Question

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:

• A. \(6 : 19\)
• B. \(5 : 24\)
• C. \(6 : 25\)
• D. \(7 : 24\)

Hint:

For regular polygons, coordinate geometry is very handy: place convenient vertices on the axes, find key points (like midpoints) using averages of coordinates, and then use distance or area formulas to compute lengths and areas.

Answer 1

The Correct Option is B

Approach: A regular hexagon splits neatly into 6 unit equilateral triangles (or, better here, into a grid of small triangles). Cut the hexagon along the two midpoints and just count what fraction of those small triangles the trapezium covers — no coordinates needed if you set them up smartly, but coordinates make the parallel sides instant.

Step 1: Let the side be \(s\). Area of the regular hexagon: \[ \text{Area}_{\text{hex}} = \frac{3\sqrt{3}}{2}s^2. \]

Step 2: Put \(B=(0,0)\) and \(C=(s,0)\) along the bottom side. Walking around the hexagon, the neighbours of \(B\) and \(C\) sit at \[ A = \left(-\tfrac{s}{2},\ \tfrac{\sqrt{3}s}{2}\right), \qquad D = \left(\tfrac{3s}{2},\ \tfrac{\sqrt{3}s}{2}\right). \]

Step 3: Midpoints: \[ P = \text{mid}(A,B) = \left(-\tfrac{s}{4},\ \tfrac{\sqrt{3}s}{4}\right), \qquad Q = \text{mid}(C,D) = \left(\tfrac{5s}{4},\ \tfrac{\sqrt{3}s}{4}\right). \]

Step 4: \(P\) and \(Q\) share the same height, so \(PQ \parallel BC\) and the trapezium \(PBCQ\) has parallel sides \[ BC = s, \qquad PQ = \tfrac{5s}{4} - \left(-\tfrac{s}{4}\right) = \tfrac{3s}{2}, \] with height \(h = \tfrac{\sqrt{3}s}{4}\) (the vertical gap). Hence \[ \text{Area}_{\text{trap}} = \tfrac{1}{2}(BC+PQ)\,h = \tfrac{1}{2}\!\left(s+\tfrac{3s}{2}\right)\!\cdot\tfrac{\sqrt{3}s}{4} = \frac{5\sqrt{3}}{16}s^2. \]

Step 5: Ratio: \[ \frac{\text{Area}_{\text{trap}}}{\text{Area}_{\text{hex}}} = \frac{\tfrac{5\sqrt{3}}{16}s^2}{\tfrac{3\sqrt{3}}{2}s^2} = \frac{5}{16}\cdot\frac{2}{3} = \frac{5}{24}. \] Final answer: \(5 : 24\).

Answer 2

Step 1: Let the side length of the regular hexagon be $s$. The area of a regular hexagon is \[ \text{Area}_{\text{hex}} = \frac{3\sqrt{3}}{2}s^2. \]
Step 2: Place the hexagon on a coordinate plane. Let \[ B = (0,0), \quad C = (s,0). \] For a regular hexagon with side $s$, the adjacent vertices are separated by $60^\circ$. So we can take \[ A = \left(-\frac{s}{2}, \frac{s\sqrt{3}}{2}\right), \quad D = \left(\frac{3s}{2}, \frac{s\sqrt{3}}{2}\right). \]
Step 3: Find midpoints $P$ and $Q$. $P$ is the midpoint of $AB$: \[ P = \left(\frac{-\frac{s}{2} + 0}{2},\, \frac{\frac{s\sqrt{3}}{2} + 0}{2}\right) = \left(-\frac{s}{4}, \frac{s\sqrt{3}}{4}\right). \] $Q$ is the midpoint of $CD$: \[ Q = \left(\frac{s + \frac{3s}{2}}{2},\, \frac{0 + \frac{s\sqrt{3}}{2}}{2}\right) = \left(\frac{5s}{4}, \frac{s\sqrt{3}}{4}\right). \]
Step 4: Area of trapezium $PBCQ$. Since $P$ and $Q$ have the same $y$-coordinate, segment $PQ$ is parallel to $BC$ (which lies on the $x$-axis). \[ \text{Height } h = \frac{s\sqrt{3}}{4}. \] Lengths of the parallel sides: \[ BC = s,\qquad PQ = x_Q - x_P = \frac{5s}{4} - \left(-\frac{s}{4}\right) = \frac{3s}{2}. \] Thus, \[ \text{Area}_{\text{trap}} = \frac{1}{2}(BC + PQ)\cdot h = \frac{1}{2}\left(s + \frac{3s}{2}\right)\cdot \frac{s\sqrt{3}}{4} = \frac{1}{2} \cdot \frac{5s}{2} \cdot \frac{s\sqrt{3}}{4} = \frac{5\sqrt{3}}{16}s^2. \]
Step 5: Ratio of areas. \[ \text{Ratio} = \frac{\text{Area}_{\text{trap}}}{\text{Area}_{\text{hex}}} = \frac{\frac{5\sqrt{3}}{16}s^2}{\frac{3\sqrt{3}}{2}s^2} = \frac{5}{16} \cdot \frac{2}{3} = \frac{10}{48} = \frac{5}{24}. \] Therefore, the ratio of the areas is \(5 : 24\).