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Three cubes of metal whose edges are in the ratio 3:4:5 are melted down in to a single cube whose diagonal is `12(sqrt(3))` cm. Find the edges of the three cubes. |
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Answer» Correct Answer - 84 `pi cm^(2)`,12 `pi m^(3)` The ratio in the edges =3:4:5 Let edges be 3x,4xand 5x respectively. Volumes of three cubes will be `27x^(3),64x^(3)` and `125x^(3)`in `cm^(3)` respectively. Now sum of volumes of these three cubes `=27x^(2),64x^(3)`and `125 x^(3)`=`216x^(3)cm^(3)` Let edge of new cube be a cm. Diagonal of new cube=`asqrt(3)cm` `A(sqrt(3))=12(sqrt(12))` volume of new cube =`(12^(3))=1728 cm^(3)` Now by the given condition `216x^(3)=1728` `x^(3)=8` x=2 Edge of I cube =`3xx2=6cm` Edge of II =`4xx2=8 cm` Edge of III cube =5`xx`2=10 cm |
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