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In a Parallelogram $ABCD, AB = 18\ cm, BC=12\ cm, AL \perp\ DC, AM \perp\ BC$ and $AL = 6.4\ cm$. Find the length of $AM$.
Given:
In a Parallelogram $ABCD, AB = 18\ cm, BC=12\ cm, AL \perp\ DC, AM \perp\ BC$ and $AL = 6.4\ cm$.
To do:
We have to find the length of $AM$.
Solution:
We know that,
Area of a parallelogram $=$ Base $\times$ Height
Therefore,
$AL\times DC=AM\times BC$
$6.4\times 18=AM\times 12$ ($AB=DC=18\ cm$, opposite sides of a parallelogram are equal)
$AM=\frac{6.4\times18}{12}$
$AM=3.2\times3$
$AM=9.6\ cm$
The length of AM is 9.6 cm.
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