Mark (✓) against the correct answer:

(b) 13 cm

Let $A B C D$ be a rhombus whose diagonals $A C$ and $B D$ intersect at a point $O$.

Let the length of the diagonal $A C$ be $24 \mathrm{~cm}$.

$A$ rea of the rhombus $=\left(\frac{1}{2} \times A C \times B D\right) \mathrm{cm}^{2}$

But the area of the rohmbus is $120 \mathrm{~cm}^{2}$ (given)

$\therefore \frac{1}{2} \times A C \times B D=120$

or $\frac{1}{2} \times 24 \times B D=120$

or $12 \times B D=120$

or $B D=\frac{120}{12}=10 \mathrm{~cm}$

$\mathrm{OB}=\frac{\mathrm{BD}}{2}=\frac{10}{2}=5 \mathrm{~cm}$

And $O A=\frac{A C}{2}=\frac{24}{2}=12 \mathrm{~cm}$

Now, in right triangle $A O B$ :

$(A B)^{2}=(O A)^{2}+(O B)^{2}$

or $(A B)^{2}=12^{2}+5^{2}$

$=144+25$

$=169$

or $A B=\sqrt{169}=13 \mathrm{~cm}$

Therefore, each side of the rhombus is $13 \mathrm{~cm}$.