The sum of the length, breadth and depth of a cuboid is 19 cm and its diagonal is $$5\sqrt{5}$$ cm. Its surface area is

Let length, breadth and height of cuboid is 'l', 'b' and 'h' respectively.

According to the question: 

l + b + h = 19 ........ (1)

Diagonal of cuboid = $$\sqrt{l^{2} + b^{2} + h^{2}} = 5\sqrt{5}$$

$$l^{2} + b^{2} + h^{2} = 125$$ ...... (2)

We know that: 

$$(l + b + h)^{2} = l^{2} + b^{2} + h^{2} + 2(lb + bh + hl)$$

From eqaution (1) and (2):

$$19^{2} = 125 + 2(lb + bh + hl)$$

$$2(lb + bh + hl) = 361 - 125 = 236$$

Surface area of cuboid = $$2(lb + bh + hl) = 236 cm^{2}$$