Question

ABCD is a rhombus, Ao= 4cm and Do= 3cm find the perimeter of the rhombus​

Answer 1

Step-by-step explanation:

ABCD is rhombus

so, AC and BD are the diagonal of rhombus which bisect each other at 90°

hence, in right triangle AOB

$ab = \sqrt{ {ao}^{2} + {bo}^{2} }$

AB= sqrt 4^2+3^2

so, AB=5cm

so perimeter of rhombus is 20cm(AB=BC=CD=DA)

Answer 2

Answer:

and DO=3cm.

We know that diagonals of a rhombus are perpendicular to each other.

⇒∠AOD=90

∘

Now, in rightangle triangle AOD

AD

2

=AO

2

+OD

2

by pythagoras theorem

⇒AD

2

=4

2

+3

2

=16+9=25sq.cm

⇒AD=

25

cm=5cm

Here perimeter of given rhombus=4×side=4×AD=4×5=20cm.