Let ABCD be the rhombus.
Diagonals AC and BD intersect at point E.
l(AC) = 30 cm …(i) and
A(ABCD) = 240 sq. cm .. .(ii)
Area of the rhombus = (1/2) × product of lengths of diagonal
∴ 240 = (1/2) × l(AC) x l(BD) …[From (ii)]
∴ 240 = (1/2) × 30 × l(BD) …[From (i)]
∴ l(BD) = (240 x 2/30)
∴ l(BD) = 8 × 2 = 16 cm …(iii)
Diagonals of a rhombus bisect each other.
∴ l(AE) = (1/2)l (AC)
= (1/2) × 30 … [From (i)]
= 15 cm …(iv) and
l(DE) = (1/2)l (BD)
= (1/2) × 16 = 8 cm
In ∆ADE, m∠AED = 90° …[Diagonals of a rhombus are perpendicular to each other]
∴ [l(AD)]2 = [l(AE)]2 + [l(DE)]2 …[Pythagoras theorem]
∴ l(AD)2 = (15)2 + (8)2 … [From (iv) and (v)]
= 225 + 64
∴ l(AD)² = 289
∴ l(AD) = √289 …[Taking square root of both sides]
∴ l(AD) = 17 cm
Perimeter of rhombus = 4 × side
= 4 × l(AD) = 4 × 17 = 68 cm
∴ The perimeter of the rhombus is 68 cm.
