Q. A regular hexagon ABCDEF has a side length of 6 cm. BF and BE are diagonals of the hexagon which intersect the diagonal AD at points P and Q respectively. What is the ratio of the area of triangle BPQ to that of the quadrilateral ABDF ?

A. \(\frac{1}{8}\)

B. \(\frac{1}{6}\)

C. \(\frac{1}{7}\)

D. \(\frac{1}{4}\)

A. \(\frac{1}{8}\)

In Δ APB, AB = 6
AP = 3
PB = \(3\sqrt{3}\)
AD = BE = 12

So, Δ PBD = 90
BQ = 1/2 BE = 6, QD = 1/2 AD = 6

In Δ PQB
opp 60 → PB = \(3\sqrt{3}\)
OPP 30 → PQ = \( \frac{3\sqrt{3}}{\sqrt{3}}\)
= 3

Area of PQB = \(\frac{1}{2} × PQ × PB\)
= \(\frac{1}{2}×3×3\sqrt{3}\)
= \(\frac{9\sqrt{3}}{2}\)

Area of Quad ABDF = Area of eq Δ BDF + Area of iso Δ ABF
= \(\frac{\sqrt{3}}{4}×{(6\sqrt{3}})^{2}+\frac{1}{2}×6×6×\sin 120°\)

Ratio = \(\frac{\frac{9\sqrt{3}}{2}}{27\sqrt{3}+18×\frac{\sqrt{3}}{2}}\)
= \(\frac{1}{8}\)