three equal cubes are placed adjacently in a row . find the ratio of the total surface area of the new cuboid to that of the sum of the surface areas of the three cubes .
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Given:-
- Three equal cubes are placed adjacently in a row .
To find:-
- Find the ratio of total surface area of the new cuboid to that of the sum of the surface area of three cubes.
Solutions:-
Let,
- a = side of each cube.
- s1 = surface area of each cube.
So, s1 = 6a²
Sum of surface area of the three cubes.
3s1 = 3 × 6a²
= 18a²
The length of the newly formed cuboids
l = 3a
it's breadth and height will be the same of each cube.
b = a
h = a
Total surface area of the new cuboids.
S2 = 2(lb + bh + hl)
= 2(3a × a + a × a + a × 3a)
= 2(7a²)
= 14a²
Required Ratio,
= S2/3s1
= 14a²/18a²
= 7/9
Hence, the total area of the new cuboid to that of the sum of the surface area of the three cubes is 7:9.
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