Question

ABC is a right-angled triangle with $\angle ABC={90}^{\circ }$. D is any point on AB and DE is perpendicular to AC. Prove that: [4 MARKS]
i) $\Delta ADE\sim \Delta ACB$
ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm, find DE and AD.
iii) Find, area of $\Delta ADE$ : area of quadrilateral BCED

Answer 1

Applying theorems: 2 Marks
Calculation: 2 Marks


i) Given: $\angle ABC={90}^{\circ }$
$DE\perp AC$

$\Rightarrow \frac{ar\left(\Delta AED\right)}{\Delta ABC}=\frac{1}{9}$
$\Rightarrow \frac{ar\left(\Delta ABC\right)}{\Delta AED}=\frac{9}{1}$
$\Rightarrow \frac{ar\left(\Delta AED\right)+ar(quad.BCED)}{ar\left(\Delta AED\right)}=\frac{9}{1}$
$\Rightarrow 1+\frac{ar(quad.BCED)}{ar\left(\Delta AED\right)}=9$
$\Rightarrow \frac{arquad.BCED}{ar\left(\Delta AED\right)}=8$
$\Rightarrow ar\left(\Delta AED\right):ar(quad.BCED)=1:8$

To prove:
$\Delta ADE\sim \Delta ACB$
Proof: Considering $\Delta ADE$ and $\Delta ACB$,
$\angle A=\angle A$ [Common]
$\angle AED=\angle ABC$ [${90}^{\circ }$]
$\therefore \Delta AED\sim \Delta ABC$ [By A.A]

ii) Given: AC = 1.3 cm BC = 5 cm AE = 4 cm
We have, $\Delta AED\sim \Delta ABC$ [Proved above]
$\Rightarrow \frac{AE}{AB}=\frac{ED}{BC}=\frac{AD}{AC}$
$\Rightarrow \frac{4}{AB}=\frac{ED}{5}=\frac{AD}{4}$ .... (1)
In $\Delta ABC$, By pythagoras theorem,
$AB=\sqrt{A{C}^2-B{C}^2}$
$=\sqrt{{13}^2-5^2}=\sqrt{169-25}$
$=\sqrt{144}=12cm$
From (1),
$\frac{4}{12}=\frac{ED}{5}=\frac{AD}{4}$
$\frac{1}{3}=\frac{ED}{5}$
$\Rightarrow ED=\frac{5}{3}=1\frac{2}{3}cm$
Also, $\frac{1}{3}=\frac{AD}{4}$
$\Rightarrow AD=\frac{4}{3}=1\frac{1}{3}cm$

iii) As $\Delta AED\sim \Delta ABC$
$\Rightarrow \frac{ar\left(\Delta AED\right)}{ar\left(\Delta ABC\right)}=\frac{A{E}^2}{A{B}^2}$ [Ratio of areas of two similar ${\Delta }^{\prime }s$ is equal to the ratio of the squares of corresponding sides]
$\Rightarrow \frac{ar\left(\Delta AED\right)}{ar\left(\Delta ABC\right)}=\frac{4^2}{{12}^2}$