Answers

2014-12-22T17:27:19+05:30
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 
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2014-12-27T18:52:50+05:30
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
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