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:

(i) $△ADE\sim △ACB$.
(ii) If $AC=13\text{cm},BC=5\text{cm}$ and $AE=4\text{cm}$. Find $DE$ and $AD$.
(iii) Find the ratio- Area of $△ADE$ : Area of quadrilateral $BCED$.

Answer 1

From the question it is given that,
$\angle ABC={90}^{\circ }$
$AB$ and $DE$ is perpendicular to $AC$

$\left(i\right)$ Consider the $△ADE$ and $△ACB$,
$\angle A=\angle A$ … [common angle for both triangle]
$\angle B=\angle E$ … [both angles are equal to ${90}^{\circ }$]
Therefore, $△ADE\sim △ACB$

$\left(ii\right)$ From $\left(i\right)$ we proved that, $△ADE\sim △ACB$
So, $\frac{AE}{AB}=\frac{AD}{AC}=\frac{DE}{BC}$ … [equation$\left(i\right)$]

Consider the $△ABC$, is a right angle triangle

From Pythagoras theorem, we have:
$A{C}^2=A{B}^2+B{C}^2$
${13}^2=A{B}^2+5^2$
$169=A{B}^2+25$
$A{B}^2=169-25$
$A{B}^2=144$
$AB=\sqrt{144}$
$AB=12\text{cm}$
Consider the equation $\left(i\right)$,
$\frac{AE}{AB}=\frac{AD}{AC}=\frac{DE}{BC}$

Taking $\frac{AE}{AB}=\frac{AD}{AC}$

$\frac{4}{12}=\frac{AD}{13}$

$\frac{1}{3}=\frac{AD}{13}$
$AD=\frac{(1\times 13)}{3}\text{cm}$
$AD=4.33\text{cm}$

Now, take $\frac{AE}{AB}=\frac{DE}{BC}$
$\frac{4}{12}=\frac{DE}{5}$
$1/3=DE/5$
$DE=\frac{(5\times 1)}{3}$
$DE=\frac{5}{3}$
$DE=1.67\text{cm}$

$\left(iii\right)$ Now, we have to find area of $△ADE$ : area of quadrilateral $BCED$,
We know that, Area of $△ADE=\frac{1}{2}\times AE\times DE$
$=\frac{1}{2}\times 4\times \frac{5}{3}$
$=\frac{10}{3}{\text{cm}}^2$
Then, area of quadrilateral $BCED=$ area of $△ABC-$ area of $△ADE$
$=\frac{1}{2}\times BC\times AB-\frac{10}{3}$
$=\frac{1}{2}\times 5\times 12-\frac{10}{3}$
$=1\times 5\times 6-\frac{10}{3}$
$=30-\frac{10}{3}$
$=\frac{(90-10)}{3}$
$=\frac{80}{3}{\text{cm}}^2$

So, the ratio of area of $△ADE$ : area of quadrilateral $BCED=\frac{\frac{10}{3}}{\frac{80}{3}}$
$=\frac{10}{3}\times \frac{3}{80}$
$=\frac{1}{8}$
Therefore, area of $△ADE$ : area of quadrilateral $BCED$ is $1:8$.