Let ABCD be the rhombus. Diagonals AC and BD intersect at point E.
l(AC) = 48 cm …(i)
l(AE) = (1/2)l (AC) …[Diagonals of a rhombus bisect each other]
= (1/2) × 48 …[From (i)]
= 24 cm …(ii)
Perimeter of rhombus = 100 cm …[Given]
Perimeter of rhombus = 4 × side
∴ 100 = 4 × l(AD)
∴ l(AD) = 100/4 = 25 cm …(iii)
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]
∴ (25)2 = (24)2 + l(DE)2 … [From (ii) and (iii)]
∴ 625 = 576 + l(DE)2
∴ l(DE)2 = 625 – 576
∴ l(DE)2 = 49
∴ l(DE) = √49 … [Taking square root of both sides]
l(DE) = 7 cm …(iv)
l(DE) = (1/2)l (BD) ….[Diagonals of a rhombus bisect each other]
∴ 7 = (1/2)l (BD) …[From (iv)]
∴ l(BD) = 7 × 2
= 14 cm …(v)
Area of a rhombus = (1/2) × product of lengths of diagonals
= (1/2) × l(AC) × l(BD)
= (1/2) × 48 × 14 … [From (i) and (v)]
= 48 × 7
= 336 sq.cm
∴ The area of the quadrilateral is 336 sq.cm.
