91Ó°ÊÓ

The one diagonal of the rhombus is 12cm and perimeter is 40 cm.

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.

The perimeter of rhombus is:

$\begin{array}{l}4×side=40\\ ⇒side=10\end{array}$

Now draw the rhombus with side $10$cm and diagonal $12$cm and labeled it $â–¡ABCD$.

Let $AO=x$.

$\begin{array}{c}DB=12\\ DO+OB=12\\ 2OB=12\\ OB=6\\ \\ AB=10\end{array}$

Now we use Pythagoras theorem in $\mathrm{Δ}AOB$.

$\begin{array}{c}A{O}^2+O{B}^2=A{B}^2\\ x^2+6^2={10}^2\\ x^2+36=100\\ x^2=64\end{array}$

Taking square root we get.

$\begin{array}{l}x=Â\pm \sqrt{64}\\ x=Â\pm 8\end{array}$

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

$\begin{array}{c}x=8\\ AO=OC\\ AO=8\\ OC=8\\ AC=AO+OC\\ AC=16\end{array}$

Therefore, the other diagonal of rhombus is $16$cm.