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The side AB of a triangle ABC is c. The median BD is of length k. If \(\angle BDA = \theta\) and \(\theta < 90^\circ\), then the area of triangle ABC is

Question and options

The side AB of a triangle ABC is c. The median BD is of length k. If ∠BDA=θ\angle BDA = \theta and θ<90∘\theta < 90^\circ, then the area of triangle ABC is
  1. k2sin⁡θ2+ksin⁡θc2+k2sin⁡2θ\dfrac{k^2 \sin \theta}{2} + k \sin \theta \sqrt{c^2 + k^2 \sin^2 \theta}
  2. k2sin⁡2θ2+ksin⁡θc2−k2sin⁡2θ\dfrac{k^2 \sin 2\theta}{2} + k \sin \theta \sqrt{c^2 - k^2 \sin^2 \theta}
  3. k2cos⁡2θ2+ksin⁡θc2−k2sin⁡2θ\dfrac{k^2 \cos 2\theta}{2} + k \sin \theta \sqrt{c^2 - k^2 \sin^2 \theta}
  4. k2cos⁡θ2+ksin⁡θc2+k2sin⁡2θ\dfrac{k^2 \cos \theta}{2} + k \sin \theta \sqrt{c^2 + k^2 \sin^2 \theta}

Answer and explanation

Correct answer: B - \(\dfrac{k^2 \sin 2\theta}{2} + k \sin \theta \sqrt{c^2 - k^2 \sin^2 \theta}\)
D is midpoint of AC; area ABC = 2*area ABD = AD*k*sin(theta). Cosine rule in ABD: c^2=AD^2+k^2-2*AD*k*cos(theta), so AD=k cos(theta)+sqrt(c^2-k^2 sin^2 theta). Area = k^2 sin(2theta)/2 + k sin(theta) sqrt(c^2-k^2 sin^2 theta).

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