MCQ
In a $\triangle\text{ABC},$ 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 $\text{Area}(\triangle\text{ADE}):\text{Area}(\Box\text{BCED})=$
- A3 : 4.
- ✓9 : 16.
- C3 : 5.
- D9 : 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 BCED is a trapezium. DE : BC = 3 : 5.
To find: Calculate the ratio of the areas of $\triangle\text{ADE}$ and the trapezium BCED.
In $\triangle\text{ADE}$ and $\triangle\text{ABC},$
$\angle\text{ADE}=\angle\text{B}$ (Corresponding angles)
$\angle\text{A}=\angle\text{A}$ (Common)
$\therefore\triangle\text{ADE}\sim\triangle\text{ABC}$ (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.
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