PQ > BC in the displayed figure, and 3AP > 4PB. What will the ratio between ABC and PQCB be?

According to the theorem,

$\frac{\text{ area of }∆ABC}{\text{ area of }∆APQ}={\left(\frac{AB}{PB}\right)}^2$

$\frac{\text{ area of triangle }ABC}{\text{ area of triangle }APQ}={\left(\frac{4}{3}\right)}^2$

$\frac{\text{ area of }∆ABC}{\text{ area of }∆APQ}=\frac{16}{9} ..........\left(1\right)$

Now,$\frac{\text{ area of }∆APQ+\text{ area of trap.PQCB }}{\text{ area of }∆APQ}=\frac{16}{9}$

$1+\frac{\text{ area of trap.PQCB }}{\text{ area of }∆APQ}=\frac{16}{9}$

$\frac{\text{ area of trap.PQCB }}{\text{ area of }∆APQ}=\frac{16}{9}-1$

$\frac{\text{ area of trap.PQCB }}{\text{ area of }∆APQ}=\frac{16-9}{9}=\frac{7}{9}$ (2)

Dividing (1) by (2),

$\frac{\text{ area of }∆ABC}{\text{ area of trap.PQCB }}=\frac{16}{7}= 16 : 7$