ΔKLM
Circle O inscribed in ΔKLM
KM, ML and KL are tangents to the circle at point A,C and B respectively
KB = 7, LC = 9, AM = 8
Tangent drawn from the external point to the circle are equal
So KB = KA =7
Similarly, LC = BL =9 and AM = MC= 8
KL = KB + BL
KL = 7 +9 =16
KM =KA + AM
KM = 7 + 8 = 15
LM = LC +MC
LM = 9 +8=17
So perimeter of triangle KLM is KM + LM + KL = 15 + 17 +16 = 48