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Question 5
You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for Δ ABC whose vertices are A (4, - 6), B (3, - 2) and C (5, 2).

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Solution


Let the vertices of the triangle be A (4, -6), B (3, -2), and C (5, 2).
Let D be the mid-point of side BC of ΔABC.
Therefore, AD is the median in ΔABC.
Coordinates of point D
=(3+52,2+22)=(4,0)
Area of a triangle
=12{x1(y2y3)+x2(y3y1)+x3(y1y2)}
Area of ΔABD
=12[(4){(2)(0)}+3{(0)(6)}
+(4){(6)(2)}]
=12(8+1816)
=3 square units
However, area cannot be negative. Therefore, area of ΔABD is 3 square units.
Area of ΔABD
=12[(4){0(2)}+4{(2)(6)}
+(5){(6)(0)}]
=12(8+3230)
=3 square units
However, area cannot be negative.
Therefore, areas of Δ ABD and Δ ACD are equal i.e. 3 square units.
Thus, median AD has divided Δ ABC in two triangles of equal areas.

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