Find the area of rhombus each side of which measures $20\ cm$ and whose diagonals is $24\ cm$.

Given: Each side of a rhombus measures $20\ cm$ and whose diagonals is $24\ cm$. 

To do: To find the area of rhombus.

$AO=OC=CB=BA=20\ cm$             ............. [side of rhombus]

$AC=24\ cm$

$AO=\frac{1}{2}\times AC=\frac{1}{2}\times 24=12\ cm$  ........[originals of rhombus bisect each other]

$\Rightarrow$ In right angled triangle $\vartriangle AOD$,

$AD^2=AO^2+OD^2$                      ........ By Pythagoras theorem

$(20)^2=( 12)^2+OD^2$

 

$\Rightarrow OD^2=400-144$

$\Rightarrow OD=256$

​

$\Rightarrow OD=16\ cm$

$\Rightarrow OB=2\times16=32\ cm$

$\Rightarrow$ Area of rhombus$=\frac{1}{2}\times product\ of\ diagonals$

$=\frac{1}{2}\times32\times24=384\ cm^2$.