Three parallel lines L1, L2, L3 are drawn in the plane such that the perpendicular distance between L1 and L2 is 3 and the perpendicular distance between L2 and L3 is also 3. A square ABCD is constructed such that A lies on L1, B lies on L3 and C lies on L2.Find the area of the square

Let the three parallel lines be represented on the coordinate plane as:

• $L_1:y=6$

• $L_2:y=3$

• $L_3:y=0$

Let the square $ABCD$ have vertices $A\in L_1$, $C\in L_2$, $B\in L_3$ in counterclockwise order.

Let $A=(x_1,6)$ and $B=(x_2,0)$. Since $ABCD$ is a square, a $90^{\circ }$ counterclockwise rotation about $B$ maps vector $\overrightarrow{BC}$ to $\overrightarrow{BA}$. Equivalently, the vector $\overrightarrow{BC}$ is orthogonal and equal in length to $\overrightarrow{AB}$.
Let $\overrightarrow{BA}=(x_1-x_2,6)$. Rotating this by $90^{\circ }$ clockwise gives the vector for $\overrightarrow{BC}$:

Since $B=(x_2,0)$, we find the coordinates of $C$:

Since $C$ lies on $L_2$ ($y=3$), its y-coordinate must be $3$:

The side length squared of the square is $A{B}^2$:

Thus, the area of the square is $45$.