Question

In rhombus ABCD, AC= 12, BD = 16,
find its perimeter​

Answer 1

Step-by-step explanation:

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

...

Answer 2

Step-by-step explanation:

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 A AOB, ZAOB = 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 x 10 = 40 cm