The medians CF and BE of a triangle ABC intersect at G. If area of quadrilateral AFGE is $36 c m^{2},$ then area $Δ G B C$ is

48 c m^{2}

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24 c m^{2}

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18 c m^{2}

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36 c m^{2}

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Solution

The correct option is D
$36 c m^{2}$

Given that the medians CF and BE of ∆ABC intersect at G. Let AH be the median from vertex A and AM be the perpendicular on side BC.

Here, $B H = H C, A E = E C a n d A F = F B N o w, A r (Δ A B H) = \frac{1}{2} \times B a s e \times A l t i t u d e o f Δ A B H = \frac{1}{2} \times B H \times A M = \frac{1}{2} \times H C \times A l t i t u t e o f Δ A H C (∵ H C = B H) = A r (Δ A C H)$

Thus, medium divides a triangle into two equal halves.

Similarly, $A r (Δ B G F) = A r (Δ A G F) . . . (i) A r (Δ A G E) = A r (Δ G E C) . . . (i i) a n d A r (Δ B G H) = A r (Δ G H C) . . (i i i) N o w, A r (Δ A B H) = A r (Δ A H C) \Rightarrow A r (Δ B G F) + A r (Δ B G H) + A r (Δ A G F) = A r (Δ A G E) + A r (Δ G E C) + A r (Δ G H C) \Rightarrow 2 A r (Δ A G F) = 2 A r (Δ A G E) [From (i), (ii) and (iii)] \Rightarrow A r (Δ A G F) = A r (Δ A G E) \dots (i v) A l s o, A r (Δ A B E) = A r (Δ B E C) \Rightarrow A r (Δ B G F) + A r (Δ A G F) + A r (Δ A G E) = A r (Δ B G H) + A r (Δ G H C) + A r (Δ G E C) \Rightarrow 2 A r (Δ B G F) = 2 A r (Δ B G H) \Rightarrow A r (Δ B G F) = A r (Δ B G H) . . (v) From (i), (ii), (iii), (iv) and (v), we get A r (Δ D G F) = A r (Δ A G F) = A r (Δ A G E) = A r (Δ G E C) = A r (Δ G H C) = A r (Δ B G H) T h u s, A r (Δ G B C) = 2 A r (Δ B G H) = 2 A r (Δ A F G) [From (iii)] = A r (Δ A F G) + A r (Δ A G E) [From (iv)] = A r (A F G E) = 36 c m^{2}$

Hence, the correct answer is option (d).

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In a triangle ABC,the medians BE and CF intersect at G.Prove that $a r (△ B C G) = a r (A F G E)$ .

c m^{2}

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