If the area of a rhombus is 96cm^2 and one of its diagonal is 16cm, find its perimeter.
Answers
Answer:
64
Step-by-step explanation:
16+16+16+16
Solution:-
• given:-
1)The area of rhombus is 96cm square .
2) one diagonal is 16 cm.
we know all side of rhombus are equal.
let, bd = 1st diagonal = 16 cm and
ac = 2cd diagonal ,
A = Area = 96cm.
=> Area of rhombus = [ac×bd]/2
=> A = [ac×bd]/2
=> 96 = [ ac × 16 ]/2
=> 96×2 = ac × 16
=> 192 = ac × 16
=> bd = 192/16
=> bd = 12 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 .