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 = $$4 \times AB$$ Â Â [$$\because$$ All sides of rhombus are equal]
= $$4 \times 13=52$$ cm
=> Ans - (A)
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