Answer:
Let ABCD be a rhombus whose diagonals BD and AC are of lengths 16 cm and 30 cm respectively. Let the diagonals BD and AC intersect each other at O. Since the diagonals of a rhombus bisect each other at right angles. Therefore \[\text{BO}=\text{OD}=\text{8 cm},\] \[\text{AO}=\text{OC}=\text{15 cm},\] \[\angle AOB=\angle BOC=\angle COD=\angle DOA={{90}^{o}}\]In right-angled triangle AOB. \[A{{B}^{2}}=O{{A}^{2}}+O{{B}^{2}}\] \[\left| \text{By Pythagoras Property} \right.\] \[\Rightarrow \] \[\text{A}{{\text{B}}^{\text{2}}}=O{{A}^{\text{2}}}+O{{B}^{\text{2}}}\] \[\Rightarrow \] \[~\text{A}{{\text{B}}^{\text{2}}}={{15}^{2}}\text{+ }{{\text{8}}^{2}}\] \[\Rightarrow \] \[A{{B}^{2}}=225+64\] \[\Rightarrow \] \[A{{B}^{2}}=289\] \[\Rightarrow \] \[AB=\text{17 cm}\] Therefore, perimeter of the rhombus ABCD \[=\text{4}\]side \[=\text{4}\] AB \[=\text{4}\times \text{17 cm}\] \[=\text{68 cm}\] Hence, the perimeter of the rhombus is 68 cm.
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