Q. 103.6( 11 Votes )

# The vertices at Δ

Answer :

Let us find area of ΔABC

Area of triangle is given by formula

Area = × [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

Where (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) are vertices of triangle

Here,

A = (x_{1}, y_{1}) = (-1, 5)

B = (x_{2}, y_{2}) = (3, 1)

C = (x_{3}, y_{3}) = (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 unit^{2} … (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 = × [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

Where (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) are vertices of triangle

Here,

D = (x_{1}, y_{1}) = (1, 3)

E = (x_{2}, y_{2}) = (4, 4)

F = (x_{3}, y_{3}) = (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 unit^{2} … (ii)

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

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