The angles A, B & C of a ΔABC are in A.P. and a:b = 1:$$\sqrt{3}$$. If c = 4 cm, then the area (in sq. cm) of this triangle is:

We have a triangle $$\triangle ABC$$ in which the three interior angles $$A,\,B,\,C$$ are in arithmetic progression (A.P.).

If three numbers are in A.P., the middle term equals the average of the other two. Hence we may write

$$A,\;A+d,\;A+2d$$

for the three angles, where $$d$$ is their common difference. Their sum must be $$180^{\circ}$$, so

$$A+(A+d)+(A+2d)=180^{\circ}$$

$$3A+3d=180^{\circ}$$

Dividing every term by $$3$$ gives

$$A+d=60^{\circ} \quad\Longrightarrow\quad B=60^{\circ}.$$

Thus

$$A=60^{\circ}-d,\qquad C=60^{\circ}+d.$$

Next, the ratio of the sides opposite these angles is given: $$a:b=1:\sqrt{3}$$. We invoke the Sine Rule, which states

$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R,$$

where $$R$$ is the circumradius. From the first two fractions we get

$$\frac{a}{b}=\frac{\sin A}{\sin B}.$$

Substituting $$a:b=1:\sqrt{3}$$ and $$B=60^{\circ}$$ yields

$$\frac{1}{\sqrt{3}}=\frac{\sin A}{\sin 60^{\circ}}.$$

Because $$\sin 60^{\circ}=\dfrac{\sqrt{3}}{2},$$ we have

$$\frac{1}{\sqrt{3}}=\frac{\sin A}{\sqrt{3}/2}.$$

Multiplying both sides by $$\dfrac{\sqrt{3}}{2}$$ gives

$$\sin A=\frac{1}{2}.$$ Hence $$A=30^{\circ}$$ (the possibility $$150^{\circ}$$ is impossible, since it would make the third angle negative).

Since $$B=60^{\circ},$$ the remaining angle is

$$C=180^{\circ}-A-B=180^{\circ}-30^{\circ}-60^{\circ}=90^{\circ}.$$

Thus $$\triangle ABC$$ is a right-angled triangle with $$C=90^{\circ}.$$ The side opposite the right angle, namely $$c=AB,$$ is the hypotenuse.

For a $$30^{\circ}\!-\!60^{\circ}\!-\!90^{\circ}$$ triangle the sides satisfy the well-known ratio

$$a:b:c = 1:\sqrt{3}:2.$$

Given $$c=4\text{ cm},$$ equating the “2” part to $$4$$ shows

$$2k = 4 \;\Longrightarrow\; k = 2.$$

Therefore

$$a = k = 2\text{ cm},\qquad b = k\sqrt{3} = 2\sqrt{3}\text{ cm}.$$

Because the legs $$a$$ and $$b$$ are perpendicular (the right angle is at $$C$$), the area $$\Delta$$ of the triangle is

$$\Delta = \frac{1}{2}\,a\,b.$$

Substituting the lengths just found,

$$\Delta = \frac{1}{2}\times 2 \times 2\sqrt{3}=2\sqrt{3}\;\text{sq.\;cm}.$$

Hence, the correct answer is Option A.