Question

the length of the diagonal of the Rhombus are 30 cm and 40 cm find the perimeter of the Rhombus​

Answer 1

Perimeter of Rhombus when diagonals are given = 4*sqrt((d1/2)^2 + (d2/2)^2)

Given:

d1 = 30 cm

d2 = 40 cm

Perimeter of Rhombus = 4*sqrt((30/2)^2 + (40/2)^2)

Perimeter of Rhombus = 4*sqrt((15)^2 + (20)^2)


Perimeter of Rhombus = 4*sqrt(225 + 400)

Perimeter of Rhombus = 4*sqrt(625)

Perimeter of Rhombus = 4*25

Perimeter of Rhombus = 100 cm —> Answer

Answer 2

Answer:

Some of the rhombus' properties:

a) The sides of a rhombus are all congruent. (the same length).

⇒AB=BC=CD=DA

b) The two diagonals are perpendicular, and they bisect each other. This means they cut each other in half.

⇒AO=OC=1/2AC,

and

BO=OD=1/2BD

Now back to our question :

Given that the two diagonals are 30,and 40 ,

⇒AO=30/2=15,

BO=40/2=20,

∠AOB=90∘

From Pythagorean theorem, we know

AB/2=AO/2+BO/2

⇒AB=√15/2+20/2

=10.7745967

SInce AB=BC=CD=DA ,

perimeter of ABCD

=10.7745967×4

=43.0983868