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.

Asked by Pragya Singh | 1 year ago |  65

##### Solution :-

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

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

Coordinates of D, E, and F are given by

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

E = ( 0+0/2, -1+$$\dfrac{3}{2}$$) = (0, 1)

F = ( 0+$$\dfrac{2}{2}$$, 3+$$\dfrac{1}{2}$$) = (1, 2)

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

Area of ΔDEF =$$\dfrac{1}{2}$${1(2-1) + 1(1-0) + 0(0-2)}

= $$\dfrac{1}{2}$$(1+1) = 1

Area of ΔDEF is 1 square units

Area of ΔABC =$$\dfrac{1}{2}$$[0(1-3) + 2{3-(-1)} + 0(-1-1)]

$$\dfrac{1}{2}$$ {8} = 4

Area of ΔABC is 4 square units

Therefore, the required ratio is 1:4.

Answered by Abhisek | 1 year ago

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