In triangle ABC, X and Y are the points on sides AB and AC, respectively, such that XY is parallel to BC. If XY : BC = 2.5 : 7, what is the ratio of the area of the trapezium BCYX to that of the ΔAXY?

$\frac{171}{25}$

The ratio of XY : BC = 2.5 : 7 = 5 : 14

Now the areas = Square of the sides

\(\frac{arAXY}{arABC}\) = \(\frac{5^2}{14^2}\)

\(\frac{arAXY}{arABC}\) = \(\frac{25}{196}\)

Now the area of trapezium XYBC = 196 - 25 = 171

So, the ratio of the area of the trapezium BCYX to that of the ΔAXY = $\frac{171}{25}$