The medians of a triangle ABC is intersect at G. Then

$a r e a (Δ A G B) = \frac{3}{4} a r e a (Δ A B C)$

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$a r e a (Δ A G B) = \frac{2}{3} a r e a (Δ A B C)$

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$a r e a (Δ A G B) = \frac{1}{3} a r e a (Δ A B C)$

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$a r e a (Δ A G B) = \frac{1}{2} a r e a (Δ A B C)$

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Solution

The correct option is C
$a r e a (Δ A G B) = \frac{1}{3} a r e a (Δ A B C)$

The medians intersect at G.

We know G divides the median in the ratio 2 : 1

(i)Area $Δ A B D = \frac{1}{2} a r e a Δ A B C$

(median divides a triangle into two $Δ^{'} s$ of equal area)

$(i i) \frac{a r e a Δ A G B}{a r e a Δ A B D} = \frac{A G}{A D} = \frac{2}{3}$ ......... $(△ A B D a n d △ A B G$ have a common vertex and their bases are in a straight line)

$∴$ area $Δ A G B = \frac{2}{3}$ area $Δ A B D$

$= \frac{2}{3} \times \frac{1}{2}$ area $Δ A B C$ (from (i))

$= \frac{1}{3}$ area $Δ A B C$

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Similar questions

A B C

G

a r (Δ A G B) = a r (Δ B G C) = a r (C G A) = \frac{1}{3} a r (Δ A B C)

\frac{1}{3}

In $△$ ABC, the medians BE and CF intersect at G. AD is a line meeting BC at D. If GD is 1.5 cm, then find the value of AD.

△ A B C

a r (△ A B C) = a r (q u a d r i l a t e r a l A F G E) .

(1, 2, 3)

(3, - 2, 7)

(4, - 3, 6)

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