A circle is inscribed in ΔABC touching AB at P, BC at Q and AC at R.  If AR = 7 cm, RC = 9 cm and AB = 15 cm, then the perimeter of ΔABC is?

30 45 48 60

48

AP = AR, BP = BQ, CR = CQ AB = 15, AP = 7, hence BP = AB - AP = 15 - 7 = 8 = BQ Perimeter of ΔABC AB + BC + CA AP + PB +BQ + QC + CR + RA 7 + 8 + 8 + 9 + 9 + 7 = 48 cm