What will be the perimeter of a rhombus whose diagonals are $16\ cm$ and $12\ cm$ respectively?

Given: A rhombus whose diagonals are $16\ cm$ and $12\ cm$ respectively.

To do: To find the perimeter of the rhombus.

Solution:

At the point $O$, where diagonals bisects each other at angle formed $90^o$.

Diagonal of rhombus$=12\ cm$ and $16\ cm$

Therefore,

Half of diagonal $BD=OB=\frac{1}{2}\times12=6\ cm$

Half of the diagonal $AC=\frac{1}{2}\times16=8\ cm$

In $vartriangle OBC$,

$OB=6\ cm,\ OC=8\ cm$ and $\angle BOC=90^o$

On using Pythagoras theorem

$a ( side)$ of rhombus$ =\sqrt{p^2+b^2}$

$a=\sqrt{6^2+8^2}=\sqrt{36+64} =\sqrt{100}=10\ cm$

As known, Perimeter of rhombus$= 4a=4\times10=40\ cm$

Thus, the perimeter of the rhombus is $40\ cm$.

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Updated on: 10-Oct-2022

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