In ∆ABC, if D is the midpoint of BC and E is the midpoint of AD, then ar(∆BED) = ?

Question:

In ABC, if D is the midpoint of BC and E is the midpoint of AD, then ar(∆BED) = ?

(a) $\frac{1}{2} \operatorname{ar}(\Delta A B C)$

(b) $\frac{1}{3} \operatorname{ar}(\Delta A B C)$

(c) $\frac{1}{4} \operatorname{ar}(\Delta A B C)$

(d) $\frac{2}{3} \operatorname{ar}(\Delta A B C)$

 

Solution:

(c) $\frac{1}{4} \operatorname{ar}(\triangle A B C)$

Since D is the mid point of BCAD is a median of ∆ABC and BE is the median of ∆ABD.
We know that a median of a triangle divides it into two triangles of equal areas.

i.e., $\operatorname{ar}(\Delta A B D)=\frac{1}{2} \operatorname{ar}(\Delta A B C)$             (i)

$\Rightarrow \operatorname{ar}(\triangle B E D)=\frac{1}{2} \operatorname{ar}(\triangle A B D)$          ...(ii)

From (i) and (ii), we have:

$\operatorname{ar}(\Delta B E D)=\frac{1}{2} \times \frac{1}{2} \times \operatorname{ar}(\Delta A B C)$

$\therefore \operatorname{ar}(\Delta B E D)=\frac{1}{4} \times \operatorname{ar}(\triangle A B C)$

 

 

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