Answer:
68 cm
Explanation:
Let ABCD be a rhombus, all sides of the rhombus have equal length and its diagonal AC and BD are intersecting each other at a point O.
As we know that the diagonals in the rhombus bisect each other at 90°.
So, AO can be written as (AC/2)
= 16/2
= 8 cm
And, DO can be written as (BD/2)
= 30/2
= 15 cm
Then, consider the triangle AOD and apply Pythagoras theorem,
AD2 = AO2 + DO2
AD2 = 82 + 152
AD2 = 64 + 225
AD2 = 289
AD = √289
AD = 17 cm
Hence, the length of each side of the rhombus is 17 cm
Now,
The perimeter of rhombus = 4 × side of the rhombus
= 4 × 17
= 68 cm
Therefore, the perimeter of the rhombus is 68 cm.