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

$=\frac{1}{2}\times 10 \times 2\sqrt{119}$

$=10\sqrt{119}\ cm^2$

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