The sum of the length and breadth of a rectangle is $$6\text{ cm}$$. A square is constructed whose side is equal to the diagonal of the rectangle. If the ratio of the areas of the square and the rectangle is $$\frac{5}{2}$$, then the area of the square in $$\text{cm}^2$$ is

Let the length and breadth of the rectangle be $$l$$ and $$b$$.

We are given that

$$l+b=6.$$

The side of the square is equal to the diagonal of the rectangle.

By the Pythagorean Theorem,

$$\text{Diagonal}^2=l^2+b^2.$$

Hence, the area of the square is $$l^2+b^2.$$

Also, the ratio of the areas of the square and the rectangle is $$\dfrac{5}{2}$$.

So, $$\dfrac{l^2+b^2}{lb}=\dfrac{5}{2},$$

or $$l^2+b^2=\dfrac{5}{2}lb.$$

Now use the identity

$$\left(l+b\right)^2=l^2+b^2+2lb.$$

Substituting the given values,

$$6^2=\dfrac{5}{2}lb+2lb.$$

$$36=\dfrac{9}{2}lb.$$

Multiplying both sides by $$2$$,

$$72=9lb.$$

$$lb=8.$$

Therefore, the area of the square is

$$l^2+b^2=\dfrac{5}{2}\times8=20.$$

Hence, the required area is $${20\text{ cm}^2}.$$