Subject: General Aptitude · Chapter: Volume and Surface Area · Exam: · Marks: · Difficulty:
The sum of perimeters of the six faces of a cuboid is 72 cm and the total surface area of the cuboid is $16 \mathrm{cm}^{2}$. 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
Answer
Answer (as printed): C
Explanation
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 $==2(l b+b h+l h)$. $$\begin{aligned} & \therefore 8(l+b+h)=72 \text { and } 2(l b+b h+l h)=16 \Rightarrow l+b+h \\ & =9 \text { and } l b+b h+l h=8 . \end{aligned}$$ Now, $(l+b+h)^{2}=l^{2}+b^{2}+h^{2}+2(l b+b h+l h)$ $$\Rightarrow 9^{2}=l^{2}+b^{2}+h^{2}+16 \Rightarrow l^{2}+b^{2}+h^{2}=81-16=65 .$$ Required length $=\sqrt{l^{2}+b^{2}+h^{2}}=\sqrt{65}=8.05 \mathrm{cm}$.