$\mathrm{ABCD}$ is a trapezium in which $\mathrm{AB}$ is parallel to $\mathrm{CD}$ . The sides $\mathrm{AD}$ and $\mathrm{BC}$ , when extended, intersect at point $E$ . If $\mathrm{AB}=2\text{ cm}$ , $\mathrm{CD}=1\text{ cm}$ , and the perimeter of $\mathrm{ABCD}$ is $6\text{ cm}$ , then the perimeter, in cm, of triangle $\mathrm{AEB}$ is:

The correct option is 3. 8.

Let's understand the logical reasoning and derivations step-by-step:

We are given a trapezium $\mathrm{ABCD}$ in which side $\mathrm{AB}$ is parallel to side $\mathrm{CD}$. The non-parallel sides $\mathrm{AD}$ and $\mathrm{BC}$, when extended, intersect at point $E$, forming the triangle $\mathrm{AEB}$.

Step 1: Use similarity of triangles
Because the line segments $\mathrm{AB}$ and $\mathrm{CD}$ are parallel, the corresponding angles formed by the transversal lines $\mathrm{AE}$ and $\mathrm{BE}$ are equal:
$\angle \mathrm{EDC}=\angle \mathrm{EAB}$
and
$\angle \mathrm{ECD}=\angle \mathrm{EBA}$
Since they also share the common angle $\angle \mathrm{DEC}$, triangle $\mathrm{DEC}$ is similar to triangle $\mathrm{AEB}$ by Angle-Angle (AA) similarity criterion (written as $\Delta \mathrm{DEC}\sim \Delta \mathrm{AEB}$).

Step 2: Set up ratios of corresponding sides
From the properties of similar triangles, the ratio of corresponding sides is constant and equal to the ratio of their bases:
$$\frac{\mathrm{DE}}{\mathrm{AE}}=\frac{\mathrm{CE}}{\mathrm{BE}}=\frac{\mathrm{CD}}{\mathrm{AB}}$$
Given that $\mathrm{AB}=2\text{ cm}$ and $\mathrm{CD}=1\text{ cm}$, we substitute these values into the ratio:
$$\frac{\mathrm{DE}}{\mathrm{AE}}=\frac{\mathrm{CE}}{\mathrm{BE}}=\frac{1}{2}$$
This gives us the relations:
$\mathrm{DE}=\frac{1}{2}\mathrm{AE}$
and
$\mathrm{CE}=\frac{1}{2}\mathrm{BE}$

Using segment addition ($\mathrm{AE}=\mathrm{AD}+\mathrm{DE}$ and $\mathrm{BE}=\mathrm{BC}+\mathrm{CE}$), we can find the lengths of the non-parallel sides $\mathrm{AD}$ and $\mathrm{BC}$ of the trapezium in terms of the triangle sides:
$\mathrm{AD}=\mathrm{AE}-\mathrm{DE}=\mathrm{AE}-\frac{1}{2}\mathrm{AE}=\frac{1}{2}\mathrm{AE}$
$\mathrm{BC}=\mathrm{BE}-\mathrm{CE}=\mathrm{BE}-\frac{1}{2}\mathrm{BE}=\frac{1}{2}\mathrm{BE}$

Step 3: Relate the perimeter of the trapezium to the triangle's sides
The perimeter of the trapezium $\mathrm{ABCD}$ is given as $6\text{ cm}$:
$$\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{AD}=6$$
Substituting the known values and side relations:
$$2+\frac{1}{2}\mathrm{BE}+1+\frac{1}{2}\mathrm{AE}=6$$
Combine the constant terms:
$$3+\frac{1}{2}(\mathrm{AE}+\mathrm{BE})=6$$
Subtract 3 from both sides:
$$\frac{1}{2}(\mathrm{AE}+\mathrm{BE})=3$$
Multiply by 2:
$$\mathrm{AE}+\mathrm{BE}=6\text{ cm}$$

Step 4: Find the perimeter of triangle AEB
The perimeter of $\Delta \mathrm{AEB}$ is the sum of its three side lengths:
$$\text{Perimeter}\left(\Delta \mathrm{AEB}\right)=\mathrm{AE}+\mathrm{BE}+\mathrm{AB}$$
Substituting $\mathrm{AE}+\mathrm{BE}=6\text{ cm}$ and $\mathrm{AB}=2\text{ cm}$:
$$\text{Perimeter}\left(\Delta \mathrm{AEB}\right)=6+2=8\text{ cm}$$