Question

In the given figure, DE ∥ BC and DE : BC = 3 : 5. Calculate the ratio of the areas of ∆ADE and the trapezium BCED.

Answer 1

It is given that DE $\parallel$ BC.
$$\begin{gathered} \therefore \angle ADE=\angle ABC(\mathrm{Corresponding}\mathrm{angles}) \\ \angle AED=\angle ACB(\mathrm{Corresponding}\mathrm{angles}) \end{gathered}$$
Applying AA similarity theorem, we can conclude that $△ADE~△ABC$.

$$\begin{gathered} \therefore \frac{ar(∆ABC)}{ar(∆ADE)}=\frac{B{C}^2}{D{E}^2} \\ \mathrm{Subtracting}1\mathrm{from}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ \frac{\mathrm{a}\mathrm{r}(∆\mathrm{A}\mathrm{B}\mathrm{C})}{\mathrm{a}\mathrm{r}(∆\mathrm{A}\mathrm{D}\mathrm{E})}-1=\frac{5^2}{3^2}-1 \\ \Rightarrow \frac{\mathrm{a}\mathrm{r}(∆\mathrm{A}\mathrm{B}\mathrm{C})-\mathrm{a}\mathrm{r}(∆\mathrm{A}\mathrm{D}\mathrm{E})}{\mathrm{a}\mathrm{r}(∆\mathrm{A}\mathrm{D}\mathrm{E})}=\frac{25-9}{9} \\ \Rightarrow \frac{\mathrm{a}\mathrm{r}\left(\mathrm{BCED}\right)}{\mathrm{a}\mathrm{r}(∆\mathrm{A}\mathrm{D}\mathrm{E})}=\frac{16}{9} \\ \mathrm{or},\frac{\mathrm{a}\mathrm{r}(∆\mathrm{A}\mathrm{D}\mathrm{E})}{\mathrm{a}\mathrm{r}\left(\mathrm{BCED}\right)}=\frac{9}{16} \end{gathered}$$