If a cuboid of dimensions 1 0 cm × 1 5 cm × 5 cm is cut to form cubes of sides 5 cm, then what is the difference between the sum of the surface areas of these cubes and the surface area of the original cuboid?
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The cuboid is divided by cutting the top into a 2x3 grid. There are 7 lines of division. Each line accounts for two cube faces that are obscured in the cuboid shape, but revealed when it is separated. Each face has area 5x5 = 25 square centimetres. Hence 14x25 = 350.
A 6 c u b e s = ( 6 ) ( 6 ) ( 5 2 ) = 9 0 0 c m 2
A c u b o i d = 2 [ ( 5 ) ( 1 0 ) + ( 1 5 ) ( 5 ) + ( 1 5 ) ( 1 0 ) ] = 5 5 0 c m 2
d i f f e r e n c e i n s u r f a c e a r e a = 9 0 0 − 5 5 0 = 3 5 0 c m 2
6 cubes are formed, each with a side of 5cm.
Let each cube's face be one unit Therefore each cube has a surface area of 6 units 6 units (surface area) X 6 cubes = 36 units
The six faces of the original cuboid (pictured above) have, 2, 2, 3, 3, 6, 6 faces = 22 faces = 22 units.
The difference in units is 36-22=14 units Each unit is 5 cm x 5cm = 25 square cm
14 units X (25 cm^2/unit) = 350 cm^2
No. of cubes formed = 5 × 5 × 5 1 0 × 1 5 × 5 = 6 cubes
Surface Area of cuboid = 2 ( 1 0 × 1 5 + 1 5 × 5 + 5 × 1 0 ) = 5 5 0 cm 2
Surface Area of each Cube = 6 × 5 2 = 1 5 0 cm 2
Sum of Surface Areas of all cubes = 6 × 1 5 0 = 9 0 0 cm 2
Required Difference = 9 0 0 − 5 5 0 = 3 5 0
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The surface area of a cuboid is given by S c u b o i d = 2 ( H W + W L + L H ) where H = height, W = width and L = length.
Substituting, we get
S c u b o i d = 2 [ 5 ( 1 0 ) + 1 0 ( 1 5 ) + 1 5 ( 5 ) ] = 5 5 0
The surface area of a cube is given by S c u b e = 6 a 2 where a = side length of the cube
Substituting, we get
S c u b e = 6 ( 5 2 ) = 6 ( 2 5 ) = 1 5 0
Since there are 6 cubes, we multiply it by 6 . Therefore, the total surface area of the 6 cubes is 6 ( 1 5 0 ) = 9 0 0 .
Finally, the difference is 9 0 0 − 5 5 0 = 3 5 0