Asked by Joshua
Three cubes of volume 1cm3, 8 cm 3 and 27cm3 are glued together at their faces. find the smallest possible surface area of the resulting configuration
Answers
Answered by
mathhelper
You have cubes of sides
1 cm, 2 cm, and 3 cm
What do I see of the 3 cm cube?
I see 6 faces minus the part of the top face obstructed by the 2 cm cube,
so : 6(3^2) - 2^2 = 50 cm^2
what do I see of the 2 cm cube?
5 sides - part of the top obstructed by the 1 cm cube
= 5(2^2) - 1^2 = 19 cm
and of course I see 5 sides of the top cube, so the total
= 50 + 19 + 5 cm^2
= 74 cm^2
1 cm, 2 cm, and 3 cm
What do I see of the 3 cm cube?
I see 6 faces minus the part of the top face obstructed by the 2 cm cube,
so : 6(3^2) - 2^2 = 50 cm^2
what do I see of the 2 cm cube?
5 sides - part of the top obstructed by the 1 cm cube
= 5(2^2) - 1^2 = 19 cm
and of course I see 5 sides of the top cube, so the total
= 50 + 19 + 5 cm^2
= 74 cm^2
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