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. \(\frac\)

B. \(\frac\)

C. \(\frac\)

D. \(\frac\)

A. \(\frac\)

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

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

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

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

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

Ratio = \(\frac{\frac{9\sqrt}}{27\sqrt+18×\frac{\sqrt}}\)
= \(\frac\)