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
Answer: Option C
Solution(By Examveda Team)
Sum of perimeters of the six faces :$$\eqalign{ & = 2\left[ {2\left( {l + b} \right) + 2\left( {b + h} \right) + 2\left( {l + h} \right)} \right] \cr & = 4\left( {2l + 2b + 2h} \right) \cr & = 8\left( {l + b + h} \right) \cr} $$
Total surface area $$ = 2\left( {lb + bh + lh} \right)$$
$$\eqalign{ & \therefore 8\left( {l + b + h} \right) = 72 \cr & \Rightarrow l + b + h = 9 \cr & 2\left( {lb + bh + lh} \right) = 16 \cr & \Rightarrow lb + bh + lh = 8 \cr} $$
Now,
$${\left( {l + b + h} \right)^2} = {l^2} + {b^2} + {h^2} + 2$$ $$\left( {lb + bh + lh} \right)$$
$$\eqalign{ & \Rightarrow {\left( 9 \right)^2} = {l^2} + {b^2} + {h^2} + 16 \cr & \Rightarrow {l^2} + {b^2} + {h^2} = 81 - 16 \cr & \Rightarrow {l^2} + {b^2} + {h^2} = 65 \cr} $$
Required length :
$$\eqalign{ & = \sqrt {{l^2} + {b^2} + {h^2}} \cr & = \sqrt {65} \cr & = 8.05\,cm \cr} $$
Related Questions on Volume and Surface Area
A. 12$$\pi$$ cm3
B. 15$$\pi$$ cm3
C. 16$$\pi$$ cm3
D. 20$$\pi$$ cm3
In a shower, 5 cm of rain falls. The volume of water that falls on 1.5 hectares of ground is:
A. 75 cu. m
B. 750 cu. m
C. 7500 cu. m
D. 75000 cu. m
A. 84 meters
B. 90 meters
C. 168 meters
D. 336 meters
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