The length of a side of a square field is 4 m. what will be the altitude of the rhombus, if the area of the rhombus is equal to the square field and one of its diagonal is 2 m?
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 $=\frac{16}{\sqrt{65}} \mathrm{~m}$