In rhombus ABCD,
l(AC) = 16 cm and l(BD) = 12 cm. Let the diagonals of rhombus ABCD intersect at point O.
l(AO) = (1/2) l(AC) …[Diagonals of a rhombus bisect each other]
∴ l(AO) = (1/2) × 16
∴ l(AO) = 8 cm
Also, l(DO) = (1/2) l(BD) …[Diagonals of a rhombus bisect each other]
∴ l(DO) = (1/2) x 12
∴ l(DO) = 6 cm
In ∆DOA, m∠DOA = 90° ..[Diagonals of a rhombus are perpendicular to each other]
[l(AD)]2 = [l(AO)]2 + [l(DO)]2 …[Pythagoras theorem]
= (8)2 + (6)2 = 64 + 36
∴ [l(AD)]2 = 100
∴ l(AD) = √100 … [Taking square root of both sides]
∴ l(AD) = 10 cm
∴ l(AB) = l(BC) = l(CD) = l(AD) = 10 cm …[Sides of a rhombus are congruent]
Perimeter of rhombus ABCD = l(AB) + l(BC) + l(CD) + l(AD)
= 10 + 10 + 10 + 10 = 40 cm
∴ The side and perimeter of the rhombus are 10 cm and 40 cm respectively.
