Question

The length of a side of a square field is $4m$. What will be the altitude of the rhombus, if the area of the rhombus is equal to the square field and one of its diagonals is $2m$?

Answer 1

We have, side of a square $=4m$ and one diagonal of a square $=2m$

Area of the rhombus$=$ Area of the square of side $4m$
$\Rightarrow \frac{1}{2}\times AC\times BD=4m^2$
$\Rightarrow \frac{1}{2}\times AC\times 2=16$
$\therefore AC=16m$

We know that the diagonals of a rhombus are perpendicular bisectors of each other.
$\Rightarrow AO=\frac{1}{2}AC$
$=8m$ and
$BO=\frac{1}{2}BD=1m$
From right angled triangle $\Delta AOB$
We have Pythagoras theorem, $A{O}^2+B{O}^2=A{B}^2$
$\Rightarrow A{B}^2=8^2+1^2$
$=64+1$
Therefore, side of a rhombus $AB=\sqrt{65}m$
Let $DX$ be the altitude.
Area of the rhombus$=AB\times DX$
$16=\sqrt{65}\times DX$
Therefore, Altitude $=DX=\frac{16}{\sqrt{65}}m$