Three metal cubes with edges 6cm, 8 cm and 10 cm respectively melt down and form a single cube. Find the volume of the new cube. Also find the edge of the new cube with its total surface area.
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Answered by
5
the volume of cube with edge 6 cm
6×6×6=216
the volume of cube with edge 8 cm
8×8×8=512
the volume of cube with edge 10 cm
10×10×10=1000
therefore,
volume of new cube
216+512+1000=1728 cu cm
edge of new cube
3√1728=12 cm
surface area of cube
6(s×s)
6(12×12)
6×144
864 sq cm
therefore,
the volume of new cube obtained is 1728 cu cm and its edge is 12 cm and surface area is 864 sq cm
6×6×6=216
the volume of cube with edge 8 cm
8×8×8=512
the volume of cube with edge 10 cm
10×10×10=1000
therefore,
volume of new cube
216+512+1000=1728 cu cm
edge of new cube
3√1728=12 cm
surface area of cube
6(s×s)
6(12×12)
6×144
864 sq cm
therefore,
the volume of new cube obtained is 1728 cu cm and its edge is 12 cm and surface area is 864 sq cm
Answered by
2
Hello friend,
Let 'a' be the side of a cube then it's volume is found as a^3.
Total volume remains constant.
Therefore,
V1 + V2 + V3 = V.
V1 = 6^3 = 216 cm^3.
V2 = 8^3 = 512 cm^3.
V3 = 10^3 = 1000 cm^3.
V = 216 + 512 + 1000 = 1727 cm^3.
Let 'L' be the length of the side of the new cube.
Therefore,
V = L^3 = 1727 cm^3.
L = cuberoot (1727) = 11.998 cm.
Total Surface area = 6×(L)^2.
= 6×(11.998)^2.
= 6×143.94
= 863.67 cm^2.
Hence, the answer!
Let 'a' be the side of a cube then it's volume is found as a^3.
Total volume remains constant.
Therefore,
V1 + V2 + V3 = V.
V1 = 6^3 = 216 cm^3.
V2 = 8^3 = 512 cm^3.
V3 = 10^3 = 1000 cm^3.
V = 216 + 512 + 1000 = 1727 cm^3.
Let 'L' be the length of the side of the new cube.
Therefore,
V = L^3 = 1727 cm^3.
L = cuberoot (1727) = 11.998 cm.
Total Surface area = 6×(L)^2.
= 6×(11.998)^2.
= 6×143.94
= 863.67 cm^2.
Hence, the answer!
SmitButle:
I m sorry to interfere but your addition is wrong. 216+512+1000=1728 .
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