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diagonals of a rhombus are perpendicular bisectors of each other so AO=OC=6cm and DO=OB=8cm in triangle AOB by pythagoras theorm AB^2=AO^2+OB^2 AB^2=6^2+8^2 AB^2=36+64 AB^2=100 AB=10 but AB is side of rhombus so side of rhombus=10cm perimeter=4 x side =4x10 =40cm answer is (C) 40 cm pls refer to the file for figure

Let ABCD be the rhombus. And the diagonals AC=12cm and BD= 16 cm Let the diagonals intersects at O. We know the diagonals of rhombus are perpendicular to each other and bisect each other. Hence triangle BOC is a right angle triangle with side OB =12/2 = 6 cm and side OC =16/2 = 8cm Hence we have BC= √(6²+8²)= 10 cm Perimeter of Rhombus= 4*10 = 40 cm