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Question:

Let ABC and ABC' be two non-congruent triangles with sides AB=4, AC=AC'=2√2 and angle B=30°. The absolute value of the difference between the areas of these triangles is

Solution:

Using cosine rule
cosB = (a² + c² - b²) / 2ac = (a² + 16 - 8) / (4a) ⇒ √3/2 = (a² + 8) / 4a ⇒ a² - 2√3a + 8 = 0
⇒ a₁ + a₂ = 4√3, a₁a₂ = 8 ⇒ |a₁ - a₂|² = (a₁ + a₂)² - 4a₁a₂ = 48 - 32 = 16 ⇒ |a₁ - a₂| = 4
∴ |Δ₁ - Δ₂| = 1/2 |a₁ - a₂| * c * sinB = 1/2 * 4 * 4 * sin30° = 4