Question

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

• A. \( k^2 \cos^2 \theta + \sin \theta \sqrt{(c^2 - k^2 \cos^2 \theta)} \)
• B. \( k^2 \sin^2 \theta + \sin \theta \sqrt{(c^2 - k^2 \sin^2 \theta)} \)
• C. \( k^2 \cos^2 \theta \sqrt{(c^2 - k^2 \sin^2 \theta)} \)
• D. \( \frac{k^2 \cos^2 \theta + \sin \theta \sqrt{(c^2 - k^2 \sin^2 \theta)}}{4} \)

Hint:

When dealing with median-related triangle problems, use known median and area formulas and apply trigonometric identities where necessary.

Answer 1

The Correct Option is B

Step 1: We use the formula for the area of a triangle involving a median. The area of triangle ABC can be computed using the formula involving the length of the median: \[ A = \frac{1}{4} \sqrt{(4k^2c^2 - (c^2 - k^2)^2)}. \] Step 2: By expanding and simplifying the expression, we get the area as: \[ A = k^2 \sin^2 \theta + \sin \theta \sqrt{(c^2 - k^2 \sin^2 \theta)}. \]