triangle ABC in which angle A= 50°,angle B=60°,BC=4.5 cm and triangle DEF in which angle E=60°,EF=4.5 cm, angle F=70°. [HINT: angle C=180° - (angle A + angle B) = 70°]
Answers
triangle ABC in which angle A= 50°,angle B=60°,BC=4.5 cm and triangle DEF in which angle E=60°,EF=4.5 cm, angle F=70°.
We know that, "The sum of interior angles of a triangle is 180°"
In Δ ABC,
∠ A + ∠ B + ∠ C = 180°
50° + 60° + ∠ C = 180°
110° + ∠ C = 180°
∠ C = 180° - 110°
∠ C = 70°
In Δ DEF,
∠ D + ∠ E + ∠ F = 180°
∠ D + 60° + 70° = 180°
130° + ∠ D = 180°
∠ D = 180° - 130°
∠ D = 50°
In Δ ABC and Δ DEF
∠ B = ∠ E = 60°
BC = EF = 4.5 cm
∴ Δ ABC ~ Δ DEF
⇒ (ar Δ ABC) / (ar Δ DEF) = BC² / EF²
⇒ (ar Δ ABC) / (ar Δ DEF) = 4.5² / 4.5²
∴ (ar Δ ABC) = (ar Δ DEF)
Answer:
In Δ ABC,
∠ A + ∠ B + ∠ C = 180°
50° + 60° + ∠ C = 180°
110° + ∠ C = 180°
∠ C = 180° - 110°
∠ C = 70°
In Δ DEF,
∠ D + ∠ E + ∠ F = 180°
∠ D + 60° + 70° = 180°
130° + ∠ D = 180°
∠ D = 180° - 130°
∠ D = 50°
In Δ ABC and Δ DEF
∠ B = ∠ E = 60°
BC = EF = 4.5 cm
∴ Δ ABC ~ Δ DEF
⇒ (ar Δ ABC) / (ar Δ DEF) = BC² / EF²
⇒ (ar Δ ABC) / (ar Δ DEF) = 4.5² / 4.5²
∴ (ar Δ ABC) = (ar Δ DEF)