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5. You have studied in Class IX, (Chapter 9, Example 3), that a median of a triangle divides it into two triangles of equal areas. Verify this result for D ABC whose vertices are A(4, – 6), B(3, –2) and C(5, 2).

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From the figure:

median triangle

The coordinates of midpoint M of side BC is:

M = \left ( \frac{3+5}{2}, \frac{-2+2}{2} \right ) = \left ( 4,0 \right )

Now, calculating the areas of the triangle ABM and ACM :

Area of triangle, ABM:

Area_{(ABM)} = \frac{1}{2}\left [ 4((-2)-0)+3(0-(-6))+4((-6)-(-2)) \right ]

= \frac{1}{2}\left [ -8+18-16 \right ] = 3\ Square\ units.

Area of triangle, ACM:

Area_{(ACM)} = \frac{1}{2}\left [ 4(0-(-2))+4(2-(-6))+5((-6)-0) \right ]

= \frac{1}{2}\left [ -8+32-30 \right ] = -3\ Square\ units.

However, the area cannot be negative, Therefore, area of \triangle ACM is 3 square units.

Clearly, the median AM divided the \triangle ABC in two equal areas.

Posted by

Divya Prakash Singh

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