The diagonals of a rhombus measure $16\ cm$ and $30\ cm$. Find its perimeter.

AcademicMathematicsNCERTClass 7

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 $68\ cm$.
raja
Updated on 10-Oct-2022 13:34:37

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