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