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# Triangle and circle

+3
884
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+2496

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

Jan 4, 2016

#14
+111325
+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

Jan 4, 2016
edited by CPhill  Jan 4, 2016
edited by CPhill  Jan 4, 2016

#1
+8574
0

Ho? Haha.. Sorry . :/

Jan 4, 2016
#2
+2496
0

i found sqrt(299/9)

Jan 4, 2016
#3
+2496
0

Ho Ho where u have been ?)

Jan 4, 2016
#4
+8574
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Haha, Missed me?!

Jan 4, 2016
#5
+2496
0

yea too much

Jan 4, 2016
#6
+8574
0

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

Jan 4, 2016
#7
+2496
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Ha :P

Jan 4, 2016
#8
+8574
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^-^.

Jan 4, 2016
#9
+8574
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Ive got a headache. :(

Jan 4, 2016
#10
+2496
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god bless you or how you saying be healthy

Jan 4, 2016
#11
+8574
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Water & Captin Crunch

Jan 4, 2016
#12
+2496
0

Water & Captin Crunch bless you :)

Jan 4, 2016
#13
+8574
0

haha :)

Jan 4, 2016
#14
+111325
+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

CPhill Jan 4, 2016
edited by CPhill  Jan 4, 2016
edited by CPhill  Jan 4, 2016
#15
+2496
+5

Thanks CPhill ! :)

Jan 4, 2016
#16
+109519
+5

Nice one Chris  :))

Jan 4, 2016
#17
+111325
+5

Thanks, Melody and Solveit......

Jan 4, 2016
#18
+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
+111325
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.......!!!!

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

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

Jan 6, 2016