The distance between the parallel sides AB and CD of a trapezium ABCD is 12 cm. If AB<CD, CD=24 cm and AD =BC=13 cm, then area of the trapezium (in cm²) is :

The correct option is 228.

Step 1: Understand the given properties of the trapezium
Let $ABCD$ be an isosceles trapezium where $AB$ is parallel to $CD$ and $AB<CD$.
We are given:
- The length of side $CD=24\text{cm}$
- Non-parallel sides $AD=BC=13\text{cm}$
- The perpendicular distance (height) between $AB$ and $CD$, $h=12\text{cm}$

Step 2: Find the length of the shorter parallel side ($AB$)
Draw perpendiculars from $A$ and $B$ to side $CD$, meeting $CD$ at points $E$ and $F$ respectively.
Here, $AE=BF=h=12\text{cm}$.
Since $ABFE$ forms a rectangle, $EF=AB$.

In right-angled triangle $AED$, by Pythagoras' theorem:
$A{D}^2=A{E}^2+D{E}^2$

Substitute the known values:
${13}^2={12}^2+D{E}^2$
$169=144+D{E}^2$
$D{E}^2=169-144=25$
$DE=5\text{cm}$

Due to symmetry in an isosceles trapezium, $FC=DE=5\text{cm}$.
Now, side $CD$ can be split into three parts:
$CD=DE+EF+FC$
$24=5+AB+5$
$24=AB+10$
$AB=24-10=14\text{cm}$

Step 3: Calculate the area of the trapezium
The formula for the area of a trapezium is:
$\text{Area}=\frac{1}{2}\times \left(\text{Sum of parallel sides}\right)\times \text{Height}$

Substitute the values of $AB$, $CD$, and $h$ into the formula:
$\text{Area}=\frac{1}{2}\times (14+24)\times 12$
$\text{Area}=\frac{1}{2}\times 38\times 12$
$\text{Area}=19\times 12=228{\text{cm}}^2$

Thus, the area of the trapezium is 228 cm².