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:
Show Hint
- \(6 : 19\)
- \(5 : 24\)
- \(6 : 25\)
- \(7 : 24\)
The Correct Option is B
Approach Solution - 1
Step 1: Let the side length of the regular hexagon be \(s\). The area of a regular hexagon is given by \[ \text{Area}_{\text{hex}} = \frac{3\sqrt{3}}{2}s^2. \] Step 2: Fix a coordinate system. Place the hexagon such that \[ B = (0,0), \quad C = (s,0). \] In a regular hexagon, consecutive sides subtend an angle of \(60^\circ\). Hence, the adjacent vertices can be taken as \[ A = \left(-\frac{s}{2}, \frac{\sqrt{3}s}{2}\right), \quad D = \left(\frac{3s}{2}, \frac{\sqrt{3}s}{2}\right). \] Step 3: Determine the midpoints \(P\) and \(Q\). The midpoint of \(AB\) is \[ P = \left(\frac{-\frac{s}{2} + 0}{2}, \frac{\frac{\sqrt{3}s}{2} + 0}{2}\right) = \left(-\frac{s}{4}, \frac{\sqrt{3}s}{4}\right). \] The midpoint of \(CD\) is \[ Q = \left(\frac{s + \frac{3s}{2}}{2}, \frac{0 + \frac{\sqrt{3}s}{2}}{2}\right) = \left(\frac{5s}{4}, \frac{\sqrt{3}s}{4}\right). \] Step 4: Find the area of trapezium \(PBCQ\). Points \(P\) and \(Q\) have the same \(y\)-coordinate, so \(PQ\) is parallel to \(BC\). The height of the trapezium is \[ h = \frac{\sqrt{3}s}{4}. \] The lengths of the parallel sides are \[ BC = s, \qquad PQ = \frac{5s}{4} - \left(-\frac{s}{4}\right) = \frac{3s}{2}. \] Hence, the area of the trapezium is \[ \text{Area}_{\text{trap}} = \frac{1}{2}(BC + PQ)\cdot h = \frac{1}{2}\left(s + \frac{3s}{2}\right)\cdot \frac{\sqrt{3}s}{4} = \frac{5\sqrt{3}}{16}s^2. \] Step 5: Compute the ratio of the areas. \[ \frac{\text{Area}_{\text{trap}}}{\text{Area}_{\text{hex}}} = \frac{\frac{5\sqrt{3}}{16}s^2}{\frac{3\sqrt{3}}{2}s^2} = \frac{5}{16}\cdot\frac{2}{3} = \frac{5}{24}. \] Therefore, the ratio of the area of trapezium \(PBCQ\) to the area of the hexagon is \[ 5 : 24. \]
Approach Solution -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 \(5 : 24\).
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