In rhombus $ \mathrm{ABCD}, \mathrm{AC}=16 $ and $ \mathrm{BD}=30 $. Find the perimeter of rhombus $ \mathrm{ABCD} $.

Given:

In rhombus \( \mathrm{ABCD}, \mathrm{AC}=16 \) and \( \mathrm{BD}=30 \).

To do:

We have to find the perimeter of the rhombus \( \mathrm{ABCD} \).

Solution:

We know that,

All the sides of a rhombus are equal.

The diagonals of the rhombus divide it into four right-angled triangles with right angles at the centre.

Therefore,

$AB^2=(\frac{AC}{2})^2+(\frac{BD}{2})^2$

$AB^2=(\frac{16}{2})^2+(\frac{30}{2})^2$

$AB^2=8^2+(15)^2$

$AB^2=64+225$

$AB^2=289$

$AB=\sqrt{289}=17$

Perimeter of the rhombus $=4\times AB$

$=4\times17$

$=68$

The perimeter of rhombus \( \mathrm{ABCD} \) is $68$. 

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Updated on: 10-Oct-2022

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