Question

two equal cubes r placed adjointly in a row . find the ratio of total surface area of a nww cuboid to that of the sum of the surface area of two cubes

Answer 1

let side of cube = a units
1) surface area of the cube = 6a^2 square units
sum of surface areas of two cubes = 12a^2---(1)
2)if two cubes are joined end to end then cuboid Dimensions are
length = l= 2a
breadth =b = a
height =h=a
total surface area of cuboid =2(lb+bh+lh)
=2(2a*a+a*a+2a*a)
=2*5a^2
=10a^2---(2)
required ratio = (2)/(1)
=(10a^2)/12a^2
=10/12
=5/6
=5:6

Answer 2

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