The lengths of the two diagonals of a rhombus are 7 cm and 24 cm. Find the length of its perimeter (in cm).

Given : ABCD is a rhombus and AC = 24 cm and BD = 7 cm

To find : Perimeter of ABCD

Solution : Diagonals of a rhombus bisect each other at right angle.

=> BE = $$\frac{7}{2}=3.5$$ cm and AE = $$\frac{24}{2}=12$$ cm

Thus, in right $$\triangle$$ AEB,

=> $$(AB)^2=(AE)^2+(BE)^2$$

=> $$(AB)^2=(12)^2+(3.5)^2$$

=> $$(AB)^2=144+12.25=156.25$$

=> $$AB=\sqrt{156.25}=12.5$$ cm

$$\therefore$$ Perimeter of rhombus ABCD = $$4\times12.5=50$$ cm

=> Ans - (D)