+0  
 
+3
675
20
avatar+2496 

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

 Jan 4, 2016

Best Answer 

 #14
avatar+98196 
+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

 Jan 4, 2016
edited by CPhill  Jan 4, 2016
edited by CPhill  Jan 4, 2016
 #1
avatar+8613 
0

Ho? Haha.. Sorry . :/

 Jan 4, 2016
 #2
avatar+2496 
0

i found sqrt(299/9)

 Jan 4, 2016
 #3
avatar+2496 
0

Ho Ho where u have been ?)

 Jan 4, 2016
 #4
avatar+8613 
0

Haha, Missed me?!

 Jan 4, 2016
 #5
avatar+2496 
0

yea too much

 Jan 4, 2016
 #6
avatar+8613 
0

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

 Jan 4, 2016
 #7
avatar+2496 
0

Ha :P

 Jan 4, 2016
 #8
avatar+8613 
0

^-^.

 Jan 4, 2016
 #9
avatar+8613 
0

Ive got a headache. :(

 Jan 4, 2016
 #10
avatar+2496 
0

god bless you or how you saying be healthy

 Jan 4, 2016
 #11
avatar+8613 
0

Water & Captin Crunch

 Jan 4, 2016
 #12
avatar+2496 
0

Water & Captin Crunch bless you :)

 Jan 4, 2016
 #13
avatar+8613 
0

haha :)

 Jan 4, 2016
 #14
avatar+98196 
+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+2496 
+5

Thanks CPhill ! :)

 Jan 4, 2016
 #16
avatar+99361 
+5

Nice one Chris  :))

 Jan 4, 2016
 #17
avatar+98196 
+5

Thanks, Melody and Solveit......

 

 

 

cool cool cool

 Jan 4, 2016
 #18
avatar+2496 
+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

 Jan 6, 2016
 #19
avatar+98196 
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

 Jan 6, 2016
edited by CPhill  Jan 6, 2016
 #20
avatar+2496 
+5

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

 Jan 6, 2016

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