In a , point D is on side AB and point E is on side AC, such that BCED is a trapezium. If DE : BC = 3 : 5, then Area( ):Area( )=

In a , point D is on side AB and point E is on side AC, such that…

MCQ

In a $\triangle\text{ABC},$ point $D$ is on side $AB$ and point $E$ is on side $AC,$ such that $\text{BCED}$ is a trapezium. If $DE : BC = 3 : 5,$ then $\text{Area}(\triangle\text{ADE}):\text{Area}(\Box\text{BCED})=$

A
$3 : 4.$
✓
$9 : 16.$
C
$3 : 5.$
D
$9 : 25.$

✓

Answer

Correct option: B.

$9 : 16.$

Given : In $\triangle\text{ABC}, D$ is on side $AB$ and point $E$ is on side $AC,$ such that $\text{BCED}$ is a trapezium. $E : BC = 3 : 5.$
To find : Calculate the ratio of the areas of $\triangle\text{ADE}$ and the trapezium $\text{BCED}$.
In $\triangle\text{ADE}$ and $\triangle\text{ABC},$
$\angle\text{ADE}=\angle\text{B} \ ($Corresponding angles$)$
$\angle\text{A}=\angle\text{A}\ ($Common$)$
$\therefore\triangle\text{ADE}\sim\triangle\text{ABC}\ (\text{AA}$ similarity$)$
We know that
$\frac{\text{Ar}(\triangle\text{ADE})}{\text{Ar}(\triangle\text{ABC})}=\frac{\text{DE}^2}{\text{BC}^2}$
$\frac{\text{Ar}(\triangle\text{ADE})}{\text{Ar}(\triangle\text{ABC})}=\frac{3^2}{5^2}$
$\frac{\text{Ar}(\triangle\text{ADE})}{\text{Ar}(\triangle\text{ABC})}=\frac{9}{25}$
Let area of $\triangle\text{ADE}=9\text{x}\text{ sq}.$ units and area of $\triangle\text{ABC}=25\text{x}\text{ sq}.$ units
$\text{Ar[trap BCED]}=\text{Ar}(\triangle\text{ABC})-\text{Ar}(\triangle\text{ADE})$
$=25\text{x}-9\text{x}$
$=16\text{x sq. units}$
Now,
$\frac{\text{Ar}(\triangle\text{ADE})}{\text{Ar}(\text{trap BCED})}=\frac{9\text{x}}{16\text{x}}$
$\frac{\text{Ar}(\triangle\text{ADE})}{\text{Ar}(\text{trap BCED})}=\frac{9}{16}$
Hence the correct answer is $B.$