Question

Given: ΔKLM Circle O inscribed in ΔKLM KB = 7, LC = 9, AM = 8 Find: Perimeter of ΔKLM

Answer 1

Add up all the lengths of the segments. kb = bl, ak = cl, am = cm.
the perimeter is 48

Answer 2

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