Question

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.​

Answer 1

SOLUTION :-

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

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

Using the mid-point formula, coordinates of D, E, and F are:

D = [(0+2)/2, (-1+1)/2] = (1, 0)

E = [(0+0)/2, (-1+3)/2] = (0, 1)

F = [(0+2)/2, (3+1)/2] = (1, 2)

We know that,

Area of triangle = ½ [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

Area of triangle DEF = ½ {(1(2 – 1) + 1(1 – 0) + 0(0 – 2)}

= ½ (1 + 1)

= 1

.°. Area of triangle DEF = 1 sq.unit

Area of triangle ABC = ½ {0(1 – 3) + 2(3 – (-1)) + 0(-1 – 1)}

= ½ (8)

= 4

.°. Area of triangle ABC = 4 sq.units

Hence, the ratio of the area of triangle DEF and ABC = 1 : 4.