The length of a side of a square field is 4 m.

Solution:

Given:

Length of the square field $=4 \mathrm{~m}$

$\therefore$ Area of the square field $=4 \times 4=16 \mathrm{~m}^{2}$

Given : Area of the rhombus $=$ Area of the square field

Length of one diagonal of the rhombus $=2 \mathrm{~m}$

$\therefore$ Side of the rhombus $=\frac{1}{2} \sqrt{d_{1}^{2}+d_{2}^{2}}$

And, area of the rhombus $=\frac{1}{2} \times\left(d_{1} \times d_{2}\right)$

$\therefore$ Area:

$16=\frac{1}{2}\left(2 \times d_{2}\right)$

$d_{2}=16 \mathrm{~m}$

Now, we need to find the length of the side of the rhombus.

$\therefore$ Side of the rhombus $=\frac{1}{2} \sqrt{2^{2}+16^{2}}=\frac{1}{2} \sqrt{260}=\frac{1}{2} \sqrt{4 \times 65}=\frac{1}{2} \times 2 \sqrt{65}=\sqrt{65} \mathrm{~m}$

Also, we know: Area of the rhombus $=$ Side $\times$ Altitude

$\therefore 16=\sqrt{65} \times$ Altitude

Altitude $=\frac{16}{\sqrt{65}} \mathrm{~m}$