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A solid cube is cut into three cuboids of same volumes. What is the ratio of the surface area of the cube to the sum of the surface areas of any two of the cuboids so formed?
Answer & Solution
Correct Answer:
Option
A
Let the side of cube is 3 m
Surface are of cube = 6 × 32 = 54
Surface are of one cuboid = 2($$l$$b + bh + h$$l$$)
= 2(1 × 3 + 3 × 3 + 3 × 1)
= 2(3 + 9 + 3)
= 30
Cube : 2 Cuboid = 54 : 2 × 30 = 9 : 10
Surface are of cube = 6 × 32 = 54
Surface are of one cuboid = 2($$l$$b + bh + h$$l$$)
= 2(1 × 3 + 3 × 3 + 3 × 1)
= 2(3 + 9 + 3)
= 30
Cube : 2 Cuboid = 54 : 2 × 30 = 9 : 10
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