Question

Diagonals of a rhombus are 20 cm and 21 cm respectively, then find the side of rhombus and its perimeter.

Answer 1

Let `square`ABCD be the rhombus.

AC = 20 cm, BD = 21 cm

AO = `1/2` AC        ...[Diagonals of a rhombus bisect each other]

= `1/2 xx 20`

= 10 cm       ...(i)

BO = `1/2` BD         ...[Diagonals of a rhombus bisect each other]

= `1/2 xx21`

= `21/2`

= 10.5 cm      ...(ii)

In ∆AOB,

∠AOB = 90°       ...[Diagonals of a rhombus are perpendicular to each other]

By Pythagoras theorem,

∴ AB² = AO² + BO²

∴ AB² = 10² + 10.5²

∴ AB² = 100 + 110.25

∴ AB² = 210.25

∴ AB = `sqrt(210.25)`

∴ AB = 14.5 cm

Perimeter of `square`ABCD = 4 × AB

= 4 × 14.5

= 58 cm