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In the figure, $C D ¯, A E ¯, B F ¯$ are one third of their respective sides. It follows that $A N 2 ¯ : N 2 N 1 ¯ : N 1 D ¯ = 3 : 3 : 1$ and similarly for lines BE and CF. Then the area of triangle $N 1 N 2 N 3$ is:

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110 Δ A B C

19 Δ A B C

17 Δ A B C

16 Δ A B C

answer is C.

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Detailed Solution

The correct answer is option 3.
Since

C D ¯, A E ¯ and B F ¯

lie on BC, AC and AB respectively

⇒ C D = 13 B C ⇒ 3 C D = B C ⇒ C D : B C = 1 : 2

Similarly,

⇒ A E = 13 A C ⇒ A E : A C = 1 : 2 a n d B F = 13 A B B F : A B = 1 : 2

Now,

Area of triangle = 12 × base × height

A E = 13 A C

Δ A B E = 13 a r Δ A B C

Accordingly,

A N 2 ¯ : N 2 N 1 ¯ : N 1 D ¯ = 3 : 3 : 1

For lines BE and CF,

a r Δ A B N 2 = 67 a r Δ A B E

From the equations,

a r Δ A B N 2 = 67 × 13 a r Δ A B C ⇒ a r Δ A B N 2 = 27 a r Δ A B C

And

a r Δ B F C = 13 a r Δ A B C and a r Δ B N 3 C = 67 a r Δ B F C

From above two equations,

a r Δ B N 3 C = 16 × 13 a r Δ A B C ⇒ a r Δ B N 3 C = 27 a r Δ A B C ⇒ a r Δ A N 1 C = 27 a r Δ A B C

To find the area of triangle

N 1 N 2 N 3

a r Δ N 1 N 2 N 3 = a r Δ A B C − a r Δ A B N 2 − a r Δ B N 3 C − a r Δ A N 1 C

But

The correct answer is option 3.

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