Question

If length of a diagonal of a rhombus is 30 cm and its area is 240 sq cm, find its perimeter.

Answer 1

Let ABCD be the rhombus.

Diagonals AC and BD intersect at point E.

l(AC) = 30 cm           …(i)

and Area of a rhombus = 240 sq. cm           …(ii)

Area of a rhombus = `1/2` (product of diagnols)

⇒ 240 = `1/2` × (30 × DB)

⇒ DB = `(240 xx 2)/30` = 16           …(iii)

Diagonals of a rhombus bisect each other.

∴ `l(AE) = 1/2 l(AC)`

= `1/2 xx 30`

= 15 cm           …(iv)

∴ `l(DE) = 1/2 l(DB)`

= `1/2 xx 16`

= 8 cm           …(v)

In ΔADE,

∠AED = 90°               …[Diagonals of a rhombus are perpendicular to each other]

AE² + DE² = AD²              …[Pythagoras theorem]

⇒ 15² + 8² = AD²              ...[From (iv) and (v)]

⇒ AD² = 225 + 64 = 289

⇒ AD = `sqrt289`             …[Taking square root of both sides]

= 17 cm

Thus, the side of the rhombus = 17 cm

Perimeter of rhombus = 4 × side

= `4 xx 17`

= 68 cm

∴ The perimeter of the rhombus is 68 cm.