Corresponding sides of two similar triangles are in the ratio of 2:3. If the area of first triangle is 48 sq.cm, then the area of second triangle is:

Let A and B be areas of the first and second triangle respectively.

From the question, we have 
Area of first triangle (A) = 48 sq.cm

As we know,

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Then,

$\frac{\mathrm{B}}{\mathrm{A}} = {\left(\frac{\mathrm{side} \mathrm{of} \mathrm{second} \mathrm{triangle}}{\mathrm{side} \mathrm{of} \mathrm{first} \mathrm{triangle}}\right)}^2$

$\frac{\mathrm{B}}{48} = {\left(\frac{3}{2}\right)}^2$

$\frac{\mathrm{B}}{48} = \frac{9}{4}$

$\mathrm{B}= \frac{48\times 9}{4}= 108 \mathrm{sq}.\mathrm{cm}$

Hence the area of the second triangle is 108 sq.cm.