Question

D is a Point on side of $\Delta ABC$, such that $BD=\frac{1}{3}BC$, E is the mid-point of AC. What fraction of the area of $\Delta ABC$ is the area of $\Delta DEC$?

• A. $\frac{3}{2}$
• B. $\frac{1}{2}$
• C. $\frac{2}{3}$
• D. $\frac{1}{3}$

Answer 1

The correct option is D $\frac{1}{3}$

Given that D is a point on side of triangle ABC such that $BD=\frac{1}{3}BC$.

And E is the mid point of AC.

Join $B$ to $E$

Then, $BE$ is the median of $\Delta ABC$

Since, median divide $\Delta ABC$ in two equal parts

$\therefore A\left(\Delta ABE\right)=A\left(\Delta ECB\right)=\frac{1}{2}A\left(\Delta ABC\right)$

Now, $A\left(\Delta DEC\right)=\frac{2}{3}A\left(EBC\right)$ ..... $\because BD=\frac{1}{3}BC$

$\therefore A\left(\Delta DEC\right)=\frac{2}{3}\times \frac{1}{2}A\left(\Delta ABC\right)$

$\Rightarrow A\left(\Delta DEC\right)=\frac{1}{3}A\left(\Delta ABC\right)$