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

1 Answer

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)

NCERT Solutions for Class 10 Chapter 7-26

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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