The sum of the length, breadth and height of a cuboid is cm and the length of its diagonal is cm. The total surface area of the cuboid is

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Solution

Q.
The sum of the length, breadth and height of a cuboid is $6 \sqrt{3}$ cm and the length of its diagonal is $2 \sqrt{3}$ cm. The total surface area of the cuboid is

1 $48 {cm}^{2}$

2 $72 {cm}^{2}$

3 $96 {cm}^{2}$

4 $108 {cm}^{2}$

Ans.
(3)

$Given: l + b + h = 6 \sqrt{3} cm$ ...(i)

$and the length of its diagonal = 2 \sqrt{3} cm \Rightarrow \sqrt{l^{2} + b^{2} + h^{2}} = 2 \sqrt{3}$

Squaring both sides, we get $l^{2} + b^{2} + h^{2} = 12$ ...(ii)

From eq. (i), ${(l + b + h)}^{2} = {(6 \sqrt{3})}^{2}$

$\Rightarrow l^{2} + b^{2} + h^{2} + 2 (l b + b h + h l) = 108$

$\Rightarrow 12 + 2 (l b + b h + h l) = 108$ $[From eq. (iii)] \Rightarrow 2 (l b + b h + h l) = 96$

Hence, total surface area of the cuboid is $96 {cm}^{2}$

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