Three cubes of volumes 1,8 and 27 are glued together at their faces. what is the smallest possible surface area of the resulting configuration?
2006-11-24 04:10:31 · 3 answers · asked by maya n 1 in Science & Mathematics ➔ Mathematics
the sides of the three cubes are 1, 2, and 3. the original surface area for all three is 1 x 6, 4 x 6, 9 x 6. 84 square units. subtract 2 x 2 x 2 where the 2 cube touches the 3 cube. subtract 2 x 1 x 2 where the 1 cube touches both of the other cubes. total subtraction of 12. that leaves a total surface area of 72 square units
2006-11-24 04:22:21 · answer #1 · answered by iberius 4 · 0⤊ 1⤋
They would all share one face with each other cube.
Volume: X, Length of a Side: Third Root of X, Area of a Side: X^2, Surface Area 6*X^2
Volume: 1, Length of a Side: 1, Area of a Side: 1, Surface Area: 6,
Volume 8, Length of a Side: 2, Area of a Side: 4, Surface Area: 24
Volume 27, Length of a Side: 3, Area of a Side: 9, Surface Area: 54
Each can only block one side at most.
6-1-1 = 4
24-4-1=19
54-1-4=49
4+19+49= 72 square feet.
2006-11-24 04:24:42 · answer #2 · answered by Skop 2 · 0⤊ 1⤋
45+20+4+3+5 = 77 square units.
2006-11-24 04:54:04 · answer #3 · answered by ag_iitkgp 7 · 0⤊ 1⤋