Find the value of $x$ in rhombus $ABCD$ as shown in the figure.
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Given: In the given figure, $ABCD$ is a rhombus and its diagonal intersects at the point $O$.

To do: To find $x$.

Solution:

In triangle $\vartriangle OCD$, 

$OD=12\ cm,\ OC=( x-2)$ and $CD=( x+6)$

$\angle COD=90^o$    [In a rhombus, diagonals form an angle of $90^o$ at the point of intersection$]

On using Pythagoras theorem,

$CD^2=OC^2+OD^2$

$\Rightarrow ( x+6)^2=( x-2)^2+12^2$

$\Rightarrow x^2+12x+36=x^2-4x+4+144$

$\Rightarrow 12x+4x=148-36$

$\Rightarrow 16x=112$

$\Rightarrow x=\frac{112}{16}$

$\Rightarrow x=7$

Thus, $x=7$.

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Updated on: 10-Oct-2022

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