Question

In ∆ABC, it is given that D is the midpoint of BC; E is the midpoint of BD and O is the midpoint of AE. Then, ar(∆BOE) = ?
(a) $\frac{1}{3}\mathrm{ar}\left(∆ABC\right)$
(b) $\frac{1}{4}\mathrm{ar}\left(∆ABC\right)$
(c) $\frac{1}{6}\mathrm{ar}\left(∆ABC\right)$
(d) $\frac{1}{8}\mathrm{ar}\left(∆ABC\right)$

Answer 1

(d) $\frac{1}{8}$ ar (∆ ABC)

Given: D is the midpoint of BC, E is the midpoint of BD and O is the mid point of AE.
Since D is the midpoint of BC, AD is the median of ∆ABC.
E is the midpoint of BC, so AE is the median of ∆ABD. O is the midpoint of AE, so BO is median of ∆ABE.
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(∆ABE ) =$\frac{1}{2}$​ ⨯​ ar(∆ABD) ...(ii)
ar(∆BOE) = ​$\frac{1}{2}$⨯​ ar(∆ABE) ...(iii)

From (i), (ii) and (iii), we have:
ar(∆BOE ) =​ $\frac{1}{2}$ar(∆ABE)
ar(∆BOE ) = $\frac{1}{2}$ ⨯​ $\frac{1}{2}$​ ⨯​ $\frac{1}{2}$​ ⨯​ ar(∆ABC)​
∴​​ ar(∆BOE )​ = $\frac{1}{8}$ ar(∆ABC)18ar (∆ ABC)