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1. The sum of perimeters of the six faces of a cuboid is 72 cm and the total surface area of the cuboid is 16 cm². Find the longest possible length that can be kept inside the cuboid :

A. 5.2 cm

B. 7.8 cm

C. 8.05 cm

D. 8.36 cm

Solution:

Sum of perimeters of the six faces :

\begin{aligned} = 2 [2 (l + b) + 2 (b + h) + 2 (l + h)] \\ = 4 (2 l + 2 b + 2 h) \\ = 8 (l + b + h) \end{aligned}

Total surface area

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

\begin{aligned} ∴ 8 (l + b + h) = 72 \\ \Rightarrow l + b + h = 9 \\ 2 (l b + b h + l h) = 16 \\ \Rightarrow l b + b h + l h = 8 \end{aligned}

Now,

{(l + b + h)}^{2} = l^{2} + b^{2} + h^{2} + 2

(l b + b h + l h)

\begin{aligned} \Rightarrow {(9)}^{2} = l^{2} + b^{2} + h^{2} + 16 \\ \Rightarrow l^{2} + b^{2} + h^{2} = 81 - 16 \\ \Rightarrow l^{2} + b^{2} + h^{2} = 65 \end{aligned}

Required length :

\begin{aligned} = \sqrt{l^{2} + b^{2} + h^{2}} \\ = \sqrt{65} \\ = 8.05 c m \end{aligned}

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