Calculate the perimeter of rhombus whose diagonals are 12 cm and 5 cm long.
A.13 cm
B.26 cm
C.39 cm
D.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} $.

Complete step-by-step answer:
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} $.