Correct Answer: $6 \sqrt{13}$

Solution : Given: The larger diagonal of a rhombus is 150% of its smaller diagonal, and its area is 432 cm ² .
Let the length of the smaller diagonal be $x$ cm.
The length of the larger diagonal = 150% of $x=\frac{3x}{2}\ \text{cm}$
We know, The area of the rhombus = $\frac{1}{2}\times \text{The first diagonal}\times \text{The second diagonal}$
According to the question,
$\frac{1}{2}\times x\times \frac{3x}{2}=432$
⇒ $x^2=144\times 4$
⇒ $x=12\times 2=24\ \text{cm}$
The length of the smaller diagonal = $x=24\ \text{cm}$
The length of the larger diagonal = $\frac{3\times 24}{2}=36\ \text{cm}$
We know, $\text{Side}^2 = \text{Half of the first diagonal}^2+\text{Half of the second diagonal}^2$
The length of the side of the rhombus
$=\sqrt{(\frac{36}{2})^2+(\frac{24}{2})^2}$
$=\sqrt{(18)^2+(12)^2}=\sqrt{324+144}=6 \sqrt{13}\ \text{cm}$
Hence, the correct answer is $6 \sqrt{13}\ \text{cm}$.