What is the radius of the circle inscribed in triangle $ABC$ if $AB = 22, AC=12,$ and $BC=14$? Express your answer in simplest radical form.

How about presenting your question properly, without the unrendered LaTex.

Let A be the point (0,0)

Let B be the point (22,0)

the line equidistance from thes points is x=11,  the centre will line on this line

now AC=12 and BC=14   Find C using locus

\(x^2+y^2=144\qquad (x-22)^2+y^2=196\\ subtract\\ (x-22)^2-x^2=196-144\\ -44x+484=52\\ x=9.\bar{81}\\ x^2=96\frac{48}{121}\\ x^2+y^2=144\\ y^2=144-96\frac{48}{121}\\ y^2=47\frac{73}{121}\\ y^2=\frac{5760}{121}\\ y^2=\frac{2^7*3^2*5}{121}\\ y^2=\frac{2^6*3^2*5*2}{121}\\ y=\frac{24\sqrt{10}}{11}\\ C(\frac{108}{11}, \frac{24\sqrt{10}}{11})\)

The centre will be the point on the line x=11 that is equidistant to  A(0,0), B(22,0)  and  C  let the point be Q(11,y)

I think I will let you take over from there.

Here is the pic

LaTex:

x^2+y^2=144\qquad (x-22)^2+y^2=196\\
subtract\\
(x-22)^2-x^2=196-144\\
-44x+484=52\\
x=9.\bar{81}\\
x^2=96\frac{48}{121}\\
x^2+y^2=144\\
y^2=144-96\frac{48}{121}\\
y^2=47\frac{73}{121}\\
y^2=\frac{5760}{121}\\
y^2=\frac{2^7*3^2*5}{121}\\
y^2=\frac{2^6*3^2*5*2}{121}\\
y=\frac{24\sqrt{10}}{11}\\
C(\frac{108}{11}, \frac{24\sqrt{10}}{11})