Question

If the area of rhombous is 48 cm sq and one of its digonal is 12 cm , find the side of rhombous

Answer 1

area of rhombus =i/2 d1 d2

here A=48cm sq

d1=12

1/2*12*d2=48

6d2=48

d2=48/6=8

in a rhombus diagonals perpendicularly bisect each other

by applying pythagoras theorem ,

4^2+6^2=s^2

16+36=s^2

s=square root of 52

s=2*square root of 13

hope u like it

Answer 2

area of rohmbus = 1/2 ( product of diagonals)
48 cm^2 = 1/2 ( 12 x d )
let another diagonal is d
now ,
d = 96/12 = 8 cm

we know , point of intersection of diagonal of rohmbus divide two equal part of both diagonals .
and intersection point contains right angle in it .
let O is the point of intersection and AO is the half of first diagonal e.g
AO = 12/2 = 6 cm
and BO is the half of second diagonal
e.g BO = 8/2 = 4

and according to Pythagoras theorem ,
triangle AOB is right angle triangle.
so,
AB^2 =AO^2 +BO^2
AB^2 =6^2 + 4^2 =52

AB = 2 root13cm= 2 x 3.6
= 7.2 cm

hence side length of rohmbus is 7.2 cm