Question:
The sum of length, breadth and height of a cuboid is $19 \mathrm{~cm}$ and its diagonal is $5 \sqrt{5} \mathrm{~cm}$. Its surface area is
(a) 361 cm2
(b) 125 cm2
(c) 236 cm2
(d) 486 cm2
Solution:
(c) 236 cm2
Let l, b and h be the length, breadth and height of the cuboid.
Then,
$l+b+h=19$
$\Rightarrow(l+b+h)^{2}=(19)^{2}$
Therefore,
$\left(l^{2}+b^{2}+h^{2}\right)+2(l b+b h+l h)=361$
$\Rightarrow(5 \sqrt{5})^{2}+2(l b+b h+l h)=361$
$\Rightarrow 2(l b+b h+l h)=(361-125)$
$\Rightarrow 2(l b+b h+l h)=236 \mathrm{~cm}^{2}$
Hence, the surface area of the cuboid is $236 \mathrm{~cm}^{2}$.