**Question:**

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

**Solution:**

Let the length, breadth and height (or depth) of the cuboid be *l* cm, *b* cm and *h* cm, respectively.

∴ *l* + *b* + *h* = 19 .....(1)

Also,

Length of the diagonal = 11 cm

$\Rightarrow \sqrt{l^{2}+b^{2}+h^{2}}=11$

$\Rightarrow l^{2}+b^{2}+h^{2}=121$ ......(2)

Squaring (1), we get

(*l* + *b* + *h*)2 = 192

⇒ *l*2 + *b*2 + *h*2 + 2(*lb* + *bh* + *hl*) = 361

⇒ 121* *+ 2(*lb* + *bh* + *hl*) = 361 [Using (2)]

⇒ 2(*lb* + *bh* + *hl*) = 361 − 121 = 240 cm2

Thus, the surface area of the cuboid is 240 cm2.

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