Question

find the perimeter of the Rhombus whose diagonals are 12 cm and 16 cm respectively​

Answer 1

$\huge\underline\mathfrak{Question-}$

Find the perimeter of the Rhombus whose diagonals are 12 cm and 16 cm respectively.

$\huge\underline\mathfrak{Answer-}$

Given ;-
Diagonals of the Rhombus = 12 cm and 16 cm respectively.

To find the perimeter ;-
Let the perimeter be "p"

Than,
Using the Formula ;

$= > p = 4a$

$= > a = \frac{ \sqrt{ {p}^{2} + {q}^{2} } }{2}$

$= > p = 2 \sqrt{ {p}^{2} + {q}^{2} }$

$= > 2 \times \sqrt{ {12}^{2} +{16}^{2} } = 40 \: cm$

Hence ,
Perimeter of the given Rhombus is 40 cm.

Answer 2

Answer:

40 cm

Step-by-step explanation:

Let ABCD be the Rhombus, AC=12cm

BD=16 cm

Let O be the meeting point of the diagonals

As diagonals of a rhombus bisect each other,

AO=OC=6cm

BO=OD=8cm

The diagonals also meet at right angle to each other,hence using pythagorus theorem,

AB^2=AO^2+BO^2

AB^2=6^2+8^2

AB^2=36+64

AB^2=100

AB=10cm

perimeter=4*(length of one side)

=4*10cm

=40cm