Question 32

# Let A(O,-1), B(0,3) and C(2,1) be three points. Let $$\triangle_1$$ be the area of the triangle ABC and $$\triangle_2$$ be the area of the triangle formed by the mid points of the sides of the triangle whose vertices are A, B and C such that $$\frac{\triangle_1}{\triangle_2} = \frac{1}{x}$$ Find the value of $$x$$.

Solution

Let the vertices of the triangle be A(0,−1),C (2,1), B(0,3)

Let D, E, F be the mid-points of the sides of this triangle.

Let $$\triangle_1$$ be the area of the triangle ABC and $$\triangle_2$$ be traiangle DEF

Coordinates of D, E, and F are given by
D = $$(\frac{0+2}{2},\frac{-1+1}{2}= (1,0)$$
E = $$(\frac{0+0}{2},\frac{-1+3}{2})= (0,1)$$
F = $$(\frac{2+0}{2},\frac{1+3}{2})= (1,2)$$

Area of a triangle $$=\frac{1}{2} [x1(y2-y3) + x2(y3-y1) + x3(y1-y2)]$$

Area of △DEF = $$=\frac{1}{2} [1(2-1) + 1(1-0) + 0(0-2)]=1 unit$$

Area of △ABC =$$=\frac{1}{2} [0(1-3) + 2(3-(-1)) + 0(-1-1)]=4 unit$$

$$\frac{\triangle_1}{\triangle_2} = \frac{1}{x} =\frac{1}{4}$$

So $$x = 4$$

Option A is correct.