The angles A, B and C of a ∆ ABC are in A . P . and a: b = 1: 3 ....

In $∆ ABC,$ angles $A, B, C$ are in $A . P .$ hence $2 ∠ B = ∠ A + ∠ C$

Also, by angle sum property $∠ A + ∠ B + ∠ C = 180 °$

$⇒ 2 ∠ B + ∠ B = 180 °$

$⇒ ∠B = 60 ° .$

Now $a: b = 1: \sqrt{3}$

By, using sine rule, we have $\dfrac{sinA}{a} = \dfrac{sinB}{b} = \dfrac{sinC}{c}$

$⇒ \dfrac{sinA}{a} = \dfrac{sinB}{b}$

$⇒ \dfrac{sinA}{\sin \left(60 °\right)} = \dfrac{a}{b}$

$⇒ sinA = \dfrac{1}{\sqrt{3}} \cdot \dfrac{\sqrt{3}}{2} = \dfrac{1}{2}$

$⇒ ∠ A = 30 °$ and hence $∠ C = 90 °$

Thus, the triangle $ABC$ is a right triangle, with hypotenuse $AB = 4cm .$

Again, by Sine rule, we have $\dfrac{\sin \left(30 °\right)}{a} = \dfrac{\sin \left(60 °\right)}{b} = \dfrac{\sin \left(90 °\right)}{4}$

$⇒ \dfrac{\left(\dfrac{1}{2}\right)}{a} = \dfrac{\left(\dfrac{\sqrt{3}}{2}\right)}{b} = \dfrac{1}{4}$

$⇒ \dfrac{1}{2a} = \dfrac{1}{4}$ and $\dfrac{\sqrt{3}}{2b} = \dfrac{1}{4}$

$⇒ a = 2cm$ and $b = 2\sqrt{3}cm$

Now, the area of $∆ ABC = \dfrac{1}{2}\left(base\right) \times \left(height\right)$

$= \dfrac{1}{2}\left(a \cdot b\right)$

$= \dfrac{1}{2} \times 2 \times 2\sqrt{3} = 2\sqrt{3}sq . cm .$