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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