A solid cube is cut into three cuboids of same volumes. Whatis the ratio of the surface area of the cube to the sum of the surface areas of any two of the cuboids so formed?
Let the side of cube be 3 cm.
Length of cuboid = 3 cm
Breadth of cuboid = 3 cm
Height of cuboid = 3/3 = 1 cm
Ratio of the surface area of the cube to the sum of the surface areas of any two of the cuboids = $$6 \times (a)^2 : 2 \times 2(lb + bh + lh)$$
= $$6 \times (3)^2 : 2 \times 2(3 \times 3 + 3 \times 1 + 1 \times 3)$$
= $$6 \times 9 : 2 \times 2(9 + 3 + 3)$$
= 54 : 60 = 9 : 10
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