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The diagonals of a rhombus measure $16\ cm$ and $30\ cm$. Find its perimeter.
Let $PQRS$ be a rhombus, all sides of the rhombus have equal length, and its diagonal $PR$ and $SQ$ are intersecting each other at a point $O$. Diagonals in rhombus bisect each other at $90^{\circ}$.
So, $PO=(\frac{PR}{2})$
$=\frac{16}{2}$
$=8\ cm$
And, $SO=(\frac{SQ}{2})$
$=\frac{30}{2}$
$=15\ cm$
Then, consider the triangle POS and apply the Pythagoras Theorem,
$PS^2=PO^2+SO^2$
$PS^2=8^2+15^2$
$PS^2=64+225$
$PS^2=289$
$PS=\sqrt{289}$
$PS=17\ cm$
Hence, the length of the side of the rhombus is $17\ cm$
Now,
The perimeter of Rhombus$=4\times$Side of the Rhombus
$=4\times17$
$=68\ cm$
$\therefore$ Perimeter of Rhombus is $68\ cm$.
Updated on: 10-Oct-2022
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