91Ó°ÊÓ

The diagonals of the rhombus are $16$ and $30$.

1. In Rhombus, diagonals are perpendicular bisector.
2. Perimeter of a rhombus is $4a$, here a is side of rhombus.
3. Pythagoras Theorem state that in a right angle triangle, the square of the hypotenuse is equal to the sum of squares of other sides.

Now we draw a rhombus with diagonals 16 and 30 and labeled $â–¡ABCD$.

And diagonal intersection point be O.

$\begin{array}{c}AO=OC\text{(byconcept1)}\\ \\ AC=30\\ AO+OC=30\\ 2AO=30\\ AO=15\\ \\ DO=DB\text{(byconcept1)}\\ \\ DB=16\\ DO+OB=16\\ 2OB=16\\ OB=8\end{array}$

In$\mathrm{Δ}AOB$.

We use Pythagoras theorem.

$\begin{array}{c}⇒x^2={15}^2+8^2\\ x^2=225+64\\ x^2=289\end{array}$

Taking square root we get.

$\begin{array}{l}x=Â\pm \sqrt{289}\\ x=Â\pm 17\end{array}$

Since, distance is always a positive quantity so we neglect negative sign.

So, the side of rhombusis $17$.

The perimeter of rhombus is:

$\begin{array}{c}\text{Perimeter}=4×\text{side}\\ =4×17\\ =68\end{array}$

Therefore, the perimeter if the rhombus is$68$.