The lengths of diagonals of a rhombus are 24cm and 10cm the perimeter of the rhombus (in cm ) is :

The diagonals of a rhombus bisect each other at right angle.

ABCD is the rhombus whose diagonals bisect at O.

Given : BD = 24 cm and AC = 10 cm

=> $$OB=\frac{BD}{2}=12$$ cm

Similarly, $$OA=5$$ cm

In $$\triangle$$ OAB

=> $$(AB)^2=(OA)^2+(OB)^2$$

=> $$(AB)^2 = (5)^2+(12)^2$$

=> $$(AB)^2=25+144=169$$

=> $$AB=\sqrt{169}=13$$ cm

$$\therefore$$ Perimeter of rhombus = $$4 \times AB$$     [$$\because$$ All sides of rhombus are equal]

= $$4 \times 13=52$$ cm

=> Ans - (A)