KVPY2020MathematicsProperties of Triangles
Let A B C be a triangle such that A B = 4 , B C = 5 and C A = 6 . Choose points D , E , F on A B , B C , C A respectively, such that A D = 2 , B E = 3 , C F = 4 . Then area   Δ D E F area   Δ A B C is
Options
- A1 4
- B3 15
- C4 15
- D7 30
Correct answer
C. 4 15
Step-by-step solution
area Δ D E F area Δ A B C = Let area Δ A B C = Δ area Δ B E D area Δ A B C = 1 2 . 6 sin B 1 2 . 20 sin B = 3 10 area Δ A D F area Δ A B C = 1 2 . 4 sin A 1 2 . 24 sin A = 1 6 area Δ C E F area Δ A B C = 1 2 . 8 sin C 1 2 . 30 sin C = 4 15 area Δ B E D = 3 10 Δ area Δ A D F = Δ 6 area Δ C E F = 4 Δ 15 area ∆ D E F = Δ - 3 10 + 1 6 + 4 15 Δ = Δ - 27 + 15 + 24 90 Δ = Δ - 66 90 Δ = 24 Δ 90 area Δ D E F area Δ A B C = 24 90 = 4 15