Three cubes of metal with edges 3cm,4cm,,5cm,respectively are melted to form a single cube. Find the lateral surface area of the new cube formed.
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Three cubes are melted into a bigger cube.
The edges of the cube are 3 cm, 4 cm and 5 cm respectively.
Consider the edge of bigger cube be k.
So that Sum of volume of 3 cubes = volume of bigger cube.
Volume of bigger cube.= k3
k3 = 33 + 43 + 53 .
k3 = 27 + 64 + 125 = 216
∴ k = 6 cm.
Lateral surface area of the new cube = 4k2 = 4x6x6 = 144 cm2 .
The edges of the cube are 3 cm, 4 cm and 5 cm respectively.
Consider the edge of bigger cube be k.
So that Sum of volume of 3 cubes = volume of bigger cube.
Volume of bigger cube.= k3
k3 = 33 + 43 + 53 .
k3 = 27 + 64 + 125 = 216
∴ k = 6 cm.
Lateral surface area of the new cube = 4k2 = 4x6x6 = 144 cm2 .
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Answer:
Three cubes are melted into a bigger cube.
The edges of the cube are 3 cm, 4 cm and 5 cm respectively.
Consider the edge of bigger cube be k.
So that Sum of volume of 3 cubes = volume of bigger cube.
Volume of bigger cube.= k3
k3 = 33 + 43 + 53 .
k3 = 27 + 64 + 125 = 216
∴ k = 6 cm.
Lateral surface area of the new cube = 4k2 = 4x6x6 = 144 cm2 .
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