# The vertices at Δ

Let us find area of ΔABC

Area of triangle is given by formula

Area = × [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

Where (x1, y1), (x2, y2) and (x3, y3) are vertices of triangle

Here,

A = (x1, y1) = (-1, 5)

B = (x2, y2) = (3, 1)

C = (x3, y3) = (5, 7)

Substituting values

Area(ΔABC) = × [(-1)(1 – 7) + 3(7 – 5) + 5(5 – 1)]

Area(ΔABC) = × [6 + 6 + 20]

Area(ΔABC) = × 32

Area(ΔABC) = 16 unit2 … (i)

Let us now find the midpoints of BC, AC and AB i.e. points D, E and F respectively

Point F is the midpoint of AB

F =

F = (1, 3)

Point D is the midpoint of BC

D =

D = (4, 4)

Point E is the midpoint of AC

E =

E = (2, 6)

Let us find area of ΔDEF

Area = × [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

Where (x1, y1), (x2, y2) and (x3, y3) are vertices of triangle

Here,

D = (x1, y1) = (1, 3)

E = (x2, y2) = (4, 4)

F = (x3, y3) = (2, 6)

Substituting values

Area(ΔDEF) = × [1(4 – 6) + 4(6 – 3) + 2(3 – 4)]

Area(ΔDEF) = × [-2 + 12 + (-2)]

Area(ΔDEF) = × 8

Area(ΔDEF) = 4 unit2 … (ii)

From (i) and (ii) we can conclude that area(ΔABC) = 4 × area(ΔDEF)

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