Question

If the area of rhombus is 96cm sq and one of its diagonals is 12cm , find its perimeter.

Answer 1

area of rhombus=1/2×d1×d2
96=1/2×12×x
96=6x
96/6=x
16=x
now half both the diagonals
they will be 6 and 8
side will be hypotenuse
the side will be 10
because 6,8 and 10 is a Pythagorean triplet
now multiply it by 4
it will be 40
please mark me as brainliest

Answer 2

Solution:-

• given:-

1)The area of rhombus is 96cm square .

2) one diagonal is 12 cm.

we know all side of rhombus are equal.

let, ac = 1st diagonal = 12 cm and

bd = 2cd diagonal ,

A = Area = 96cm.

=> Area of rhombus = [ac×bd]/2

=> A = [ac×bd]/2

=> 96 = [ 12 × bd ]/2

=> 96×2 = 12 × bd

=> 192 = 12 × bd

=> bd = 192/12

=> bd = 16 cm

•we know,

ac = ao + oc and bd = bo + od

• diagonals of a rhombus bisect each other at an angle of 90° or right angle.

means ao = oc = 12/2 = 6 cm &

bo = od = 16/2 = 8 cm.

in Δ aod

by Pythagoras theorem

=> (ad)² = (ao)² + (od) ²

=> (ad)² = (6)² + (8) ²

=> (ad)² = 36 + 64

=> (ad)² = 100

=> ad = √100

=> ad = 10 cm

•Perimeter of rhombus = 4 × sides

•Perimeter of rhombus = 4 × 10

•Perimeter of rhombus = 40 cm

Hence all side of rhombus are 10 cm and perimeter of rhombus is 40 cm

i hope it helps you .