The side AB of a triangle ABC is c. The median BD is of length k. If BDA = and < 90^, then the area of triangle ABC is Option 1: (k^2 )/2 + k √(c^2 + k^2 ^2 ) Option 2: (k^2 2)/2 + k √(c^2 - k^2 ^2 ) Option 3: (k^2 2)/2 + k √(c^2 - k^2 ^2 ) Option 4: (k^2 )/2 + k √(c^2 + k^2 ^2 )

Answer: (k^2 2)/2 + k √(c^2 - k^2 ^2 ).

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

(k^2 2)/2 + k √(c^2 - k^2 ^2 )

(k^2 )/2 + k √(c^2 + k^2 ^2 )

(k^2 2)/2 + k √(c^2 - k^2 ^2 )

(k^2 )/2 + k √(c^2 + k^2 ^2 )

Geometry

Triangles

Hard

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

$\dfrac{k^2 \sin \theta}{2} + k \sin \theta \sqrt{c^2 + k^2 \sin^2 \theta}$

$\dfrac{k^2 \sin 2\theta}{2} + k \sin \theta \sqrt{c^2 - k^2 \sin^2 \theta}$

$\dfrac{k^2 \cos 2\theta}{2} + k \sin \theta \sqrt{c^2 - k^2 \sin^2 \theta}$

$\dfrac{k^2 \cos \theta}{2} + k \sin \theta \sqrt{c^2 + k^2 \sin^2 \theta}$

✅ Correct Option: 2

Solution figure for IPMAT Indore 2024 MCQ question 10 (Geometry)

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

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IPMAT Indore Past Year Questions (Topic-Wise)

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