# The perimeter of a rhombus is 180 cm and one of its diagonals is 72 cm.

Question:

The perimeter of a rhombus is 180 cm and one of its diagonals is 72 cm.

Solution:

Let $A B C D$ be a rhombus whose diagonals $A C$ and $B D$ intersect at a point $O$.

Let the length of the diagonal AC be $72 \mathrm{~cm}$ and the side of the rhombus be $x \mathrm{~cm}$.

$P$ erimeter of the rhombus $=4 x \mathrm{~cm}$

But it is given that the perimeter of the rhombus is $180 \mathrm{~cm}$.

$\therefore 4 x=180$

$\Rightarrow x=\frac{180}{4}$

$\Rightarrow x=45$

Hence, the length of the side of the rhombus is $45 \mathrm{~cm}$.

We know that the diagonals of the rhombus bisect each other at right angles.

$\therefore A O=\frac{1}{2} A C$

$\Rightarrow A O=\left(\frac{1}{2} \times 72\right) \mathrm{cm}$

$\Rightarrow A O=36 \mathrm{~cm}$

From right $\Delta \mathrm{AOB}$, we have :

$B O^{2}=A B^{2}-A O^{2}$

$\Rightarrow B O^{2}=\left\{(45)^{2}-(36)^{2}\right\}$

$\Rightarrow B O^{2}=(2025-1296)$

$\Rightarrow B O^{2}=729$

$\Rightarrow B O=\sqrt{729}$

$\Rightarrow B O=27 \mathrm{~cm}$

$\therefore B D=2 \times B O$

$B D=(2 \times 27) \mathrm{cm}$

$B D=54 \mathrm{~cm}$

Hence, the length of the other diagonal is $54 \mathrm{~cm}$.

Area of the rhombus $=\left(\frac{1}{2} \times 72 \times 54\right) \mathrm{cm}^{2}$

$=1944 \mathrm{~cm}^{2}$