Three cubes of side $ 6 $ are glued together to make a rectangular box. The surface area of the rectangular box is how much less than the total surface area of the three separate cubes?
Answer
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Hint: Here use the formula for the area of the cube and the cuboid. Here, the three cubes combined to form the shape of the cuboid and so use the formula for the surface area of the cuboid.
Complete step-by-step answer:
When we glued three such cubes then,
The length of the cuboid become, $ l = 3 \times 6 = 18 $ units
The breadth, $ b = 6 $ units
The height, $ h = 6 $ units
Now, the surface area of the cuboid $ = 2(lb + bh + hl) $
Place the values in the above equation-
The surface area of the cuboid
$
= 2\left[ {(18)(6) + (6)(6) + (6)(18)} \right] \\
= 2\left[ {108 + 36 + 108} \right] \\
= 2[252] \\
= 504\,{\text{uni}}{{\text{t}}^2}{\text{ }}.......{\text{(1)}} \\
$
Now, area of one cube is $ = 6{l^2} $
Place Length, $ l = 6{\text{ units}} $
Therefore, area of one cube
$
= 6{(6)^2} \\
= 6(36) \\
= 216{\text{ uni}}{{\text{t}}^2} \\
$
Therefore area of three cubes $ = 3 \times Area{\text{ of one cube}} $
Area of three cubes is
$
= 3 \times 216 \\
= 648{\text{ uni}}{{\text{t}}^2}{\text{ }}.......{\text{(2)}} \\
$
The total difference between the surface area of the rectangular box and the total surface area of the three separate cubes is the difference of the equation between equations $ (2){\text{ and (1)}} $
Therefore, the difference is $ = 648 - 504 $
Difference is $ = 144{\text{ uni}}{{\text{t}}^2} $
Hence, the required solution is - The total difference between the surface area of the rectangular box and the total surface area of the three separate cubes is $ = 144{\text{ uni}}{{\text{t}}^2} $
Note: Always check the given units. Also, all the terms should have the same system of units MKS system or CGS system. A Cube is a three dimensional shape, it can either be solid or the hollow. It has six-square shaped faces of equal size. The cuboid is the box-like three dimensional shape can be either solid or hollow and has six rectangular faces. The faces opposite to each other are equal.
Complete step-by-step answer:
When we glued three such cubes then,
The length of the cuboid become, $ l = 3 \times 6 = 18 $ units
The breadth, $ b = 6 $ units
The height, $ h = 6 $ units
Now, the surface area of the cuboid $ = 2(lb + bh + hl) $
Place the values in the above equation-
The surface area of the cuboid
$
= 2\left[ {(18)(6) + (6)(6) + (6)(18)} \right] \\
= 2\left[ {108 + 36 + 108} \right] \\
= 2[252] \\
= 504\,{\text{uni}}{{\text{t}}^2}{\text{ }}.......{\text{(1)}} \\
$
Now, area of one cube is $ = 6{l^2} $
Place Length, $ l = 6{\text{ units}} $
Therefore, area of one cube
$
= 6{(6)^2} \\
= 6(36) \\
= 216{\text{ uni}}{{\text{t}}^2} \\
$
Therefore area of three cubes $ = 3 \times Area{\text{ of one cube}} $
Area of three cubes is
$
= 3 \times 216 \\
= 648{\text{ uni}}{{\text{t}}^2}{\text{ }}.......{\text{(2)}} \\
$
The total difference between the surface area of the rectangular box and the total surface area of the three separate cubes is the difference of the equation between equations $ (2){\text{ and (1)}} $
Therefore, the difference is $ = 648 - 504 $
Difference is $ = 144{\text{ uni}}{{\text{t}}^2} $
Hence, the required solution is - The total difference between the surface area of the rectangular box and the total surface area of the three separate cubes is $ = 144{\text{ uni}}{{\text{t}}^2} $
Note: Always check the given units. Also, all the terms should have the same system of units MKS system or CGS system. A Cube is a three dimensional shape, it can either be solid or the hollow. It has six-square shaped faces of equal size. The cuboid is the box-like three dimensional shape can be either solid or hollow and has six rectangular faces. The faces opposite to each other are equal.
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