Finding the radius of incircle of a triangle The question is -


AB and AC are the circular arcs of center O' and O of radius 5 units respectively. OO' = 6 unit. In $\Delta$ABC, AB = AC and a circle is inscribed in the triangle. The radius of the circle can be written as $\frac{a}{\sqrt b +1}$ where a, b are integers, a+b=?

My try:
At first,

*

*AO = 5

*AO' = 5

*OC = 5

*O'B = 5

*OO' = 6

*OB = OO' - O'B = 6-5 = 1

*O'C = 1

*BC = OO' - (OB+O'C) = 6-2 = 4

So,BC = 4,

*

*In $\Delta$AOO' ,(using Law of cosines) AO$^2$ = AO'$^2$ + OO'$^2$ - 2AO'$*$OO'$*$$\cos$$\angle$AO'O
So $\angle$AO'O = 53.13°


*$\Delta$AO'B is Isosceles , i know AO' = 5, BO' = 5 and $\angle$AO'B = 53.13°
So base AB = 2×AO'×$\sin$(0.5×$\angle$AO'B) = 4.4721.
So AC = 4.4721
Now i know the length of AB, AC and BC, and i can find the radius of incircle of $\Delta$ABC
so the radius= $\frac{\sqrt{s(s-AB)(s-AC)(s-BC)}}{s}$
$s = (AB+AC+BC)/2 = (4.4721+4.4721+4)/2 = 6.4721.$
Also
${\sqrt{s(s-AB)(s-AC)(s-BC)}} = 8$ but the value of s is 6.4721, how can I express $\frac{8}{6.4721} $ as $\frac{a}{\sqrt b +1}$ ??

$AO = AO' = 5$ so perp from $A$ to $OO'$ will meet at the midpoint of $OO'$, say $H$. Then, $OH = O'H = 3$
So, $AH = \sqrt{AO^2 - OH^2} = \sqrt{5^2 - 3^2} = 4$
Now, $BH = BC/2 = 2$ and $AB = AC = \sqrt{AH^2 + BH^2} = \sqrt{4^2 + 2^2} = 2 \sqrt5$
Sub-perimeter of $\triangle ABC$,
$s = \frac 12 (4 + 2 \cdot 2 \sqrt5) = 2 (1 + \sqrt5)$
Area of $~\triangle ABC, [ABC] = \frac 12 \cdot BC \cdot AH = 8$
If $r$ is the inradius of $\triangle ABC$, we know $[ABC] = r\cdot s$ $$\implies r = \dfrac{4}{1 + \sqrt5}$$
$ \therefore ~a + b = 9$
