The perimeter of rhombus $ \mathrm{ABCD} $ is $ 116 . $ If $ \mathrm{AC}=42 $, find $ \mathrm{BD} $.
Given:
Perimeter of the rhombus $ABCD = 116\ cm$
Diagonal $AC$ of rhombus $ABCD = 42\ cm$
To do:
We have to find \( \mathrm{BD} \).
Solution:
We know that,
All the sides of a rhombus are equal.
Let each side of the rhombus be $s$.
This implies,
$4s = 116$
$s = \frac{116}{4}$
$s=29\ cm$
The diagonals of the rhombus divide it into four right-angled triangles with right angles at the centre.
Therefore,
$AB^2=(\frac{AC}{2})^2+(\frac{BD}{2})^2$
$29^2=(\frac{42}{2})^2+(\frac{BD}{2})^2$
$841=21^2+(\frac{BD}{2})^2$
$841-441=(\frac{BD}{2})^2$
$400\times4=BD^2$
$BD=\sqrt{1600}$
$BD=40\ cm$
The length of \( \mathrm{BD} \) is $40\ cm$.
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