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# The diagonals of rhombus measure 16cm and 30cm. Find its perimeter.  Hint – Here first we draw a rhombus after that to determine its perimeter we will use the concept that Diagonals of rhombus bisect perpendicular to each other and Perimeter of rhombus $\;4 \times side$. AC and BD are diagonals of a rhombus.
AC=16, BD=30
We know that diagonals of rhombus are perpendicular bisectors of each other.
$\therefore$ AO = 8 and BO = 15
In right angled $\Delta AOB$
$\begin{array}{*{20}{l}} { \Rightarrow \;\;{{\left( {AB} \right)}^2} = {{\left( {AO} \right)}^2} + {{\left( {BO} \right)}^2}} \\ { \Rightarrow \;\;{{\left( {AB} \right)}^2} = {{\left( 8 \right)}^2} + {{\left( {15} \right)}^2}} \\ { \Rightarrow \;\;{{\left( {AB} \right)}^2} = 64 + 225} \\ { \Rightarrow \;\;{{\left( {AB} \right)}^2} = 289} \\ { \Rightarrow \;\;AB = 17cm} \end{array}$
We know that all four sides of rhombus are equal.
$\therefore$ Perimeter of rhombus ABCD $= {\text{ }}4 \times AB$
$\begin{array}{*{20}{l}} {\; = {\text{ }}4 \times 17} \\ { = {\text{ }}68cm} \end{array}$

Note – In this problem, first we calculate the side of the rhombus with help of Pythagoras theorem then calculate the perimeter of the rhombus by using the concept which is given above.
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