three equal cubes are placed adjacently in a row find the ratio of total surface of the new cuboid to the sum of the surface areas of the three cubes ?
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Given, three cube are placed adjacent in a row to form a cuboid.
Let the side of cube be a, three cube placed in row then breadth of cuboid be a, length of cuboid is 3a, height of cuboid is a.
Sum of total surface area of three cubes =6a
2
+6a
2
+6a
2
=18a
2
Total surface area of resulting cuboid =2(lb+bh+lh)
=2(3a×a+a×a+3a×a)
=2(7a
2
)
=14a
2
Ratio of total surface area of cuboid to that of cube =
18a
2
14a
2
=
9
7
=7:9
hope you got it the answer ☺️.........
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