Question

If $D$ and $E$ are the mi-points of the sides $AB$ and $AC$, respectively of the $△ABC$ what percent of the whole triangular region is the area of the trapezium $BCED$?

• A. $50$%
• B. $25$%
• C. $75$%
• D. $60$%

Answer 1

The correct option is C $75$%

REF.Image.

Taking origin at A

and position vectors of B

and C as $\left(\overrightarrow{b}\right)$ and $\left(\overrightarrow{c}\right)$

$\Rightarrow$ D and E will have

p.v.s $\frac{\overrightarrow{b}}{2}$ and $\frac{\overrightarrow{c}}{2}$ resp.

Area of $△ABC=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|=\frac{1}{2}|\overrightarrow{b}\times \overrightarrow{c}|$

Area of $△ADE=\frac{1}{2}|\overrightarrow{AD}\times \overrightarrow{AE}|=\frac{1}{2}|\frac{\overrightarrow{b}}{2}\times \frac{\overrightarrow{c}}{2}|=\frac{1}{4}\left(\frac{1}{2}|\overrightarrow{b}\times \overrightarrow{c}|\right)$

$=\frac{1}{4}\left(△ABC\right)$

$\Rightarrow$ area of trapezium $=(1-\frac{1}{4})\left(△ABC\right)$

$=\frac{3}{4}\times ar\left(△ABC\right)$

$\Rightarrow 75\%\Rightarrow \left(C\right)$