Question

area of a rhombus is 28 cm and one of its diagonal is 4 cm find its perimeter​

Answer 1

Answer:

4 root53

Step-by-step explanation:

area of rhombus = product of diagonals /2

thus 28= pq /2 (p and q are two diagonals)

28 = p×4/2

28=2p

p = 14

now the diagonals bisect at right angles

thus the side of rhombus would be the hypotenuse

thus 2²+ 7²= h²

or 4+49 =h²

or h = sq rt53

thus perimeter = 4 root53

Answer 2

Answer:

As given

The area of rhombus =28cm^2

One of its diagonal =4cm

Let other diagonal be x

Then

1/2 × product of diagonal=28

1/2 × 4 × x=28

or 2x=28

or x=28/2

or x=14cm

Let ABCD be a rhombus

In triangle AOB

OA= 1/2 × AC = 1/2 × 4= 2 cm

OB =1/2 × BD =1/2 × 14 =7 cm

As we know

Diagonals of rhombus bisect each other at 90°

By pythagoras property

AB^2 =OA^2 + OB^2

AB = √2^2 + 7^2

AB = √4 +49

AB =√53cm

Then

Its perimeter =4 × side = 4 × √53 = 4√53 cm