The sum of the length, breadth and depth of cuboid is 19 cm and the length of its diagonal is 11 cm. Find the surface area of the cuboid.

Given,

l + b + h = 19 cm ……..(1)

Diagonal = 11 cm.

By formula,

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

⇒ 11 = $\sqrt{l^2 + b^2 + h^2}$

⇒ 11² = l² + b² + h²

⇒ 121 = l² + b² + h² ………(2)

By formula,

(l + b + h)² = l² + b² + h² + 2(lb + bh + hl)

Substituting the values from equation (1) and (2) in above equation, we get :

⇒ (19)² = 121 + 2(lb + bh + hl)

⇒ 361 = 121 + 2(lb + bh + hl)

⇒ 2(lb + bh + hl) = 361 - 121

⇒ 2(lb + bh + hl) = 240

We know that,

Surface area of the cuboid = 2(lb + bh + hl)

Hence, surface area of the cuboid = 240 cm².