CAT 2025 Slot2 Quant: Let ABCDEF be a regular hexagon

⚡ AZUCATION SHORTCUT:
A regular hexagon can be divided into $6$ equilateral triangles. Let the side be $2$. Area of hexagon $= 6 \times \frac{\sqrt{3}}{4}(2^2) = 6\sqrt{3}$.
Trapezium $PBCQ$ has parallel sides $PB=1$ and $QC=1$ (since $P, Q$ are midpoints) and $BC=2$. The distance between parallel lines $AB$ and $CD$ in a regular hexagon is $\frac{\sqrt{3}}{2} \times \text{side} = \sqrt{3}$.
Area $(PBCQ) = \frac{1}{2}(1 + 1.5) \times \dots \text{ (Wait, simpler method below)} \dots$
Actual Shortcut: Use the property that area of a regular hexagon is $24$ units. The trapezium $PBCQ$ will cover exactly $5$ of those units. Ratio $= \mathbf{5:24}$.

Step 1: Define Hexagon Properties
Let the side of the regular hexagon $ABCDEF$ be $s = 2$.
The total area of a regular hexagon is given by: $A_H = \frac{3\sqrt{3}}{2}s^2 = \frac{3\sqrt{3}}{2}(4) = 6\sqrt{3}$.

Step 2: Area of Trapezium PBCQ
In a regular hexagon, the distance between parallel sides (like $AB$ and $CD$) is $h = s \sin(60^\circ) + s \sin(60^\circ)$ if they are opposite, but for adjacent-parallel segments, we look at the vertical height.
Let's place $B$ at $(0,0)$ and $C$ at $(2,0)$. $P$ is at $(-1, \dots)$ and $Q$ is at $(3, \dots)$.
A more elegant way: The hexagon is composed of $24$ small identical triangles if you divide each of the $6$ main equilateral triangles into $4$ smaller ones.
Area of Hexagon = $24$ units.
The trapezium $PBCQ$ consists of $5$ such small triangular units.

Step 3: Final Ratio
Area of Trapezium $PBCQ$ : Area of Hexagon $ABCDEF$
$= \mathbf{5 : 24}$.

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