Lengths of diagonals of a rhombus ABCD are 16 cm and 12 cm. Find the side and perimeter of the rhombus.
Answers
solution :
Given : ABCD is a rhombus .
AC and BD are the diagonals .
AC = 16 cm , BD = 12 cm
Proof :
We know that , diagonals in a rhombus bisect each other
perpendicularly.
Let O is the intersecting point of diagonals AC and BD.
OA = AC/2 = 16/2 = 8 cm
OB = BD/2 = 12/2 = 6 cm
NOw ,
IN Δ AOB , ∠AOB = 90°
By Pythagorean Theorem ,
AB² = OA² + OB²
⇒ AB² = 8² + 6²
= 64 + 36
= 100
∴ AB = √100 = 10 cm
perimeter of the Rhombus = 4 × side
P = 4 × AB
P = 4 × 10 = 40 cm
...
we know that the diagonals of a Rhombus bisect each other perpendicularly the one side of the diagonal is 16 cm and the other is 12 CM so 16 cm divided by 2 equals 8 and 12 divided by 2 equals to 6 square root of 8 + square root of 6 equals to hundred root over of 100 equals to 10 and therefore the length of the side of rhombus is 10 cm