If P is the incenter of ΔJKL, then we have,
PN = PM = PO
To find PJ first we need to calculate the value of x
Since,
PN = 22
So,
PM = 22
Now, in ΔPJM we have,
PJ^2 = PM^2 + MJ^2
PJ^2= (22)^2 +(2x+3)^2...(1)
Also, In ΔPJO we have,
PJ^2 = PO^2 + OJ^2
PJ^2 = (22)^2 (5x - 45)^2...(2)
From equation (1) and (2) we obtain,
(5x - 45)^2+(22)^2=(2x+3)^2 +(22)^2
25x^2 — 450x + 2025 = 4x^2 + 12x + 9
25x^2 — 4x^2— 450x — 12x + 2025 — 9 = 0
21x^2 — 462x + 2016 = 0
x^2 - 22x+96 = 0
x^2 — 16x — 6x + 96 = 0
x(x - 16) -6(x - 16) = 0
(x - 6)(x - 16) = 0
Either x=6 or x=16
At x = 6
JO = 5 x 6 - 45
= 30 - 45
= -15
Which cannot be possible because edge length cannot be negative
So the value of x = 16
Hence, the length of PJ is
PJ^2 = 484 + (80 - 45)^2
PJ^2 = 484 + 1225
PJ^2 = 1709
PJ = √1709
PJ = 41.33
Therefore, the length of PJ is 41.3 units