In a triangle ABC, AB = 12, BC = 18, CA = 25. A semicircle is inscribed in D ABC such that the diameter of the semicircle lies on . If O is the centre of teh circle, then the length AO = ___

Oh, the diameter of the semicircle lies on the line AC

I think the answer is 10 but my solution is very complex.If correct,I will post it.

Here's my solution.It seems very complex.

Let us assume A as the origin and AC as +ve x-axis.Then A=(0,0) and C=(25,0)

Let B=(x,y) then AB=12 and BC=18x²+y²=144 and (x-25)²+y²=324.

So,we get,x=89/10 and y=

Now,since centre lies on AC,Let O=(k,0) be the centre of semicircle.Then we know that AB and AC are tangents to semicircle and so,d=R,the radius.

Slope of AB is rt6479/89 and that of AC is rt6479/89-250=-rt6479/161.

So,Their eqns. are rt6479x-89y=0 and rt6479x+161y=25rt6479.

Using the condition of tangency,we have,

Irt6479kI/120=R and Irt6479k-25rt6479I/180=R

Dividing these eqns.,we get,IkI/Ik-25I=2/3

k/25-k=2/3 as k<25

k=10.

Hence,the length AO=10units. 

allamraju had asked for a simpler method. It would help to refer to your figure

You have to note that BC and BA are the common tangents to the constructed semi-circle. So, the line joining B and the centre O of the semicircle is actually the  bisector of .

So, now using the angle-bisector theorem,

Oh..It appears very simple now.But I didn't get the right idea then.The only thing for which I could be happy is that I got the right answer after doing hefty calculations.

What guided me was knowing that hpudipeddi was in Xth Class

The semi-circle's diameter lies on AC