# What is the perimeter of a rhombus whose diagonals are 16 and 30?

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Dec 25, 2016

$68$

#### Explanation:

Some of the rhombus' properties:

a) The sides of a rhombus are all congruent. (the same length).
$\implies A B = B C = C D = D A$

b) The two diagonals are perpendicular, and they bisect each other. This means they cut each other in half.
$\implies A O = O C = \frac{1}{2} A C , \mathmr{and} B O = O D = \frac{1}{2} B D$

Now back to our question :

Given that the two diagonals are $30 , \mathmr{and} 16$,
$\implies A O = \frac{30}{2} = 15 , B O = \frac{16}{2} = 8 , \angle A O B = {90}^{\circ}$

From Pythagorean theorem, we know
$A {B}^{2} = A {O}^{2} + B {O}^{2}$
$\implies A B = \sqrt{{15}^{2} + {8}^{2}} = \sqrt{289} = 17$

SInce $A B = B C = C D = D A$,
perimeter of $A B C D = 17 \cdot 4 = 68$

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