The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

Make a rhombus PQRS, all sides of the rhombus have equal length, and its diagonal PR and SQ are intersecting each other at point O. Diagonals in the rhombus bisect each other at 90^o.

So, $PO =\frac{PR}{2}$

= $\frac{16}{2}$

$= 8 cm$

Also, $QO = \frac{SQ}{2}$

= $\frac{30}{2}$

$= 15 cm$

In $△$POQ and by Pythagoras property,

$P{Q}^2 = P{O}^2 + Q{O}^2$

$P{Q}^{2 }= 8^2 + {15}^2$

$P{Q}^2 = 64 + 225$

$P{Q}^2 = 289$

$PQ = \sqrt{289}$

$PQ = 17 cm$

So the length of the side of the rhombus is 17 cm

Now,

The perimeter of the rhombus = 4 × side of the rhombus

$= 4 \times 17$

$= 68 cm$

∴  the rhombus has a perimeter of 68 cm.