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The medians BE and CF of a triangle ABC intersect at G. Prove that the area of ∆GBC = area of the quadrilateral AFGE.

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प्रश्न

The medians BE and CF of a triangle ABC intersect at G. Prove that the area of ∆GBC = area of the quadrilateral AFGE.

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उत्तर

Given: The medians BE and CF of a triangle ABC intersect at G

To prove: ar (ΔGBC) = ar (AFCE)

Proof: As median CF divides a triangle into triangle of equal area. So, ar (ΔBCF) = ar (ΔACF)

ar (ΔGBF) + ar (ΔGBC) = ar (AFGE) + ar (ΔGCE)  ...(I)

Now, median BE divides a triangle into two triangle of equal area.

So, ar (ΔGBF) + ar (AFGE) = ar (ΔGCE) + ar (ΔGBC)  ...(II)

Now, subtracting (II) from (I), we get

ar (ΔGBC) – ar (AFGE) = ar (ΔAFGE) – ar (ΔGBC)

ar (ΔGBC) + ar (ΔGBC) = ar (ΔAFGE) + ar (ΔAFGE)

2ar (ΔGBC) = 2ar (AFGE)

Hence, ar (ΔGBC) = ar (AFGE).

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अध्याय 9: Areas of Parallelograms & Triangles - Exercise 9.4 [पृष्ठ ९४]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 9
अध्याय 9 Areas of Parallelograms & Triangles
Exercise 9.4 | Q 3. | पृष्ठ ९४

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