+0  
 
+3
339
20
avatar+2493 

circle in isosles triangle with length 18;18;10 ho to find radius of circle

Solveit  Jan 4, 2016

Best Answer 

 #14
avatar+78744 
+15

This one is a little difficult, Solveit

 

Look at the following illustration........

 

On CB, draw BE = BF = 5

 

And  arccos(CBA)  = 5/18 

 

And we can bisect this angle and form kite BFDE........where  triangles BFD and BDE are congruent right triangles by  SAS.......note that DF = DE   and  DEB and DFB are right angles, so any inscribed circle will have a radius that is tangent to CB  and AB at  E and F

 

And (1/2) of  angle CBA  = (1/2)arccos(5/18)

 

The x coordinate for the center of the circle = 0....   and the y coordinate lies on the bisector (BD) of base angle CBA  at  point D and is given by :

 

tan [ (1/2)arccos(5/18)]  = y / 5

 

5 tan [ (1/2)arccos(5/18)]  =  y  =  DF   = 3.75904705778

 

And this is the approximate radius of the inscribed circle......and DF  = DE  = 3.75904705778

 

 

 

 

cool cool cool

CPhill  Jan 4, 2016
edited by CPhill  Jan 4, 2016
edited by CPhill  Jan 4, 2016
Sort: 

20+0 Answers

 #1
avatar+8621 
0

Ho? Haha.. Sorry . :/

Hayley1  Jan 4, 2016
 #2
avatar+2493 
0

i found sqrt(299/9)

Solveit  Jan 4, 2016
 #3
avatar+2493 
0

Ho Ho where u have been ?)

Solveit  Jan 4, 2016
 #4
avatar+8621 
0

Haha, Missed me?!

Hayley1  Jan 4, 2016
 #5
avatar+2493 
0

yea too much

Solveit  Jan 4, 2016
 #6
avatar+8621 
0

Haha, no need to cry, I'm here!

Hayley1  Jan 4, 2016
 #7
avatar+2493 
0

Ha :P

Solveit  Jan 4, 2016
 #8
avatar+8621 
0

^-^.

Hayley1  Jan 4, 2016
 #9
avatar+8621 
0

Ive got a headache. :(

Hayley1  Jan 4, 2016
 #10
avatar+2493 
0

god bless you or how you saying be healthy

Solveit  Jan 4, 2016
 #11
avatar+8621 
0

Water & Captin Crunch

Hayley1  Jan 4, 2016
 #12
avatar+2493 
0

Water & Captin Crunch bless you :)

Solveit  Jan 4, 2016
 #13
avatar+8621 
0

haha :)

Hayley1  Jan 4, 2016
 #14
avatar+78744 
+15
Best Answer

This one is a little difficult, Solveit

 

Look at the following illustration........

 

On CB, draw BE = BF = 5

 

And  arccos(CBA)  = 5/18 

 

And we can bisect this angle and form kite BFDE........where  triangles BFD and BDE are congruent right triangles by  SAS.......note that DF = DE   and  DEB and DFB are right angles, so any inscribed circle will have a radius that is tangent to CB  and AB at  E and F

 

And (1/2) of  angle CBA  = (1/2)arccos(5/18)

 

The x coordinate for the center of the circle = 0....   and the y coordinate lies on the bisector (BD) of base angle CBA  at  point D and is given by :

 

tan [ (1/2)arccos(5/18)]  = y / 5

 

5 tan [ (1/2)arccos(5/18)]  =  y  =  DF   = 3.75904705778

 

And this is the approximate radius of the inscribed circle......and DF  = DE  = 3.75904705778

 

 

 

 

cool cool cool

CPhill  Jan 4, 2016
edited by CPhill  Jan 4, 2016
edited by CPhill  Jan 4, 2016
 #15
avatar+2493 
+5

Thanks CPhill ! :)

Solveit  Jan 4, 2016
 #16
avatar+91045 
+5

Nice one Chris  :))

Melody  Jan 4, 2016
 #17
avatar+78744 
+5

Thanks, Melody and Solveit......

 

 

 

cool cool cool

CPhill  Jan 4, 2016
 #18
avatar+2493 
+5

i found way esear in the internet they devided (area)triangle into three triangles and their height will be radius of circles

r*18/2+r*18/2+r*10/2=sqrt(299)*10/2

23r=sqrt(299)*10/2

r=3.7590470577805614

Solveit  Jan 6, 2016
 #19
avatar+78744 
0

Thanks, Solveit......that IS easier.......!!!

 

That's the thing about some problems - the  simplest way is usually the best, but there may be other [more complex] methods, too......looks like I chose one of those.......!!!!

 

 

cool cool cool

CPhill  Jan 6, 2016
edited by CPhill  Jan 6, 2016
 #20
avatar+2493 
+5

don t worry you are not alone my teacher did the same :)

Solveit  Jan 6, 2016

7 Online Users

avatar
avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details