Question

To the given figure, BC is parallel to DE Area of triangle $ABC=25c{m}^2,$

Area of trapezium BCED $=24c{m}^2$ and DE =14 cm.

The length of BC = ___ cm.

Answer 1

In $\Delta ADE,BC||DE$

Area of $\Delta ABC=25c{m}^2$

and area of trapezium BCED $=24c{m}^2$

Area of $\Delta ADE=25+24=49c{m}^2$

DE =14 cm, Let BC = x cm

Now in $\Delta ABC$ and $\Delta ADE,$

$\angle ABC=\angle ADE$ (corresponding angles)

$\angle A=\angle A$ (common)

$\therefore \Delta ABC\sim \Delta ADE$ (AA postulate)

$\therefore \frac{area\Delta ABC}{area\Delta ADE}=\frac{B{C}^2}{D{E}^2}\Rightarrow \frac{25}{49}=\frac{B{C}^2}{(14{)}^2}$

$\Rightarrow B{C}^2=\frac{25\times 14\times 14}{49}=100=(10{)}^2$

$\therefore BC=10cm.$