Question

In a \(\triangle ABC\), if \(b = 2\), \(\angle B = 30^\circ\), then the area of the circumcircle of \(\triangle ABC\) (in sq units) is

• A. \(\pi\)
• B. \(2\pi\)
• C. \(4\pi\)
• D. \(6\pi\)

Hint:

Sine rule: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R\), where \(R\) is circumradius.

Answer 1

The Correct Option is C

To find the area of the circumcircle of a triangle \( \triangle ABC\), we can use the formula for circumradius \( R \) in terms of side \( b \) and angle \( B \). The formula is given by:

\(R = \frac{b}{2 \sin B}\)

Given:

• \(b = 2\)
• \(\angle B = 30^\circ\)

First, we calculate \( \sin B \):

\(\sin 30^\circ = \frac{1}{2}\)

Substitute these values into the formula for \( R \):

\(R = \frac{2}{2 \times \frac{1}{2}} = \frac{2}{1} = 2\)

Now, the area of the circumcircle is given by the formula:

\(\text{Area} = \pi R^2\)

Substituting \(R = 2\) into the formula for the area:

\(\text{Area} = \pi \times 2^2 = 4\pi\)

Therefore, the correct answer is \(4\pi\) square units, which matches the given correct answer.