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The sides of a rhombus measures \( 12 \mathrm{~cm} \) and its diagonal is \( 10 \mathrm{~cm} \). What is the area of the rhombus?
Given:
Length of each side of the rhombs$=12\ cm$.
Length of one of the diagonals$=10\ cm$.
To do:
We have to find the area of the rhombus.
Solution:
Let ABCD be the rhombus in which O is the intersecting point of the diagonals.
We know that,
Diagonals of a rhombus bisect each other at right angles.
Therefore,
$\angle BOC=90^o$, $BC=12\ cm$ and $OC=5\ cm$
$\triangle BOC$ is a right-angle triangle.
This implies, using Pythagoras theorem,
$BC^2=BO^2+OC^2$
$(12)^2=BO^2+(5)^2$
$BO^2=(144-25)\ cm^2$
$BO^2=119\ cm^2$
$BO=\sqrt{119}\ cm$
$BD=2(BO)=2(\sqrt{119})\ cm$
Area of the rhombus $=\frac{1}{2}\times AC \times BD$
$=\frac{1}{2}\times 10 \times 2\sqrt{119}$
$=10\sqrt{119}\ cm^2$