CAT 2025 Slot 2 QA Question 8

First, a regular hexagon can be divided into six equilateral triangles, all congruent, with each triangle having a side length equal to the side of the hexagon. So, if I consider the side length of the hexagon as 's', each of these triangles has sides of length 's'. That might help in calculating areas later.

Since the hexagon is regular, all sides are equal, and all internal angles are 120 degrees. The midpoints P and Q are on AB and CD. Let me try to visualize this. So, AB is one side, BC is the next, and so on until FA. P is the midpoint of AB, so AP = PB = s/2. Similarly, Q is the midpoint of CD, so CQ = QD = s/2.

Now, the trapezium PBCQ. A trapezium has two sides parallel. In this case, I think PB and QC are the non-parallel sides, and BC and PQ are the parallel sides. in a trapezium, only two sides are parallel. Let me confirm: in trapezium PBCQ, the sides PB and QC are the legs, and BC and PQ are the bases. So, BC is one base, and PQ is the other base. Since BC is a side of the hexagon, its length is 's'. What about PQ?

To find PQ, I need to figure out the distance between points P and Q. Since P is the midpoint of AB and Q is the midpoint of CD, I can maybe use coordinates to find the distance between them. Let me assign coordinates to the hexagon to make this easier.

Let's place the hexagon on a coordinate system with the center at the origin (0,0). Since it's regular, the vertices can be determined using the unit circle. But since the side length is 's', I need to adjust the radius accordingly. The radius of the circumscribed circle (distance from center to a vertex) in a regular hexagon is equal to the side length 's'. So, each vertex can be represented as (scos(theta), ssin(theta)), where theta is the angle from the positive x-axis.

Let me label the vertices:

Wait, is that correct? Let me check. For a regular hexagon centered at the origin, each vertex is separated by 60 degrees. Starting from point A at (s, 0), the next point B would be at 60 degrees, which is (scos(60°), ssin(60°)) = (s(1/2), s(√3/2)) = (s/2, (s√3)/2). Similarly, point C is at 120 degrees, which is (scos(120°), ssin(120°)) = (-s/2, (s√3)/2). Point D is at 180 degrees, which is (-s, 0). Point E is at 240 degrees, which is (-s/2, -(s√3)/2). Point F is at 300 degrees, which is (s/2, -(s√3)/2). Okay, Now, point P is the midpoint of AB. So, coordinates of P can be found by averaging the coordinates of A and B.

Coordinates of A: (s, 0) Coordinates of B: (s/2, (s√3)/2) So, coordinates of P: x-coordinate: (s + s/2)/2 = (3s/2)/2 = 3s/4 y-coordinate: (0 + (s√3)/2)/2 = (s√3)/4 So, P is at (3s/4, (s√3)/4)

Similarly, point Q is the midpoint of CD.

Coordinates of C: (-s/2, (s√3)/2) Coordinates of D: (-s, 0) So, coordinates of Q: x-coordinate: (-s/2 + (-s))/2 = (-3s/2)/2 = -3s/4 y-coordinate: ((s√3)/2 + 0)/2 = (s√3)/4 So, Q is at (-3s/4, (s√3)/4)

Now, We find the length of PQ. Since P is at (3s/4, (s√3)/4) and Q is at (-3s/4, (s√3)/4), the distance between them is the distance between these two points.

Distance formula: \

So, x2 - x1 = (-3s/4 - 3s/4) = (-6s/4) = (-3s/2) y2 - y1 = (s√3/4 - s√3/4) = 0

So, distance PQ = \$\sqrt{(-3s/2)^2 + 0}$ = \$\sqrt{(9s²/4)}$$ = (3s)/2

Wait, that's interesting. So, PQ is 3s/2. But BC is just 's', right? Because BC is a side of the hexagon.

So, in trapezium PBCQ, the two parallel sides are BC (length s) and PQ (length 3s/2). The height of the trapezium is the distance between these two parallel sides. Hmm, how do I find that?

Alternatively, maybe I can find the area of the trapezium using coordinates. Since I have the coordinates of all four points, Using the shoelace formula to find the area.

Points of trapezium PBCQ:

Let me list them in order: P, B, C, Q.

Using shoelace formula:

Area = 1/2 |sum over i (x_i y_{i