Question

In a triangle ABC, if \(\sin\frac{A}{2} = \dfrac{1}{4}\sqrt{\dfrac{5}{\sqrt{5}}}, a = 2, c = 5\), and \(b\) is an integer, then the area (in sq. units) of triangle ABC is

• A. \(\dfrac{\sqrt{297}}{4}\)
• B. \(\dfrac{\sqrt{231}}{4}\)
• C. \(\dfrac{\sqrt{385}}{4}\)
• D. \(\dfrac{\sqrt{185}}{4}\)

Hint:

Use sine and cosine identities along with area formulas such as \(\dfrac{1}{2}bc \sin A\) to compute area in triangle-based problems.

Answer 1

The Correct Option is B

Using sine rule and angle bisector theorem, find angle \(A\), then use Heron’s formula or sine rule based formula for area: \[ \text{Area} = \dfrac{abc}{4R},\text{ or use } \dfrac{1}{2}bc \sin A. \] Calculations yield exact area as \(\dfrac{\sqrt{231}}{4}\).