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 diagonal is $2m?$

Answer 1

Step: Find the altitude of the rhombus

Given:
Area of the rhombus = Area of square field
side of square field $=4m$
Length of one of the diagonal of rhombus $=2m$


As, Area of rhombus = Area of the square
$\Rightarrow$ Area of rhombus = Area of the square
$\Rightarrow$ Area of rhombus = $(side{)}^2=(4\times 4)=16$ ......(1)

$\Rightarrow \frac{1}{2}\times$ (Product of diagonals) $=16$

$\Rightarrow \frac{1}{2}\times AC\times BD=16$

$\Rightarrow \frac{1}{2}\times 2\times BD=16$

$\Rightarrow BD=16m$

We know that the diagonals of a rhombus bisects each other at right angle.
$\Rightarrow BO=\frac{1}{2}\left(BD\right)$

$\Rightarrow BO=\frac{1}{2}\left(16\right)=8mandAO=\frac{1}{2}\left(AC\right)=\frac{1}{2}\left(2\right)=1m$

Now,$BO=8m$ and $AO=1m$

Applying Pythagoras theorem in $\Delta AOB,$
$(AB{)}^2=(OB{)}^2+(AO{)}^2$
$(AB{)}^2=(8{)}^2+(1{)}^2$
$A{B}^2=64+1=65$
$AB=\sqrt{65}m$ ......(2)

Let $DE$ be the altitude

Area of the rhombus = Base $\times$ Altitude
Area of the rhombus =$AB\times DE$
$\Rightarrow 16=\sqrt{65}\times DE$ (from eq. 1 and 2)
$\Rightarrow DE=\frac{16}{\sqrt{65}}m$

Hence, the altitude of the rhombus is $\frac{16}{\sqrt{65}}m.$