A doubt on Circumcenter and Incenter

[dawnvarghese]

Hi Evry Bdy
Cud any body Clarify the relation ship between the InRadius and CirCumRadius if any .....
Also For an Equilateral Triangle is the InRadius Half or 1/3 rd of the Circumradius????
and How can we use this relationship vis a vis with the Centriod and Median length and the ratio in which Centriod divides the Median......
Dawn
Experience is the NAme Evry one gives to their Mistakes.

[CATCALL]

Originally Posted by dawnvarghese

Hi Evry Bdy
Cud any body Clarify the relation ship between the InRadius and CirCumRadius if any .....
Also For an Equilateral Triangle is the InRadius Half or 1/3 rd of the Circumradius????
and How can we use this relationship vis a vis with the Centriod and Median length and the ratio in which Centriod divides the Median......
Dawn
Experience is the NAme Evry one gives to their Mistakes.

in a triangle given sides a,b,c and corresponding angles A,B,C :-

a/sinA = b/sinB = C/sinC = 2R

area = sqrt(s*(s-a)*(s-b)*(s-c)) = r*s = abc/4R , where s=(a+b+c)/2 , R- circumradius , r- inradius

for an equilateral traingle,

height= sqrt(3)*a/2

inradius, r=height/3
circumradius, R= 2*height/3

therfore, R=2r

also in an equilateral triangle , the altitude and the median are the same things .....and also the centroid divides the median in the ratio 2:1, i,e the distance from one vertex to centroid(R) and the distance from centroid to opp side (r) are in the ratio 2:1
and as the height is given by sqrt(3)*a/2....therfore
r=height/3, and
R=2*height/3

hope it clarifies ur doubt

cheers !!!
vishu.

[dawnvarghese]

THanks ViSHU

[raul_v20]

area of circle = r * s = abc/4R
( r - inradius , R - circumradius s- semiperimeter)


thus
R = abc / 4(r*s)

[reachmonil]

U could have asked the doubt on any of the existing thread, y start a new thread?