Question

In the given figure, $AB$ and $DE$ are perpendicular to $BC$.
(i) Prove that $△ABC\sim △DEC$
(ii) If $AB=6cm:DE=4cm$ and $AC=15cm$, calculate $CD$.
(iii) Find the ratio of the area of $△ABC$ : area of $△DEC$.

Answer 1

(i) From $\Delta ABC$ and $\Delta DEC$,
$\angle ABC=\angle DEC={90}^{\circ }$ (Given)
and $\angle ACB=\angle DCE=$ Common
$\therefore$ $\Delta ABC\sim \Delta DEC$ (By $A-A$ similarity)
(ii) In $\Delta$ABC and $\Delta$DEC,
$\Delta ABC\sim \Delta DEC$ (proved in (i) part)
$\therefore \frac{AB}{DE}=\frac{AC}{CD}$
Given: $AB=6cm,DE=4cm,AC=15cm$,
$\therefore \frac{6}{4}=\frac{15}{CD}$
$\Rightarrow 6\times CD=15\times 4$
$\Rightarrow CD=\frac{60}{6}$
$\Rightarrow CD=10cm$.
(iii) $\frac{Areaof\Delta ABC}{Areaof\Delta DEC}=\frac{A{B}^2}{D{E}^2}$ $(\because \Delta ABC\sim \Delta DEC)$
$=\frac{(6{)}^2}{(4{)}^2}=\frac{36}{16}=\frac{9}{4}$
$\therefore$ Area of $\Delta ABC:$ Area of $\Delta DEC=9:4$.