Draw in altitude AD which will pass through center O,
and bisect the base into 2 segments each
units in length,
since the triangle is isoceles. So segment BD is
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.
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:
Simplify the first equation:
Multiply through by LCD=4
Simplify the second equation:
Multiply through by LCD=4
So now our system to solve becomes:
We solve the second equation of the system for 4hČ:
And substitute that for 4hČ in the first equation of the system:
and simplify:
Divide through by 4
Now we solve
for 2h by the principle of square roots,
(we only take positive square roots:
Substitute for 2h in
Square both sides:
Simplify:
, with denominator rationalized, if you like, as
or approximately:
r=6.623632224
Edwin