Question 479621
<pre>
{{{drawing(400,400,-8,8,-2,14,

locate(-.15,13.3,A), locate(-2.5,0,B), locate(2.5,0,C),
triangle(-2.5,0,2.5,0,0,12.75735082), circle(0,6.133718598,6.623632224),
circle(0,6.133718598,.05), locate(.2,6.4,O), locate(-2.2,6.4,13),
locate(1.5,6.4,13), locate(0,.6,5)  



 )}}}

Draw in altitude AD which will pass through center O,
and bisect the base into 2 segments each {{{5/2}}} units in length,
since the triangle is isoceles.  So segment BD is {{{5/2}}} units.
We will also draw in radius OB, and label it r.  Radius
OA also has length r. We will label the length of OD as h. 


{{{drawing(400,400,-8,8,-2,14,
triangle(-2.5,0,2.5,0,0,12.75735082), circle(0,6.133718598,6.623632224),
red(line(-2.5,0,0,6.133718598),line(0,sqrt(651)/2)),
locate(-1.1,3.3,r), locate(.2,6.4,O), locate(.1,.6,D),
locate(-2.2,6.4,13), locate(-1.4,1.4,5/2),locate(.2,9.4,r),

locate(.2,3,h),



locate(-.15,13.3,A), locate(-2.5,0,B), locate(2.5,0,C)


 )}}}


ADB and ODB are both right triangles, so using the
Pythagorean theorem

BDČ + ADČ = ABČ and BDČ + ODČ = OBČ

Since AD = OA + OD, we have:

BDČ + (OA + OD)Č = ABČ and BDČ + ODČ = OBČ

In terms of the lengths of the sides we have this
system of equations to solve:

{{{system((5/2)^2 + (r + h)^2 = 13^2, (5/2)^2 + h^2 = r^2)}}}

Simplify the first equation:

{{{(5/2)^2 + (r + h)^2 = 13^2}}}
{{{25/4+r^2+2rh+h^2=169}}}
Multiply through by LCD=4
{{{25+4r^2+8rh+4h^2=676}}}
{{{4r^2+8hr+4h^2=651}}}

Simplify the second equation:

{{{(5/2)^2 + h^2 = r^2}}}
{{{25/4+h^2=r^2}}}
Multiply through by LCD=4
{{{25+4h^2=4r^2}}}

So now our system to solve becomes:

{{{system(4r^2+8hr+4h^2=651, 25+4h^2=4r^2)}}}

We solve the second equation of the system for 4hČ:

{{{25+4h^2=4r^2)}}}
{{{4h^2=4r^2-25}}}

And substitute that for 4hČ in the first equation of the system:

{{{4r^2+8hr+4h^2=651}}}
{{{4r^2+8hr+4r^2-25=651}}}
and simplify:
{{{8r^2+8hr-25=651}}}
{{{8r^2+8hr=676}}}
Divide through by 4
{{{2r^2+2hr=169}}}

Now we solve

{{{4h^2=4r^2-25}}} for 2h by the principle of square roots,
(we only take positive square roots:

{{{2h = sqrt(4r^2-25)}}}

Substitute for 2h in

{{{2r^2+2hr=169}}}
{{{2r^2+(sqrt(4r^2-25))r=169}}}
{{{(sqrt(4r^2-25))r=169-2r^2}}}
Square both sides:
{{{(4r^2-25)r^2=(169-2r^2)^2}}}
{{{4r^4-25r^2=169^2-676r^2 + 4r^4}}}
Simplify:
{{{651r^2=169^2}}}
{{{r^2=169^2/651}}}

{{{r=sqrt(169^2/651)}}}

{{{r=169/sqrt(651)}}}, with denominator rationalized, if you like, as

{{{r=(169sqrt(651))/651}}}
or approximately:
r=6.623632224

Edwin</pre>