Question

In the given figure, $DE\parallel BC$.
(i) Prove that $△ADE$ and $△ABC$ are similar.
(ii) Given taht $AD=\frac{1}{2}BD$, calculate $DE$ if $BC=4.5cm$.
(iii) If area of $△ABC=18c{m}^2$, find the area of trapezium $DBCE$.

Answer 1

(i) From the question it is given that, $DE\parallel BC$
We have to prove that, $△ADE$ and $△ABC$ are similar
$\angle A=\angle A$ … [common angle for both triangles]
$\angle ADE=\angle ABC$ … [because corresponding angles are equal]
Therefore, $△ADE\sim △ABC$ … [$AA$ axiom]
(ii) From (i) we proved that, $△ADE\sim △ABC$
Then, $AD/AB=AB/AC=DE/BC$
So, $AD/(AD+BD)=DE/BC$
$\left(\frac{1}{2}BD\right)/\left(\right(\frac{1}{2}BD)+BD)=DE/4.5$
$\left(\frac{1}{2}BD\right)/\left(\right(3/2\left)BD\right)=DE/4.5$
$\frac{1}{2}\times \left(2/3\right)=DE/4.5$
$1/3=DE/4.5$
Therefore, $DE=4.5/3$
$DE=1.5cm$
(iii) From the question it is given that, area of $△ABC=18c{m}^2$
Then, area of $△ADE$/area of $△ABC=D{E}^2/B{C}^2$ area of $△ADE/18=(DE/BC{)}^2$
area of $△ADE/18=(AD/AB{)}^2$
area of $△ADE/18=(1/3{)}^2=1/9$
area of $△ADE=18\times 1/9$
area of $△ADE=2$
So, area of trapezium $DBCE=$ area of $△ABC-$ area of $△ADE$
$=18-2$
$=16c{m}^2$.