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: Pure ratio reasoning, no coordinates. The line $PQ$ joining the two midpoints cuts the hexagon into a triangle-fraction grid; build the trapezium area from the basic equilateral triangle of side $s$ and a small corner triangle.

Step 1: Take side $s=2$ so midpoints land on whole numbers; then $AP=PB=1$ and $CQ=QD=1$. The hexagon area is $\dfrac{3\sqrt3}{2}(2)^2 = 6\sqrt3$.

Step 2: Drop the trapezium $PBCQ$ onto the standard hexagon. Look at the big triangle $ABC$ inside the hexagon: with $B$ at the bottom corner, $A$ up-left and $C$ to the right, triangle $PBCQ$ is triangle-shaped region trimmed by joining the midpoints $P$ (on $AB$) and $Q$ (on $CD$).

Step 3: Compute directly with the trapezium formula using the two parallel chords. The lower chord is $BC = s = 2$. The upper chord $PQ$ runs between the midpoint of $AB$ and the midpoint of $CD$; by the geometry of the hexagon this length is $PQ = \tfrac{3}{2}s = 3$. The perpendicular distance between these two parallel chords is half the height of the band $ABCD$, namely $\tfrac{\sqrt3}{4}s = \tfrac{\sqrt3}{2}$. Then \[ \text{Area}_{PBCQ} = \frac12(BC+PQ)\times h = \frac12(2+3)\times\frac{\sqrt3}{2} = \frac{5\sqrt3}{4}. \]

Step 4: Form the ratio against $6\sqrt3$: \[ \frac{\text{Area}_{PBCQ}}{\text{Area}_{\text{hex}}} = \frac{5\sqrt3/4}{6\sqrt3} = \frac{5}{24}. \] Final answer: $5 : 24$.