Question

The medians CF and BE of a triangle ABC intersect at G. If area of quadrilateral AFGE is 36 cm², then area AGBC is
0 48 cm2
0 24 cm2
018 cm2
0 36 cm2​

Answer 1

Answer:

area (△FBC) = 1/2 area (△ABC) .....(i)

[ Median divides the triangle into two triangles of equal area ]

area (△EBC) = 1/2 area (△ABC) ......(ii)

From equation (i) and (ii),

area (△FBC) = area (△EBC)

Subtract area (△BGC) from both sides

area(△FBC) - area(△BGC) = area (△EBC) - area (△BGC)

therefore, area(△FGB) = area (△EGC) ....(iii)

area(△ABE) = area (△BEC) [Since, BE is median]

area(△BFG) + area (Quadrilateral AFGE) = area (△BGC) + area (△GEC)

⇒ area (Quadrilateral AFGE) = area (△BGC) [ From equation (iii)]

Step-by-step explanation:

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