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.
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
⇒ l2 + b2 + h2 + 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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