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

Solveit
Jan 4, 2016

#14**+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

#14**+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

CPhill
Jan 4, 2016

#18**+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