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Answer :

The vertices of triangle ABC are A(4,-6), B (3, – 2) and C (5, 2).

Let D be the mid-point of side BC of ∆ ABC. Therefore, AD is the median in ∆ ABC.

So, coordinates of D are ( \( \frac{3+5}{2} \ , \ \frac{-2+2}{2} \) )

Therefore, coordinates of D = (4,0)

Now,

Area of ∆ ABD = \( \frac{1}{2} \ | \ [4(-2-0) \ + \ 3(0+6) \ + \ 4(-6+2)] | \)

= \( \frac{1}{2} \ | \ [ -8 + 18 - 16] \ | \)

= \( |-3| \ = \ 3 \) sq. units

Similarly,

Area of ∆ ACD = \( \frac{1}{2} \ | \ [4(0-2) \ + \ 4(2+6) \ + \ 5(-6-0)] \ | \)

= \( \frac{1}{2} \ | \ [ -8 + 32 - 30] \ | \)

= \( |-3| \ = \ 3 \) sq. units

\(\therefore \) Clearly, Area of ∆ ABD = Area of ∆ ACD = 3 sq. units.

Hence, median of a triangle divides it into two triangles of equal areas.

- Find the area of the triangle whose vertices are: (i) (2, 3), (-1, 0), (2, -4) (ii) (-5, -1), (3, -5), (5, 2)
- In each of the following find the value of ‘k’, for which the points are collinear. (i) (7, -2), (5, 1), (3, -k) (ii) (8, 1), (k, -4), (2, -5)
- Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
- Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).

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