Question

In figure, if DEIIBC, find the ratio of ar $\left(\Delta ADE\right)$ and are $\left(\Delta DECB\right)$.

Answer 1

Given, DE || BC, DE= 6cm and BC = 12cm
In $\Delta ABC$ and $\Delta ADE$,
$\angle ABC=\angle ADE$ [corresponding angle]
$\angle ACB=\angle AED$ [corresponding angle]
And $\angle A=\angle A$ [common side]
$\therefore$ $\Delta ABC\sim \Delta AED$ [by AAA similarity criterion]
Then, $\frac{ar\left(\Delta ADE\right)}{ar\left(\Delta ABC\right)}=\frac{(DE{)}^2}{(BC{)}^2}$
$=\frac{(6{)}^2}{(12{)}^2}={\left(\frac{1}{2}\right)}^2$
$\Rightarrow$ $\frac{ar\left(\Delta ADE\right)}{ar\Delta ABC}={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$
Let ar $\left(\Delta ADE\right)=k$, then ar $\left(\Delta ABC\right)=4k$
Now, $ar\left(DECB\right)=ar\left(ABC\right)-ar\left(ADE\right)=4k-k=3k$
$\therefore$ Required ratio = ar(ADE) : ar(DECB) = k : 3k = 1 : 3