CAT 2025

Slot 2

QA

Question & SolutionQues & Sol

Let ABCDEF be a regular hexagon and P and Q be the midpoints of AB and CD, respectively. Then, the ratio of the areas of trapezium PBCQ and hexagon ABCDEF is

1. Concept Used

• Topic: Geometry – Regular Hexagon, Area of Trapezium
• Formula: $$\text{Area of Trapezium} = \frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{Height}$$

2. Calculation

Let the side of the regular hexagon ABCDEF be $a$. Label the vertices in order: A, B, C, D, E, F. P is the midpoint of AB and Q is the midpoint of CD.

Step 1: Area of the regular hexagon.

The standard formula for the area of a regular hexagon with side $a$ is: $$\text{Area of hexagon} = \frac{3\sqrt{3}}{2}a^2$$

Step 2: Identify the parallel sides of trapezium PBCQ.

In a regular hexagon with side $a$, BC is a side of the hexagon, so $BC = a$.

P is the midpoint of AB, so $PB = \frac{a}{2}$.

Q is the midpoint of CD, so $CQ = \frac{a}{2}$.

Now, PQ connects the midpoint of AB to the midpoint of CD. Using the property of a regular hexagon, the full diagonal AD (spanning from vertex A to vertex D, passing through the center) has length $2a$. The segment PQ is the average of BC and AD (since P and Q are midpoints of AB and CD respectively): $$PQ = \frac{BC + AD}{2} = \frac{a + 2a}{2} = \frac{3a}{2}$$

So the two parallel sides of trapezium PBCQ are $BC = a$ and $PQ = \frac{3a}{2}$.

Step 3: Find the height of trapezium PBCQ.

In a regular hexagon, the perpendicular distance between two opposite parallel sides (e.g., BC and EF) equals $\sqrt{3}a$.

The perpendicular distance between BC and AD (the line through the midpoints P and Q) is half of the full distance between BC and the opposite parallel side: $$\text{Distance between BC and AD} = \sqrt{3}a$$

Since PQ lies halfway between BC and the far side (at the midpoints of AB and CD), the height of the trapezium PBCQ is: $$h = \frac{\sqrt{3}a}{4}$$

(This is because the full vertical span from BC to AD is $\sqrt{3}a$, and the midpoint line PQ sits at half of that span, i.e., $\frac{\sqrt{3}a}{2}$ from BC to the full line AD. But since PQ is at the midpoint of AB and CD segments, the height from BC to PQ is $\frac{\sqrt{3}a}{4}$.)

Step 4: Compute the area of trapezium PBCQ. $$\text{Area of PBCQ} = \frac{1}{2} \times (BC + PQ) \times h = \frac{1}{2} \times \left(a + \frac{3a}{2}\right) \times \frac{\sqrt{3}a}{4}$$ $$= \frac{1}{2} \times \frac{5a}{2} \times \frac{\sqrt{3}a}{4} = \frac{5\sqrt{3}a^2}{16}$$

Step 5: Compute the ratio. $$\text{Ratio} = \frac{\text{Area of PBCQ}}{\text{Area of hexagon}} = \frac{\dfrac{5\sqrt{3}a^2}{16}}{\dfrac{3\sqrt{3}a^2}{2}} = \frac{5\sqrt{3}a^2}{16} \times \frac{2}{3\sqrt{3}a^2} = \frac{10}{48} = \frac{5}{24}$$

So the ratio of the area of trapezium PBCQ to the area of hexagon ABCDEF is $5 : 24$.

3. Solution

Answer = Option B ✅

The ratio of the area of trapezium PBCQ to the area of hexagon ABCDEF is $\mathbf{5 : 24}$.