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 $40\ cm$.

Tutorialspoint

Updated on: 10-Oct-2022

31 Views

Kickstart Your Career

Get certified by completing the course

Get Started