Question

the area rhombus is 96 square metre if one of its diagonal is 16 m find its perimeter. plz answer it fast plz plz

Answer 1

A = pq÷2          ∵( p=diagonal,q=diagonal)
96=16q÷2
96 =8q
12 =q

∴ one side is =10      by pythagoras theorem
perimeter of rhombus= 4a                (a=side)
                                    = 4×10
                                     = 40

Answer 2

`Area of rhombus =  1 / 2 x d1 x d2
 96 = 1 / 2 x 16 x d2
 96 = 8 x d2
 96 / 8 = d2
12 = d2

A rhombus makes 90 when its diagonals intersect with each other
So the triangle AOD is right angled triangle 
the diagonal of Rhombus bisect each other
Diagonal BD (longer one) = 16 cm
Diagonal AC (shorter one) = 12 cm
SO , OD = 1 / 2 x 16 = 8 cm
AO =  1/ 2 x 12 = 6 cm
So , (AD)^2 = (DO)^2 + (AO)^2
      (AD)^2 = (8)^2 + (6)^2
      (AD)^2 = 64 + 36
       (AD)^2 = 100
        (AD) = root 100
         AD = 10
Perimeter of rhombus = 4 x side
                                    = 4 x 10
                                     = 40 cm

Hope this helps you 
Please mark this answer as brainest