Question

In ∆ABC, if D is the midpoint of BC and E is the midpoint of AD, then ar(∆BED) = ?
(a) $\frac{1}{2}\mathrm{ar}\left(∆ABC\right)$
(b) $\frac{1}{3}\mathrm{ar}\left(∆ABC\right)$
(c) $\frac{1}{4}\mathrm{ar}\left(∆ABC\right)$
(d) $\frac{2}{3}\mathrm{ar}\left(∆ABC\right)$

Answer 1

(c) $\frac{1}{4}$ar (∆ ABC )

Since D is the mid point of BC, AD 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., ar(∆ABD ) = $\frac{1}{2}$ ar(∆ABC) ...(i)

⇒ ar(∆BED) = $\frac{1}{2}$​ ar(∆ABD) ...(ii)

From (i) and (ii), we have:
ar(∆BED) = $\frac{1}{2}$⨯ $\frac{1}{2}$​ ⨯​ ar(∆ABC)
∴​ ar(∆BED)​ = $\frac{1}{4}$⨯ ar(∆ABC)
ar(∆ABC)