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# Calculate the perimeter of rhombus whose diagonals are 12 cm and 5 cm long.A.13 cmB.26 cmC.39 cmD.52 cm  Hint: Rhombus is a quadrilateral whose four sides have the same length and the diagonals intersect at ${90^ \circ }$. The perimeter of the rhombus $= 2\sqrt {{d_1}^2 + {d_2}^2}$.

Given,
Length of diagonal ${d_1}$ = 12 cm
Length of diagonal ${d_2}$ = 5 cm
We know that,
Perimeter of rhombus $= 2\sqrt {{d_1}^2 + {d_2}^2}$
Where ${d_1}$ and ${d_2}$ are the lengths of the diagonals.
Substitute the values of the
$= 2\sqrt {{{12}^2} + {5^2}}$
$= 2\sqrt {144 + 25}$
$= 2\sqrt {169}$
$= 2 \times 13$
$= 26cm$
Hence option B is correct.

Note: Generally the perimeter of the rhombus of given side of length S $= 4 \times S$. We know that the relation length of side and lengths of the diagonals that is $S = \dfrac{{\sqrt {{d_1}^2 + {d_2}^2} }}{2}$$\therefore$ Perimeter of rhombus of given lengths of diagonals $= 2\sqrt {{d_1}^2 + {d_2}^2}$.
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