Two identical cubes are placed adjointly in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of two cubes.

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Solution

Dimensions of cube $=$ a $,$ a $,$ a
Dimensions of cuboid $=$ a $,$ a $,$ 2a
Surface area of cuboid $= 2 (l b + b h + h l)$
$S_{c u b o i d} = 2 (a^{2} + 2 a^{2} + 2 a^{2}) = 10 a^{2}$
$S_{c u b e s} = 6 a^{2} \times 2 = 12 a^{2}$
$\frac{S_{c u b o i d}}{S_{c u b e s}} = \frac{10 a^{2}}{12 a^{2}} = \frac{5}{6}$ .

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Two identical cubes are placed adjointly in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of two cubes.

Dimensions of cube=a,a,aDimensions of cuboid = a,a,2aSurface area of cuboid =2(lb+bh+hl)Scuboid=2(a2+2a2+2a2)=10a2Scubes=6a2×2=12a2ScuboidScubes=10a212a2=56.