Answer :
length = l
breadth = b
height = h
but here we have l+b+h = 19 cm
and for diagonal = 11
like this,
∴ √ l^2 + b^1 + h^2 = 11
∴ l^2 + b^1 + h^2 = 121
so,
[ l + b + h ]^2 = l^2 + b^1 + h^2 + 2[lb + bh + lh]
∴ 19 ^2 = 121 + 2[lb + bh + lh]
∴ 361 - 121= 2[lb + bh + lh]
∴ 240 = 2[lb + bh + lh]
∴ surface area of cuboid box is 240 cm^2
breadth = b
height = h
but here we have l+b+h = 19 cm
and for diagonal = 11
like this,
∴ √ l^2 + b^1 + h^2 = 11
∴ l^2 + b^1 + h^2 = 121
so,
[ l + b + h ]^2 = l^2 + b^1 + h^2 + 2[lb + bh + lh]
∴ 19 ^2 = 121 + 2[lb + bh + lh]
∴ 361 - 121= 2[lb + bh + lh]
∴ 240 = 2[lb + bh + lh]
∴ surface area of cuboid box is 240 cm^2