Trapezium With Quadrilateral

Let $X_1$ , $X_2$ , $\cdots$ , $X_n$ be the points on an $\mathbb R^2$ plane. In the following, we denote the area of an $n$ -sided polygon $X_1X_2\cdots X_n$ by $(X_1 X_2 \cdots X_n)$ .

$\textbf{Solution}$

Let $AB=5k$ , $DE = 7k$ , $BC=2h$ and $CD = 3h$ for some positive constants $h$ and $k$ (as labelled in the figure).

In the figure, we construct a point $G$ on $AD$ such that joining $CG$ yields the result $CG$ // $BA$ . Then we have $CG$ // $DE$ . Now, it is readily seen that

$\begin{aligned} &(1) & \triangle CGF &\sim \triangle EDF,\ \text{ and } \\ &(2) & \triangle DAB &\sim \triangle DGC. \end{aligned}$

For (2), the proportionality of the corresponding sides gives $CG = 3k$ ; and then, for (1), it gives $GF : DF = CG : ED = 3:7$ . Since $\triangle CFG$ and $\triangle CFD$ have the same height relative to their respective bases $GF$ and $DF$ , we have $(CGF) = (CDF) \cdot \frac{GF}{DF} = 63 \times \frac{3}{7} = 27 \text{cm}^2$ and so $(GCD) = (CDF) + (CGF) = 63 + 27 = 90\ \text{cm}^2$ .

For (2), we have $(ABD) = (GCD) \left(\frac{AB}{GC}\right)^2 = 90 \cdot \left(\frac{5}{3}\right)^2 = 250\ \text{cm}^2$ .

Finally, we have $(ABCF) = (ABD) - (CDF) = 250 - 63 = 187\ \text{cm}^2$ . $\Box$