+0  
 
+3
638
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+94545 
+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+94545 
+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+95350 
+5

Nice one Chris  :))

 Jan 4, 2016
 #17
avatar+94545 
+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+94545 
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

27 Online Users

avatar
avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.