Correct Answer - Option 4 : 16 cm
Given:
Difference between two diagonals of a rhombus = 14 cm
Perimeter of the rhombus = 68 cm
Concept used:
The two diagonals bisect each other at 90° in a rhombus.
Formula used:
Area of rhombus = (1/2) × Product of the two diagonals
Perimeter = 4 × (Side of the rhombus)
Calculations:
Let the two diagonals of the rhombus be a and b and the side of the rhombus be x
⇒ a – b = 14 cm
⇒ a = 14 + b ----(i)
Perimeter of the rhombus = 4x
⇒ 4x = 68
⇒ x = 17 cm
If the two diagonals of rhombus ABCD intersect at O then,
⇒ AO2 + BO2 = AB2
⇒ (a/2)2 + (b/2)2 = 172
⇒ (a2/4) + (b2/4) = 289
⇒ (a2 + b2) = 1156
Using (i),
⇒ [(14 + b)2 + b2] = 1156
⇒ (196 + b2 + 28b + b2) = 1156
⇒ 2b2 + 28b – 960 = 0
⇒ b2 + 14b – 480 = 0
⇒ b2 + 30b – 16b – 480 = 0
⇒ b(b + 30) – 16(b + 30) = 0
⇒ (b – 16)(b + 30) = 0
⇒ b = 16 cm
⇒ b = -30 cm (negative value not possible)
Diagonals, a = 30 cm and b = 16 cm
∴ The length of the shorter diagonal of the rhombus is 16 cm
