

TheSimpliFire

15k77 gold badges3333 silver badges7474 bronze badges

asked Apr 21 at 13:07

Nikol DimitrovaNikol Dimitrova

31099 bronze badges

$endgroup$

share|cite|improve this question

edited Apr 21 at 16:13

TheSimpliFire

15k77 gold badges3333 silver badges7474 bronze badges

asked Apr 21 at 13:07

Nikol DimitrovaNikol Dimitrova

31099 bronze badges

share|cite|improve this question

edited Apr 21 at 16:13

TheSimpliFire

15k77 gold badges3333 silver badges7474 bronze badges

asked Apr 21 at 13:07

Nikol DimitrovaNikol Dimitrova

31099 bronze badges

edited Apr 21 at 16:13

TheSimpliFire

15k77 gold badges3333 silver badges7474 bronze badges

edited Apr 21 at 16:13

TheSimpliFire

15k77 gold badges3333 silver badges7474 bronze badges

asked Apr 21 at 13:07

Nikol DimitrovaNikol Dimitrova

31099 bronze badges

asked Apr 21 at 13:07

Nikol DimitrovaNikol Dimitrova

31099 bronze badges

2 Answers
2

active

oldest

votes

5

$begingroup$

Let $h$ be the altitude of the triangle, whose line splits $sqrt170$ into $x$ and $sqrt170-x$. We know that the total area of the triangle is $$frac12cdot7cdot11=frac12h(sqrt170-x)+frac12hximpliestextaltitude=h=frac77sqrt170.$$

share|cite|improve this answer

edited Apr 21 at 13:56

answered Apr 21 at 13:32

TheSimpliFireTheSimpliFire

15k77 gold badges3333 silver badges7474 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @NikolDimitrova Of course, that was a silly typo
  $endgroup$
  – TheSimpliFire
  Apr 21 at 13:56
  

  
  
  
  

add a comment
 | 

4

$begingroup$

I assume you mean that $CH$ is the altitude from $C$ to the hypotenuse.

Since $triangle ABC$ is a right triangle, and since
$triangle AHC$ and $triangle HBC$ each share one angle with $triangle ABC$,
the three triangles are similar. We have
$$
fracCHAC = fracBCAB
$$
and (equivalently)
$$
fracCHBC = fracACAB.
$$

The radii of the three incircles are proportional to $AB,$ $AC,$ and $BC,$
so their sum is
beginalign
fraca+b-c2 + fraca(a+b-c)2c + fracb(a+b-c)2c
&= frac(a+b+c)(a+b-c)2c
\
&= frac(a+b)^2 - c^22c \
&= fracabc,
endalign
which is indeed equal to the altitude.

share|cite|improve this answer

edited Apr 21 at 14:20

answered Apr 21 at 14:04

David KDavid K

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$endgroup$

add a comment
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5

$begingroup$

Let $h$ be the altitude of the triangle, whose line splits $sqrt170$ into $x$ and $sqrt170-x$. We know that the total area of the triangle is $$frac12cdot7cdot11=frac12h(sqrt170-x)+frac12hximpliestextaltitude=h=frac77sqrt170.$$

share|cite|improve this answer

edited Apr 21 at 13:56

answered Apr 21 at 13:32

TheSimpliFireTheSimpliFire

15k77 gold badges3333 silver badges7474 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @NikolDimitrova Of course, that was a silly typo
  $endgroup$
  – TheSimpliFire
  Apr 21 at 13:56
  

  
  
  
  

add a comment
 | 

5

$begingroup$

Let $h$ be the altitude of the triangle, whose line splits $sqrt170$ into $x$ and $sqrt170-x$. We know that the total area of the triangle is $$frac12cdot7cdot11=frac12h(sqrt170-x)+frac12hximpliestextaltitude=h=frac77sqrt170.$$

share|cite|improve this answer

edited Apr 21 at 13:56

answered Apr 21 at 13:32

TheSimpliFireTheSimpliFire

15k77 gold badges3333 silver badges7474 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @NikolDimitrova Of course, that was a silly typo
  $endgroup$
  – TheSimpliFire
  Apr 21 at 13:56
  

  
  
  
  

add a comment
 | 

$begingroup$

Let $h$ be the altitude of the triangle, whose line splits $sqrt170$ into $x$ and $sqrt170-x$. We know that the total area of the triangle is $$frac12cdot7cdot11=frac12h(sqrt170-x)+frac12hximpliestextaltitude=h=frac77sqrt170.$$

share|cite|improve this answer

edited Apr 21 at 13:56

answered Apr 21 at 13:32

TheSimpliFireTheSimpliFire

15k77 gold badges3333 silver badges7474 bronze badges

$endgroup$

share|cite|improve this answer

edited Apr 21 at 13:56

answered Apr 21 at 13:32

TheSimpliFireTheSimpliFire

15k77 gold badges3333 silver badges7474 bronze badges

answered Apr 21 at 13:32

TheSimpliFireTheSimpliFire

15k77 gold badges3333 silver badges7474 bronze badges

answered Apr 21 at 13:32

TheSimpliFireTheSimpliFire

15k77 gold badges3333 silver badges7474 bronze badges

4

$begingroup$

I assume you mean that $CH$ is the altitude from $C$ to the hypotenuse.

Since $triangle ABC$ is a right triangle, and since
$triangle AHC$ and $triangle HBC$ each share one angle with $triangle ABC$,
the three triangles are similar. We have
$$
fracCHAC = fracBCAB
$$
and (equivalently)
$$
fracCHBC = fracACAB.
$$

The radii of the three incircles are proportional to $AB,$ $AC,$ and $BC,$
so their sum is
beginalign
fraca+b-c2 + fraca(a+b-c)2c + fracb(a+b-c)2c
&= frac(a+b+c)(a+b-c)2c
\
&= frac(a+b)^2 - c^22c \
&= fracabc,
endalign
which is indeed equal to the altitude.

share|cite|improve this answer

edited Apr 21 at 14:20

answered Apr 21 at 14:04

David KDavid K

60.6k44 gold badges4848 silver badges137137 bronze badges

$endgroup$

add a comment
 | 

4

$begingroup$

I assume you mean that $CH$ is the altitude from $C$ to the hypotenuse.

Since $triangle ABC$ is a right triangle, and since
$triangle AHC$ and $triangle HBC$ each share one angle with $triangle ABC$,
the three triangles are similar. We have
$$
fracCHAC = fracBCAB
$$
and (equivalently)
$$
fracCHBC = fracACAB.
$$

The radii of the three incircles are proportional to $AB,$ $AC,$ and $BC,$
so their sum is
beginalign
fraca+b-c2 + fraca(a+b-c)2c + fracb(a+b-c)2c
&= frac(a+b+c)(a+b-c)2c
\
&= frac(a+b)^2 - c^22c \
&= fracabc,
endalign
which is indeed equal to the altitude.

share|cite|improve this answer

edited Apr 21 at 14:20

answered Apr 21 at 14:04

David KDavid K

60.6k44 gold badges4848 silver badges137137 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

I assume you mean that $CH$ is the altitude from $C$ to the hypotenuse.

Since $triangle ABC$ is a right triangle, and since
$triangle AHC$ and $triangle HBC$ each share one angle with $triangle ABC$,
the three triangles are similar. We have
$$
fracCHAC = fracBCAB
$$
and (equivalently)
$$
fracCHBC = fracACAB.
$$

The radii of the three incircles are proportional to $AB,$ $AC,$ and $BC,$
so their sum is
beginalign
fraca+b-c2 + fraca(a+b-c)2c + fracb(a+b-c)2c
&= frac(a+b+c)(a+b-c)2c
\
&= frac(a+b)^2 - c^22c \
&= fracabc,
endalign
which is indeed equal to the altitude.

share|cite|improve this answer

edited Apr 21 at 14:20

answered Apr 21 at 14:04

David KDavid K

60.6k44 gold badges4848 silver badges137137 bronze badges

$endgroup$

share|cite|improve this answer

edited Apr 21 at 14:20

answered Apr 21 at 14:04

David KDavid K

60.6k44 gold badges4848 silver badges137137 bronze badges

answered Apr 21 at 14:04

David KDavid K

60.6k44 gold badges4848 silver badges137137 bronze badges

answered Apr 21 at 14:04

David KDavid K

60.6k44 gold badges4848 silver badges137137 bronze badges