two cubes with area 64 cm^3 each are joined end to end what is the surface area of the resulting cuboid?
Answers
Step-by-step explanation:
The question is not very clear
If you mean that the VOLUME of the cube is 64 cm³ then
Volume of a cube = a³
Where a = side of the cube
=> a³ = 64
=> a = ∛64
=> a = 4 cm
Now the dimensions of the resulting cuboid formed by joining the ends of the cubes are
Length = 4+4 = 8 cm
Breadth = 4 cm
Height = 4 cm
Surface area of the cuboid = 2 * [ (l*b) + (b*h) + (l*h) ]
Where l, b, and h, are the length, breadth, and height of the cuboid respectively
Substhe values and solving, we get
Surface area = 2*[ (4*8) + (4*4) + (8*4) ] = 2 * [ 32 + 32 + 16 ] = 2 * 80 = 160 cm²
Therefore, the surface area = 160 cm²
Now, if you mean the surface area of the cube = 64 cm², then
Surface area of a cube = 6 * a²
Where a = side of the cube
=> 6 * a² = 64
=> a² = 64/6
=> a = √(64/6) = 8/√6 = 8/2.4 = 10/3 = 3.3 cm
Now, by joining the cubes together, the new dimensions are
Length = 3.3 + 3.3 = 6.6
Breadth = 3.3
Height = 3.3
The surface area of the cuboid = 2 * [ (3.3*6.6) + (3.3*3.3) + (6.6*3.3) ]
= 2 * [ 21.78 + 21.78 + 10.89]
= 2 * 54.45 = 108.9 cm²
Therefore, the surface area = 108.9 cm²
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