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Solution:

(b) 216 cm2

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 $24 \mathrm{~cm}$ and the side of the rhombus be $15 \mathrm{~cm}$.

We know that the diagonals of the rhombus bisect each other at right angles.

$\therefore A O=\frac{1}{2} A C$

$\Rightarrow A O=\left(\frac{1}{2} \times 24\right) \mathrm{cm}$

$\Rightarrow A O=12 \mathrm{~cm}$

From right $\Delta A O B$, we have :

$B O^{2}=A B^{2}-A O^{2}$

$\Rightarrow B O^{2}=\left\{(15)^{2}-(12)^{2}\right\}$

$\Rightarrow B O^{2}=(225-144)$

$\Rightarrow B O^{2}=81$

$\Rightarrow B O=\sqrt{81}$

$\Rightarrow B O=9 \mathrm{~cm}$

$\therefore B D=2 \times B O$

$B D=(2 \times 9) \mathrm{cm}$

$B D=18 \mathrm{~cm}$

Hence, the length of the other diagonal is $18 \mathrm{~cm} .$

Area of the rhombus $=\left(\frac{1}{2} \times 24 \times 18\right) \mathrm{cm}^{2}$

$=216 \mathrm{~cm}^{2}$