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Answer :

The vertices of the triangle ABC are A (0, -1), B (2, 1), C (0, 3).

Let D, E, F be the mid-points of the sides AB,BC,CA of this triangle ABC respectively.

Coordinates of D, E, and F are given by :

D = \( (\frac{0+2}{2} \ , \ \frac{-1+1}{2}) \) = (1,0)

E = ( \( \frac{0+2}{2} \ , \ \frac{1+3}{2} \) ) = (1,2)

F = ( \( \frac{0+0}{2} \ , \ \frac{3-1}{2} \) ) = (0,1)

So, Area of \( ∆ \ DEF \ = \ \frac{1}{2} \ | \ [ 1(2-1) + 1(1-0) + 0(0-2)] \ | \ = \ 1 \)

∴ Area of ∆ DEF = 1 sq. units

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

\(\therefore \) Area of \(\triangle \) ABC = 4 sq. units

So, \( ∆ \ DEF \ : \ ∆ \ ABC \ = \ 1 \ : \ 4 \).

- Find the area of the triangle whose vertices are: (i) (2, 3), (-1, 0), (2, -4) (ii) (-5, -1), (3, -5), (5, 2)
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- Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).
- You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for \(\triangle \) ABC whose vertices are A (4, – 6), B (3, – 2) and C (5, 2).

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