Answer:
[tex]68\text{cm}[/tex]
Step-by-step explanation:
[tex]\text{Solution:}\\\text{Let ABCD be a rhombus. Let AC and BD be its diagonals and they intersect}\\\text{at O.}\\\text{Let BD = 30cm}[/tex]
[tex]\text{Now, }\\\text{Area of Rhombus = }\dfrac{1}{2}\times(\text{Product of Diagonal lengths})=\dfrac{1}{2}\times\text{AC}\times\text{BD}\\\text{or, }240=\dfrac{1}{2}\times\text{AC}\times30\\\\\text{or, }240=15\text{AC}\\\text{or, AC}=8\text{cm}[/tex]
[tex]\text{All sides of rhombus are equal. So let AB = BC = CD = AD = }a.[/tex]
[tex]\text{Diagonals of rhombus bisect each other perpendicularly.}\\\therefore\ \text{OB = OD = 15cm and OA = OC = 8cm}[/tex]
[tex]\text{Using Pythagoras Theorem in triangle OBC,}\\\text{CB}^2=\text{OC}^2+\text{OB}^2\\\text{or, }a^2=8^2+15^2\\\text{or, }a^2=289\\\text{or, }a=17[/tex]
[tex]\text{Now perimeter of rhombus}=4a=4(17)=68\text{cm}[/tex]
