D is the mid-point of side BC of $Δ A B C$ and E is the mid point of BD. If O is the midpoint of AE.Then Area of ( $Δ B O E$ ) .

\frac{1}{8} A r e a (Δ A B C)

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\frac{1}{6} A r e a (Δ A B C)

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\frac{1}{4} A r e a (Δ A B C)

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\frac{1}{3} A r e a (Δ A B C)

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Solution

The correct option is A $\frac{1}{8} A r e a (Δ A B C)$

Since AD is the median,
$∴ 2 a r (Δ A B D) = a r (Δ A B C)$ ...(i)
$S i m i l a r l y, 2 a r (Δ A B E) = a r (Δ A B D) . . . (i i)$
$A n d 2 a r (Δ B O E) = a r (Δ A B E) . . . (i i i)$
$N o w, a r (Δ A B C) = 2 a r (Δ A B D)$
$= 2 [2 a r (Δ A B E)]$ from (ii)
$= 4 [2 a r (Δ B O E)]$ from (iii)
$= 8 a r (Δ B O E)$
$∴ \frac{1}{8} a r (Δ A B C) = a r (Δ B O E)$

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Δ A B C

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D is the mid-point of side BC of $Δ A B C$ and E is the mid point of BD. If O is the midpoint of AE, then area of ( $Δ B O E$ ) is equal to

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\frac{1}{8}

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