Q33 of 50 Page 1

If A (4, 2), B (7, 6) and C (1, 4) are the vertices of a Δ ABC and AD is its median, prove that the median AD divides Δ ABC into two triangles of equal areas.

Given A (4, 2), B (7, 6) and C (1, 4) are the vertices of Δ ABC.


Mid- point of BC, D = (4, 5)



We know that area of triangle = 1/2 |x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) |


Consider Δ ABD,


⇒ Area = 1/2 [4 (6 – 5) + 7 (5 – 2) + 4 (2 – 6)]


= 1/2 [4 + 21 – 16]


Area of Δ ABD = 9/2 square units … (1)


Now Δ ADC,


⇒ Area = 1/2 [4 (5 – 4) + 4 (4 – 2) + 1 (2 – 5)]


= 1/2 [4 + 8 – 3]


Area of Δ ADC = 9/2 square units … (2)


From (1) and (2),


ar (Δ ABD) = ar (Δ ADC)


∴ Median AD divides Δ ABC into two triangles of equal area.


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