The area of a rhombus is given by:
$ \text{Area} = \dfrac{1}{2} \times d_1 \times d_2 $
where $d_1$ and $d_2$ are the diagonals of the rhombus.

Given: Area = 96
So,
$96 = \dfrac{1}{2} \times d_1 \times d_2$
$\Rightarrow d_1 \times d_2 = 96 \times 2 = 192$
Each half-diagonal forms a right-angled triangle with the side of the rhombus.
Given side of rhombus = 10 units.
Using:
$ \left( \dfrac{d_1}{2} \right)^2 + \left( \dfrac{d_2}{2} \right)^2 = 10^2 $
$ \Rightarrow \dfrac{d_1^2}{4} + \dfrac{d_2^2}{4} = 100 $
$ \Rightarrow \dfrac{d_1^2 + d_2^2}{4} = 100 $
$ \Rightarrow d_1^2 + d_2^2 = 400 $
Using identity: $ (a + b)^2 = a^2 + b^2 + 2ab $
$ (d_1 + d_2)^2 = d_1^2 + d_2^2 + 2d_1d_2 $
Substituting values:
$ (d_1 + d_2)^2 = 400 + 2 \times 192 $
$ = 400 + 384 = 784 $
$ \Rightarrow d_1 + d_2 = \sqrt{784} = 28 $
Cost per meter = ₹125
Total length of wires = $d_1 + d_2 = 28$ meters
Total cost = $28 \times 125 = ₹3500$
The total cost of laying electric wires along the diagonals is ₹3500.
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: